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How can I round a number resulting from a calculation to 2 decimal places? [duplicate]

I'd like to round at most two decimal places, but only if necessary.
Input:
10
1.7777777
9.1
Output:
10
1.78
9.1
How can I do this in JavaScript?

Use Math.round() :
Math.round(num * 100) / 100
Or to be more specific and to ensure things like 1.005 round correctly, use Number.EPSILON :
Math.round((num + Number.EPSILON) * 100) / 100

If the value is a text type:
parseFloat("123.456").toFixed(2);
If the value is a number:
var numb = 123.23454;
numb = numb.toFixed(2);
There is a downside that values like 1.5 will give "1.50" as the output. A fix suggested by #minitech:
var numb = 1.5;
numb = +numb.toFixed(2);
// Note the plus sign that drops any "extra" zeroes at the end.
// It changes the result (which is a string) into a number again (think "0 + foo"),
// which means that it uses only as many digits as necessary.
It seems like Math.round is a better solution. But it is not! In some cases it will not round correctly:
Math.round(1.005 * 100)/100 // Returns 1 instead of expected 1.01!
toFixed() will also not round correctly in some cases (tested in Chrome v.55.0.2883.87)!
Examples:
parseFloat("1.555").toFixed(2); // Returns 1.55 instead of 1.56.
parseFloat("1.5550").toFixed(2); // Returns 1.55 instead of 1.56.
// However, it will return correct result if you round 1.5551.
parseFloat("1.5551").toFixed(2); // Returns 1.56 as expected.
1.3555.toFixed(3) // Returns 1.355 instead of expected 1.356.
// However, it will return correct result if you round 1.35551.
1.35551.toFixed(2); // Returns 1.36 as expected.
I guess, this is because 1.555 is actually something like float 1.55499994 behind the scenes.
Solution 1 is to use a script with required rounding algorithm, for example:
function roundNumber(num, scale) {
if(!("" + num).includes("e")) {
return +(Math.round(num + "e+" + scale) + "e-" + scale);
} else {
var arr = ("" + num).split("e");
var sig = ""
if(+arr[1] + scale > 0) {
sig = "+";
}
return +(Math.round(+arr[0] + "e" + sig + (+arr[1] + scale)) + "e-" + scale);
}
}
It is also at Plunker.
Note: This is not a universal solution for everyone. There are several different rounding algorithms. Your implementation can be different, and it depends on your requirements. See also Rounding.
Solution 2 is to avoid front end calculations and pull rounded values from the backend server.
Another possible solution, which is not a bulletproof either.
Math.round((num + Number.EPSILON) * 100) / 100
In some cases, when you round a number like 1.3549999999999998, it will return an incorrect result. It should be 1.35, but the result is 1.36.

I found this on MDN. Their way avoids the problem with 1.005 that was mentioned.
function roundToTwo(num) {
return +(Math.round(num + "e+2") + "e-2");
}
console.log('1.005 => ', roundToTwo(1.005));
console.log('10 => ', roundToTwo(10));
console.log('1.7777777 => ', roundToTwo(1.7777777));
console.log('9.1 => ', roundToTwo(9.1));
console.log('1234.5678 => ', roundToTwo(1234.5678));

MarkG's answer is the correct one. Here's a generic extension for any number of decimal places.
Number.prototype.round = function(places) {
return +(Math.round(this + "e+" + places) + "e-" + places);
}
Usage:
var n = 1.7777;
n.round(2); // 1.78
Unit test:
it.only('should round floats to 2 places', function() {
var cases = [
{ n: 10, e: 10, p:2 },
{ n: 1.7777, e: 1.78, p:2 },
{ n: 1.005, e: 1.01, p:2 },
{ n: 1.005, e: 1, p:0 },
{ n: 1.77777, e: 1.8, p:1 }
]
cases.forEach(function(testCase) {
var r = testCase.n.round(testCase.p);
assert.equal(r, testCase.e, 'didn\'t get right number');
});
})

You should use:
Math.round( num * 100 + Number.EPSILON ) / 100
No one seems to be aware of Number.EPSILON.
Also it's worth noting that this is not a JavaScript weirdness like some people stated.
That is simply the way floating point numbers works in a computer. Like 99% of programming languages, JavaScript doesn't have home made floating point numbers; it relies on the CPU/FPU for that. A computer uses binary, and in binary, there isn't any numbers like 0.1, but a mere binary approximation for that. Why? For the same reason than 1/3 cannot be written in decimal: its value is 0.33333333... with an infinity of threes.
Here come Number.EPSILON. That number is the difference between 1 and the next number existing in the double precision floating point numbers. That's it: There is no number between 1 and 1 + Number.EPSILON.
EDIT:
As asked in the comments, let's clarify one thing: adding Number.EPSILON is relevant only when the value to round is the result of an arithmetic operation, as it can swallow some floating point error delta.
It's not useful when the value comes from a direct source (e.g.: literal, user input or sensor).
EDIT (2019):
Like #maganap and some peoples have pointed out, it's best to add Number.EPSILON before multiplying:
Math.round( ( num + Number.EPSILON ) * 100 ) / 100
EDIT (december 2019):
Lately, I use a function similar to this one for comparing numbers epsilon-aware:
const ESPILON_RATE = 1 + Number.EPSILON ;
const ESPILON_ZERO = Number.MIN_VALUE ;
function epsilonEquals( a , b ) {
if ( Number.isNaN( a ) || Number.isNaN( b ) ) {
return false ;
}
if ( a === 0 || b === 0 ) {
return a <= b + EPSILON_ZERO && b <= a + EPSILON_ZERO ;
}
return a <= b * EPSILON_RATE && b <= a * EPSILON_RATE ;
}
My use-case is an assertion + data validation lib I'm developing for many years.
In fact, in the code I'm using ESPILON_RATE = 1 + 4 * Number.EPSILON and EPSILON_ZERO = 4 * Number.MIN_VALUE (four times the epsilon), because I want an equality checker loose enough for cumulating floating point error.
So far, it looks perfect for me.
I hope it will help.

In general, decimal rounding is done by scaling: round(num * p) / p
Naive implementation
Using the following function with halfway numbers, you will get either the upper rounded value as expected, or the lower rounded value sometimes depending on the input.
This inconsistency in rounding may introduce hard to detect bugs in the client code.
function naiveRound(num, decimalPlaces = 0) {
var p = Math.pow(10, decimalPlaces);
return Math.round(num * p) / p;
}
console.log( naiveRound(1.245, 2) ); // 1.25 correct (rounded as expected)
console.log( naiveRound(1.255, 2) ); // 1.25 incorrect (should be 1.26)
// testing edge cases
console.log( naiveRound(1.005, 2) ); // 1 incorrect (should be 1.01)
console.log( naiveRound(2.175, 2) ); // 2.17 incorrect (should be 2.18)
console.log( naiveRound(5.015, 2) ); // 5.01 incorrect (should be 5.02)
In order to determine whether a rounding operation involves a midpoint value, the Round function multiplies the original value to be rounded by 10 ** n, where n is the desired number of fractional digits in the return value, and then determines whether the remaining fractional portion of the value is greater than or equal to .5. This "Exact Testing for Equality" with floating-point values are problematic because of the floating-point format's issues with binary representation and precision. This means that any fractional portion of a number that is slightly less than .5 (because of a loss of precision) will not be rounded upward.
In the previous example, 5.015 is a midpoint value if it is to be rounded to two decimal places, the value of 5.015 * 100 is actually 501.49999999999994. Because .49999999999994 is less than .5, it is rounded down to 501 and finally the result is 5.01.
Better implementations
Exponential notation
By converting the number to a string in the exponential notation, positive numbers are rounded as expected.
But, be aware that negative numbers round differently than positive numbers.
In fact, it performs what is basically equivalent to "round half up" as the rule, you will see that round(-1.005, 2) evaluates to -1 even though round(1.005, 2) evaluates to 1.01. The lodash _.round method uses this technique.
/**
* Round half up ('round half towards positive infinity')
* Negative numbers round differently than positive numbers.
*/
function round(num, decimalPlaces = 0) {
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
// test rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // 0
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1
console.log( round(-2.175, 2) ); // -2.17
console.log( round(-5.015, 2) ); // -5.01
If you want the usual behavior when rounding negative numbers, you would need to convert negative numbers to positive before calling Math.round(), and then convert them back to negative numbers before returning.
// Round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
Approximate rounding
To correct the rounding problem shown in the previous naiveRound example, we can define a custom rounding function that performs a "nearly equal" test to determine whether a fractional value is sufficiently close to a midpoint value to be subject to midpoint rounding.
// round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
var p = Math.pow(10, decimalPlaces);
var n = num * p;
var f = n - Math.floor(n);
var e = Number.EPSILON * n;
// Determine whether this fraction is a midpoint value.
return (f >= .5 - e) ? Math.ceil(n) / p : Math.floor(n) / p;
}
// test rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
Number.EPSILON
There is a different purely mathematical technique to perform round-to-nearest (using "round half away from zero"), in which epsilon correction is applied before calling the rounding function.
Simply, we add the smallest possible float value (= 1.0 ulp; unit in the last place) to the product before rounding. This moves to the next representable float value, away from zero, thus it will offset the binary round-off error that may occur during the multiplication by 10 ** n.
/**
* Round half away from zero ('commercial' rounding)
* Uses correction to offset floating-point inaccuracies.
* Works symmetrically for positive and negative numbers.
*/
function round(num, decimalPlaces = 0) {
var p = Math.pow(10, decimalPlaces);
var n = (num * p) * (1 + Number.EPSILON);
return Math.round(n) / p;
}
// rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
After adding 1 ulp, the value of 5.015 * 100 which is 501.49999999999994 will be corrected to 501.50000000000006, this will rounded up to 502 and finally the result is 5.02.
Note that the size of a unit in last place ("ulp") is determined by (1) the magnitude of the number and (2) the relative machine epsilon (2^-52). Ulps are relatively larger at numbers with bigger magnitudes than they are at numbers with smaller magnitudes.
Double rounding
Here, we use the toPrecision() method to strip the floating-point round-off errors in the intermediate calculations. Simply, we round to 15 significant figures to strip the round-off error at the 16th significant digit. This technique to preround the result to significant digits is also used by PHP 7 round function.
The value of 5.015 * 100 which is 501.49999999999994 will be rounded first to 15 significant digits as 501.500000000000, then it will rounded up again to 502 and finally the result is 5.02.
// Round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
var p = Math.pow(10, decimalPlaces);
var n = (num * p).toPrecision(15);
return Math.round(n) / p;
}
// rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
Arbitrary-precision JavaScript library - decimal.js
// Round half away from zero
function round(num, decimalPlaces = 0) {
return new Decimal(num).toDecimalPlaces(decimalPlaces).toNumber();
}
// rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
<script src="https://cdnjs.cloudflare.com/ajax/libs/decimal.js/10.2.1/decimal.js" integrity="sha512-GKse2KVGCCMVBn4riigHjXE8j5hCxYLPXDw8AvcjUtrt+a9TbZFtIKGdArXwYOlZvdmkhQLWQ46ZE3Q1RIa7uQ==" crossorigin="anonymous"></script>
Solution 1: string in exponential notation
Inspired by the solution provided by KFish here: https://stackoverflow.com/a/55521592/4208440
A simple drop in solution that provides accurate decimal rounding, flooring, and ceiling to a specific number of decimal places without adding a whole library. It treats floats more like decimals by fixing the binary rounding issues to avoid unexpected results: for example, floor((0.1+0.7)*10) will return the expected result 8.
Numbers are rounded to a specific number of fractional digits. Specifying a negative precision will round to any number of places to the left of the decimal point.
// Solution 1
var DecimalPrecision = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var shift = function(value, exponent) {
value = (value + 'e').split('e');
return +(value[0] + 'e' + (+value[1] + (exponent || 0)));
};
var n = shift(num, +decimalPlaces);
return shift(Math[type](n), -decimalPlaces);
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces);
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
return decimalAdjust('ceil', num, decimalPlaces);
},
// Decimal floor
floor: function(num, decimalPlaces) {
return decimalAdjust('floor', num, decimalPlaces);
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return decimalAdjust('trunc', num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision.round(0.5)); // 1
console.log(DecimalPrecision.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision.round(5.12, 1) === 5.1);
console.log(DecimalPrecision.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision.round(1.005, 2) === 1.01);
console.log(DecimalPrecision.round(39.425, 2) === 39.43);
console.log(DecimalPrecision.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision.round(1262.48, -1) === 1260);
console.log(DecimalPrecision.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision.toFixed(1.005, 2) === "1.01");
Solution 2: purely mathematical (Number.EPSILON)
This solution avoids any string conversion / manipulation of any kind for performance reasons.
// Solution 2
var DecimalPrecision2 = (function() {
if (Number.EPSILON === undefined) {
Number.EPSILON = Math.pow(2, -52);
}
if (Math.sign === undefined) {
Math.sign = function(x) {
return ((x > 0) - (x < 0)) || +x;
};
}
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces || 0);
var n = (num * p) * (1 + Number.EPSILON);
return Math.round(n) / p;
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces || 0);
var n = (num * p) * (1 - Math.sign(num) * Number.EPSILON);
return Math.ceil(n) / p;
},
// Decimal floor
floor: function(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces || 0);
var n = (num * p) * (1 + Math.sign(num) * Number.EPSILON);
return Math.floor(n) / p;
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return (num < 0 ? this.ceil : this.floor)(num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return this.round(num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision2.round(0.5)); // 1
console.log(DecimalPrecision2.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision2.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision2.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision2.round(5.12, 1) === 5.1);
console.log(DecimalPrecision2.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision2.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision2.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision2.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision2.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision2.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision2.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision2.round(1.005, 2) === 1.01);
console.log(DecimalPrecision2.round(39.425, 2) === 39.43);
console.log(DecimalPrecision2.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision2.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision2.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision2.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision2.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision2.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision2.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision2.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision2.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision2.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision2.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision2.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision2.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision2.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision2.round(1262.48, -1) === 1260);
console.log(DecimalPrecision2.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision2.toFixed(1.005, 2) === "1.01");
Solution 3: double rounding
This solution uses the toPrecision() method to strip the floating-point round-off errors.
// Solution 3
var DecimalPrecision3 = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7,
1e8, 1e9, 1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
/* Not in lookup table */
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
// Eliminate binary floating-point inaccuracies.
var stripError = function(num) {
if (Number.isInteger(num))
return num;
return parseFloat(num.toPrecision(15));
};
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var p = intpow10(decimalPlaces || 0);
var n = stripError(num * p);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces);
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
return decimalAdjust('ceil', num, decimalPlaces);
},
// Decimal floor
floor: function(num, decimalPlaces) {
return decimalAdjust('floor', num, decimalPlaces);
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return decimalAdjust('trunc', num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision3.round(0.5)); // 1
console.log(DecimalPrecision3.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision3.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision3.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision3.round(5.12, 1) === 5.1);
console.log(DecimalPrecision3.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision3.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision3.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision3.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision3.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision3.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision3.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision3.round(1.005, 2) === 1.01);
console.log(DecimalPrecision3.round(39.425, 2) === 39.43);
console.log(DecimalPrecision3.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision3.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision3.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision3.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision3.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision3.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision3.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision3.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision3.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision3.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision3.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision3.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision3.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision3.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision3.round(1262.48, -1) === 1260);
console.log(DecimalPrecision3.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision3.toFixed(1.005, 2) === "1.01");
Solution 4: double rounding v2
This solution is just like Solution 3, however it uses a custom toPrecision() function.
// Solution 4
var DecimalPrecision4 = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7,
1e8, 1e9, 1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
/* Not in lookup table */
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
var toPrecision = function(num, significantDigits) {
// Return early for ±0, NaN and Infinity.
if (!num || !Number.isFinite(num))
return num;
// Compute shift of the decimal point (sf - leftSidedDigits).
var shift = significantDigits - 1 - Math.floor(Math.log10(Math.abs(num)));
// Return if rounding to the same or higher precision.
var decimalPlaces = 0;
for (var p = 1; num != Math.round(num * p) / p; p *= 10) decimalPlaces++;
if (shift >= decimalPlaces)
return num;
// Round to "shift" fractional digits
var scale = intpow10(Math.abs(shift));
return shift > 0 ?
Math.round(num * scale) / scale :
Math.round(num / scale) * scale;
};
// Eliminate binary floating-point inaccuracies.
var stripError = function(num) {
if (Number.isInteger(num))
return num;
return toPrecision(num, 15);
};
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var p = intpow10(decimalPlaces || 0);
var n = stripError(num * p);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces);
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
return decimalAdjust('ceil', num, decimalPlaces);
},
// Decimal floor
floor: function(num, decimalPlaces) {
return decimalAdjust('floor', num, decimalPlaces);
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return decimalAdjust('trunc', num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision4.round(0.5)); // 1
console.log(DecimalPrecision4.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision4.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision4.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision4.round(5.12, 1) === 5.1);
console.log(DecimalPrecision4.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision4.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision4.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision4.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision4.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision4.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision4.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision4.round(1.005, 2) === 1.01);
console.log(DecimalPrecision4.round(39.425, 2) === 39.43);
console.log(DecimalPrecision4.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision4.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision4.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision4.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision4.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision4.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision4.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision4.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision4.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision4.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision4.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision4.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision4.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision4.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision4.round(1262.48, -1) === 1260);
console.log(DecimalPrecision4.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision4.toFixed(1.005, 2) === "1.01");
Benchmarks
http://jsbench.github.io/#31ec3a8b3d22bd840f8e6822e681a3ac
Here is a benchmark comparing the operations per second in the solutions above on Chrome 109.0.0.0. Rounding functions using the Number.EPSILON is at least 10x-20x faster. Obviously all browsers differ, so your mileage may vary.
(Note: More is better)
Thanks #Mike for adding a screenshot of the benchmark.

This question is complicated.
Suppose we have a function, roundTo2DP(num), that takes a float as an argument and returns a value rounded to 2 decimal places. What should each of these expressions evaluate to?
roundTo2DP(0.014999999999999999)
roundTo2DP(0.0150000000000000001)
roundTo2DP(0.015)
The 'obvious' answer is that the first example should round to 0.01 (because it's closer to 0.01 than to 0.02) while the other two should round to 0.02 (because 0.0150000000000000001 is closer to 0.02 than to 0.01, and because 0.015 is exactly halfway between them and there is a mathematical convention that such numbers get rounded up).
The catch, which you may have guessed, is that roundTo2DP cannot possibly be implemented to give those obvious answers, because all three numbers passed to it are the same number. IEEE 754 binary floating point numbers (the kind used by JavaScript) can't exactly represent most non-integer numbers, and so all three numeric literals above get rounded to a nearby valid floating point number. This number, as it happens, is exactly
0.01499999999999999944488848768742172978818416595458984375
which is closer to 0.01 than to 0.02.
You can see that all three numbers are the same at your browser console, Node shell, or other JavaScript interpreter. Just compare them:
> 0.014999999999999999 === 0.0150000000000000001
true
So when I write m = 0.0150000000000000001, the exact value of m that I end up with is closer to 0.01 than it is to 0.02. And yet, if I convert m to a String...
> var m = 0.0150000000000000001;
> console.log(String(m));
0.015
> var m = 0.014999999999999999;
> console.log(String(m));
0.015
... I get 0.015, which should round to 0.02, and which is noticeably not the 56-decimal-place number I earlier said that all of these numbers were exactly equal to. So what dark magic is this?
The answer can be found in the ECMAScript specification, in section 7.1.12.1: ToString applied to the Number type. Here the rules for converting some Number m to a String are laid down. The key part is point 5, in which an integer s is generated whose digits will be used in the String representation of m:
let n, k, and s be integers such that k ≥ 1, 10k-1 ≤ s < 10k, the Number value for s × 10n-k is m, and k is as small as possible. Note that k is the number of digits in the decimal representation of s, that s is not divisible by 10, and that the least significant digit of s is not necessarily uniquely determined by these criteria.
The key part here is the requirement that "k is as small as possible". What that requirement amounts to is a requirement that, given a Number m, the value of String(m) must have the least possible number of digits while still satisfying the requirement that Number(String(m)) === m. Since we already know that 0.015 === 0.0150000000000000001, it's now clear why String(0.0150000000000000001) === '0.015' must be true.
Of course, none of this discussion has directly answered what roundTo2DP(m) should return. If m's exact value is 0.01499999999999999944488848768742172978818416595458984375, but its String representation is '0.015', then what is the correct answer - mathematically, practically, philosophically, or whatever - when we round it to two decimal places?
There is no single correct answer to this. It depends upon your use case. You probably want to respect the String representation and round upwards when:
The value being represented is inherently discrete, e.g. an amount of currency in a 3-decimal-place currency like dinars. In this case, the true value of a Number like 0.015 is 0.015, and the 0.0149999999... representation that it gets in binary floating point is a rounding error. (Of course, many will argue, reasonably, that you should use a decimal library for handling such values and never represent them as binary floating point Numbers in the first place.)
The value was typed by a user. In this case, again, the exact decimal number entered is more 'true' than the nearest binary floating point representation.
On the other hand, you probably want to respect the binary floating point value and round downwards when your value is from an inherently continuous scale - for instance, if it's a reading from a sensor.
These two approaches require different code. To respect the String representation of the Number, we can (with quite a bit of reasonably subtle code) implement our own rounding that acts directly on the String representation, digit by digit, using the same algorithm you would've used in school when you were taught how to round numbers. Below is an example which respects the OP's requirement of representing the number to 2 decimal places "only when necessary" by stripping trailing zeroes after the decimal point; you may, of course, need to tweak it to your precise needs.
/**
* Converts num to a decimal string (if it isn't one already) and then rounds it
* to at most dp decimal places.
*
* For explanation of why you'd want to perform rounding operations on a String
* rather than a Number, see http://stackoverflow.com/a/38676273/1709587
*
* #param {(number|string)} num
* #param {number} dp
* #return {string}
*/
function roundStringNumberWithoutTrailingZeroes (num, dp) {
if (arguments.length != 2) throw new Error("2 arguments required");
num = String(num);
if (num.indexOf('e+') != -1) {
// Can't round numbers this large because their string representation
// contains an exponent, like 9.99e+37
throw new Error("num too large");
}
if (num.indexOf('.') == -1) {
// Nothing to do
return num;
}
if (num[0] == '-') {
return "-" + roundStringNumberWithoutTrailingZeroes(num.slice(1), dp)
}
var parts = num.split('.'),
beforePoint = parts[0],
afterPoint = parts[1],
shouldRoundUp = afterPoint[dp] >= 5,
finalNumber;
afterPoint = afterPoint.slice(0, dp);
if (!shouldRoundUp) {
finalNumber = beforePoint + '.' + afterPoint;
} else if (/^9+$/.test(afterPoint)) {
// If we need to round up a number like 1.9999, increment the integer
// before the decimal point and discard the fractional part.
// We want to do this while still avoiding converting the whole
// beforePart to a Number (since that could cause loss of precision if
// beforePart is bigger than Number.MAX_SAFE_INTEGER), so the logic for
// this is once again kinda complicated.
// Note we can (and want to) use early returns here because the
// zero-stripping logic at the end of
// roundStringNumberWithoutTrailingZeroes does NOT apply here, since
// the result is a whole number.
if (/^9+$/.test(beforePoint)) {
return "1" + beforePoint.replaceAll("9", "0")
}
// Starting from the last digit, increment digits until we find one
// that is not 9, then stop
var i = beforePoint.length - 1;
while (true) {
if (beforePoint[i] == '9') {
beforePoint = beforePoint.substr(0, i) +
'0' +
beforePoint.substr(i+1);
i--;
} else {
beforePoint = beforePoint.substr(0, i) +
(Number(beforePoint[i]) + 1) +
beforePoint.substr(i+1);
break;
}
}
return beforePoint
} else {
// Starting from the last digit, increment digits until we find one
// that is not 9, then stop
var i = dp-1;
while (true) {
if (afterPoint[i] == '9') {
afterPoint = afterPoint.substr(0, i) +
'0' +
afterPoint.substr(i+1);
i--;
} else {
afterPoint = afterPoint.substr(0, i) +
(Number(afterPoint[i]) + 1) +
afterPoint.substr(i+1);
break;
}
}
finalNumber = beforePoint + '.' + afterPoint;
}
// Remove trailing zeroes from fractional part before returning
return finalNumber.replace(/0+$/, '')
}
Example usage:
> roundStringNumberWithoutTrailingZeroes(1.6, 2)
'1.6'
> roundStringNumberWithoutTrailingZeroes(10000, 2)
'10000'
> roundStringNumberWithoutTrailingZeroes(0.015, 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes('0.015000', 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes(1, 1)
'1'
> roundStringNumberWithoutTrailingZeroes('0.015', 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes(0.01499999999999999944488848768742172978818416595458984375, 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes('0.01499999999999999944488848768742172978818416595458984375', 2)
'0.01'
> roundStringNumberWithoutTrailingZeroes('16.996', 2)
'17'
The function above is probably what you want to use to avoid users ever witnessing numbers that they have entered being rounded wrongly.
(As an alternative, you could also try the round10 library which provides a similarly-behaving function with a wildly different implementation.)
But what if you have the second kind of Number - a value taken from a continuous scale, where there's no reason to think that approximate decimal representations with fewer decimal places are more accurate than those with more? In that case, we don't want to respect the String representation, because that representation (as explained in the spec) is already sort-of-rounded; we don't want to make the mistake of saying "0.014999999...375 rounds up to 0.015, which rounds up to 0.02, so 0.014999999...375 rounds up to 0.02".
Here we can simply use the built-in toFixed method. Note that by calling Number() on the String returned by toFixed, we get a Number whose String representation has no trailing zeroes (thanks to the way JavaScript computes the String representation of a Number, discussed earlier in this answer).
/**
* Takes a float and rounds it to at most dp decimal places. For example
*
* roundFloatNumberWithoutTrailingZeroes(1.2345, 3)
*
* returns 1.234
*
* Note that since this treats the value passed to it as a floating point
* number, it will have counterintuitive results in some cases. For instance,
*
* roundFloatNumberWithoutTrailingZeroes(0.015, 2)
*
* gives 0.01 where 0.02 might be expected. For an explanation of why, see
* http://stackoverflow.com/a/38676273/1709587. You may want to consider using the
* roundStringNumberWithoutTrailingZeroes function there instead.
*
* #param {number} num
* #param {number} dp
* #return {number}
*/
function roundFloatNumberWithoutTrailingZeroes (num, dp) {
var numToFixedDp = Number(num).toFixed(dp);
return Number(numToFixedDp);
}

Consider .toFixed() and .toPrecision():
http://www.javascriptkit.com/javatutors/formatnumber.shtml

One can use .toFixed(NumberOfDecimalPlaces).
var str = 10.234.toFixed(2); // => '10.23'
var number = Number(str); // => 10.23

Here is a simple way to do it:
Math.round(value * 100) / 100
You might want to go ahead and make a separate function to do it for you though:
function roundToTwo(value) {
return(Math.round(value * 100) / 100);
}
Then you would simply pass in the value.
You could enhance it to round to any arbitrary number of decimals by adding a second parameter.
function myRound(value, places) {
var multiplier = Math.pow(10, places);
return (Math.round(value * multiplier) / multiplier);
}

A precise rounding method. Source: Mozilla
(function(){
/**
* Decimal adjustment of a number.
*
* #param {String} type The type of adjustment.
* #param {Number} value The number.
* #param {Integer} exp The exponent (the 10 logarithm of the adjustment base).
* #returns {Number} The adjusted value.
*/
function decimalAdjust(type, value, exp) {
// If the exp is undefined or zero...
if (typeof exp === 'undefined' || +exp === 0) {
return Math[type](value);
}
value = +value;
exp = +exp;
// If the value is not a number or the exp is not an integer...
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0)) {
return NaN;
}
// Shift
value = value.toString().split('e');
value = Math[type](+(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp));
}
// Decimal round
if (!Math.round10) {
Math.round10 = function(value, exp) {
return decimalAdjust('round', value, exp);
};
}
// Decimal floor
if (!Math.floor10) {
Math.floor10 = function(value, exp) {
return decimalAdjust('floor', value, exp);
};
}
// Decimal ceil
if (!Math.ceil10) {
Math.ceil10 = function(value, exp) {
return decimalAdjust('ceil', value, exp);
};
}
})();
Examples:
// Round
Math.round10(55.55, -1); // 55.6
Math.round10(55.549, -1); // 55.5
Math.round10(55, 1); // 60
Math.round10(54.9, 1); // 50
Math.round10(-55.55, -1); // -55.5
Math.round10(-55.551, -1); // -55.6
Math.round10(-55, 1); // -50
Math.round10(-55.1, 1); // -60
Math.round10(1.005, -2); // 1.01 -- compare this with Math.round(1.005*100)/100 above
// Floor
Math.floor10(55.59, -1); // 55.5
Math.floor10(59, 1); // 50
Math.floor10(-55.51, -1); // -55.6
Math.floor10(-51, 1); // -60
// Ceil
Math.ceil10(55.51, -1); // 55.6
Math.ceil10(51, 1); // 60
Math.ceil10(-55.59, -1); // -55.5
Math.ceil10(-59, 1); // -50

None of the answers found here is correct. stinkycheeseman asked to round up, but you all rounded the number.
To round up, use this:
Math.ceil(num * 100)/100;

This may help you:
var result = Math.round(input*100)/100;
For more information, you can have a look at Math.round(num) vs num.toFixed(0) and browser inconsistencies

For me Math.round() was not giving correct answer. I found toFixed(2) works better.
Below are examples of both:
console.log(Math.round(43000 / 80000) * 100); // wrong answer
console.log(((43000 / 80000) * 100).toFixed(2)); // correct answer

Use this function Number(x).toFixed(2);

If you are using the Lodash library, you can use the round method of Lodash like following.
_.round(number, precision)
For example:
_.round(1.7777777, 2) = 1.78

+(10).toFixed(2); // = 10
+(10.12345).toFixed(2); // = 10.12
(10).toFixed(2); // = 10.00
(10.12345).toFixed(2); // = 10.12

Try this lightweight solution:
function round(x, digits){
return parseFloat(x.toFixed(digits))
}
round(1.222, 2);
// 1.22
round(1.222, 10);
// 1.222

There are a couple of ways to do that. For people like me, Lodash's variant
function round(number, precision) {
var pair = (number + 'e').split('e')
var value = Math.round(pair[0] + 'e' + (+pair[1] + precision))
pair = (value + 'e').split('e')
return +(pair[0] + 'e' + (+pair[1] - precision))
}
Usage:
round(0.015, 2) // 0.02
round(1.005, 2) // 1.01
If your project uses jQuery or Lodash, you can also find the proper round method in the libraries.

2017
Just use native code .toFixed()
number = 1.2345;
number.toFixed(2) // "1.23"
If you need to be strict and add digits just if needed it can use replace
number = 1; // "1"
number.toFixed(5).replace(/\.?0*$/g,'');

Another simple solution (without writing any function) may to use toFixed() and then convert to float again:
For example:
var objNumber = 1201203.1256546456;
objNumber = parseFloat(objNumber.toFixed(2))

Since ES6 there is a 'proper' way (without overriding statics and creating workarounds) to do this by using toPrecision
var x = 1.49999999999;
console.log(x.toPrecision(4));
console.log(x.toPrecision(3));
console.log(x.toPrecision(2));
var y = Math.PI;
console.log(y.toPrecision(6));
console.log(y.toPrecision(5));
console.log(y.toPrecision(4));
var z = 222.987654
console.log(z.toPrecision(6));
console.log(z.toPrecision(5));
console.log(z.toPrecision(4));
then you can just parseFloat and zeroes will 'go away'.
console.log(parseFloat((1.4999).toPrecision(3)));
console.log(parseFloat((1.005).toPrecision(3)));
console.log(parseFloat((1.0051).toPrecision(3)));
It doesn't solve the '1.005 rounding problem' though - since it is intrinsic to how float fractions are being processed.
console.log(1.005 - 0.005);
If you are open to libraries you can use bignumber.js
console.log(1.005 - 0.005);
console.log(new BigNumber(1.005).minus(0.005));
console.log(new BigNumber(1.005).round(4));
console.log(new BigNumber(1.005).round(3));
console.log(new BigNumber(1.005).round(2));
console.log(new BigNumber(1.005).round(1));
<script src="https://cdnjs.cloudflare.com/ajax/libs/bignumber.js/2.3.0/bignumber.min.js"></script>

The easiest approach would be to use toFixed and then strip trailing zeros using the Number function:
const number = 15.5;
Number(number.toFixed(2)); // 15.5
const number = 1.7777777;
Number(number.toFixed(2)); // 1.78

One way to achieve such a rounding only if necessary is to use Number.prototype.toLocaleString():
myNumber.toLocaleString('en', {maximumFractionDigits:2, useGrouping:false})
This will provide exactly the output you expect, but as strings. You can still convert those back to numbers if that's not the data type you expect.

MarkG and Lavamantis offered a much better solution than the one that has been accepted. It's a shame they don't get more upvotes!
Here is the function I use to solve the floating point decimals issues also based on MDN. It is even more generic (but less concise) than Lavamantis's solution:
function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);
value = +value;
exp = +exp;
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
Use it with:
round(10.8034, 2); // Returns 10.8
round(1.275, 2); // Returns 1.28
round(1.27499, 2); // Returns 1.27
round(1.2345678e+2, 2); // Returns 123.46
Compared to Lavamantis's solution, we can do...
round(1234.5678, -2); // Returns 1200
round("123.45"); // Returns 123

It may work for you,
Math.round(num * 100)/100;
to know the difference between toFixed and round. You can have a look at Math.round(num) vs num.toFixed(0) and browser inconsistencies.

Keep the type as integer for later sorting or other arithmetic operations:
Math.round(1.7777777 * 100)/100
1.78
// Round up!
Math.ceil(1.7777777 * 100)/100
1.78
// Round down!
Math.floor(1.7777777 * 100)/100
1.77
Or convert to a string:
(1.7777777).toFixed(2)
"1.77"

This is the simplest, more elegant solution (and I am the best of the world;):
function roundToX(num, X) {
return +(Math.round(num + "e+"+X) + "e-"+X);
}
//roundToX(66.66666666,2) => 66.67
//roundToX(10,2) => 10
//roundToX(10.904,2) => 10.9
Modern syntax alternative with fallback values
const roundToX = (num = 0, X = 20) => +(Math.round(num + `e${X}`) + `e-${X}`)

var roundUpto = function(number, upto){
return Number(number.toFixed(upto));
}
roundUpto(0.1464676, 2);
toFixed(2): Here 2 is the number of digits up to which we want to round this number.

See AmrAli's answer for a more thorough run through and performance breakdown of all the various adaptations of this solution.
var DecimalPrecision = (function(){
if (Number.EPSILON === undefined) {
Number.EPSILON = Math.pow(2, -52);
}
if(Number.isInteger === undefined){
Number.isInteger = function(value) {
return typeof value === 'number' &&
isFinite(value) &&
Math.floor(value) === value;
};
}
this.isRound = function(n,p){
let l = n.toString().split('.')[1].length;
return (p >= l);
}
this.round = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
if(n<0)
o *= -1;
return Math.round((n + r) * o) / o;
}
this.ceil = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
return Math.ceil((n + r) * o) / o;
}
this.floor = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
return Math.floor((n + r) * o) / o;
}
return this;
})();
console.log(DecimalPrecision.round(1.005));
console.log(DecimalPrecision.ceil(1.005));
console.log(DecimalPrecision.floor(1.005));
console.log(DecimalPrecision.round(1.0049999));
console.log(DecimalPrecision.ceil(1.0049999));
console.log(DecimalPrecision.floor(1.0049999));
console.log(DecimalPrecision.round(2.175495134384,7));
console.log(DecimalPrecision.round(2.1753543549,8));
console.log(DecimalPrecision.round(2.1755465135353,4));
console.log(DecimalPrecision.ceil(17,4));
console.log(DecimalPrecision.ceil(17.1,4));
console.log(DecimalPrecision.ceil(17.1,15));

Two digits of precision and wrong calculating [duplicate]

I'd like to round at most two decimal places, but only if necessary.
Input:
10
1.7777777
9.1
Output:
10
1.78
9.1
How can I do this in JavaScript?

Use Math.round() :
Math.round(num * 100) / 100
Or to be more specific and to ensure things like 1.005 round correctly, use Number.EPSILON :
Math.round((num + Number.EPSILON) * 100) / 100

If the value is a text type:
parseFloat("123.456").toFixed(2);
If the value is a number:
var numb = 123.23454;
numb = numb.toFixed(2);
There is a downside that values like 1.5 will give "1.50" as the output. A fix suggested by #minitech:
var numb = 1.5;
numb = +numb.toFixed(2);
// Note the plus sign that drops any "extra" zeroes at the end.
// It changes the result (which is a string) into a number again (think "0 + foo"),
// which means that it uses only as many digits as necessary.
It seems like Math.round is a better solution. But it is not! In some cases it will not round correctly:
Math.round(1.005 * 100)/100 // Returns 1 instead of expected 1.01!
toFixed() will also not round correctly in some cases (tested in Chrome v.55.0.2883.87)!
Examples:
parseFloat("1.555").toFixed(2); // Returns 1.55 instead of 1.56.
parseFloat("1.5550").toFixed(2); // Returns 1.55 instead of 1.56.
// However, it will return correct result if you round 1.5551.
parseFloat("1.5551").toFixed(2); // Returns 1.56 as expected.
1.3555.toFixed(3) // Returns 1.355 instead of expected 1.356.
// However, it will return correct result if you round 1.35551.
1.35551.toFixed(2); // Returns 1.36 as expected.
I guess, this is because 1.555 is actually something like float 1.55499994 behind the scenes.
Solution 1 is to use a script with required rounding algorithm, for example:
function roundNumber(num, scale) {
if(!("" + num).includes("e")) {
return +(Math.round(num + "e+" + scale) + "e-" + scale);
} else {
var arr = ("" + num).split("e");
var sig = ""
if(+arr[1] + scale > 0) {
sig = "+";
}
return +(Math.round(+arr[0] + "e" + sig + (+arr[1] + scale)) + "e-" + scale);
}
}
It is also at Plunker.
Note: This is not a universal solution for everyone. There are several different rounding algorithms. Your implementation can be different, and it depends on your requirements. See also Rounding.
Solution 2 is to avoid front end calculations and pull rounded values from the backend server.
Another possible solution, which is not a bulletproof either.
Math.round((num + Number.EPSILON) * 100) / 100
In some cases, when you round a number like 1.3549999999999998, it will return an incorrect result. It should be 1.35, but the result is 1.36.

I found this on MDN. Their way avoids the problem with 1.005 that was mentioned.
function roundToTwo(num) {
return +(Math.round(num + "e+2") + "e-2");
}
console.log('1.005 => ', roundToTwo(1.005));
console.log('10 => ', roundToTwo(10));
console.log('1.7777777 => ', roundToTwo(1.7777777));
console.log('9.1 => ', roundToTwo(9.1));
console.log('1234.5678 => ', roundToTwo(1234.5678));

MarkG's answer is the correct one. Here's a generic extension for any number of decimal places.
Number.prototype.round = function(places) {
return +(Math.round(this + "e+" + places) + "e-" + places);
}
Usage:
var n = 1.7777;
n.round(2); // 1.78
Unit test:
it.only('should round floats to 2 places', function() {
var cases = [
{ n: 10, e: 10, p:2 },
{ n: 1.7777, e: 1.78, p:2 },
{ n: 1.005, e: 1.01, p:2 },
{ n: 1.005, e: 1, p:0 },
{ n: 1.77777, e: 1.8, p:1 }
]
cases.forEach(function(testCase) {
var r = testCase.n.round(testCase.p);
assert.equal(r, testCase.e, 'didn\'t get right number');
});
})

You should use:
Math.round( num * 100 + Number.EPSILON ) / 100
No one seems to be aware of Number.EPSILON.
Also it's worth noting that this is not a JavaScript weirdness like some people stated.
That is simply the way floating point numbers works in a computer. Like 99% of programming languages, JavaScript doesn't have home made floating point numbers; it relies on the CPU/FPU for that. A computer uses binary, and in binary, there isn't any numbers like 0.1, but a mere binary approximation for that. Why? For the same reason than 1/3 cannot be written in decimal: its value is 0.33333333... with an infinity of threes.
Here come Number.EPSILON. That number is the difference between 1 and the next number existing in the double precision floating point numbers. That's it: There is no number between 1 and 1 + Number.EPSILON.
EDIT:
As asked in the comments, let's clarify one thing: adding Number.EPSILON is relevant only when the value to round is the result of an arithmetic operation, as it can swallow some floating point error delta.
It's not useful when the value comes from a direct source (e.g.: literal, user input or sensor).
EDIT (2019):
Like #maganap and some peoples have pointed out, it's best to add Number.EPSILON before multiplying:
Math.round( ( num + Number.EPSILON ) * 100 ) / 100
EDIT (december 2019):
Lately, I use a function similar to this one for comparing numbers epsilon-aware:
const ESPILON_RATE = 1 + Number.EPSILON ;
const ESPILON_ZERO = Number.MIN_VALUE ;
function epsilonEquals( a , b ) {
if ( Number.isNaN( a ) || Number.isNaN( b ) ) {
return false ;
}
if ( a === 0 || b === 0 ) {
return a <= b + EPSILON_ZERO && b <= a + EPSILON_ZERO ;
}
return a <= b * EPSILON_RATE && b <= a * EPSILON_RATE ;
}
My use-case is an assertion + data validation lib I'm developing for many years.
In fact, in the code I'm using ESPILON_RATE = 1 + 4 * Number.EPSILON and EPSILON_ZERO = 4 * Number.MIN_VALUE (four times the epsilon), because I want an equality checker loose enough for cumulating floating point error.
So far, it looks perfect for me.
I hope it will help.

In general, decimal rounding is done by scaling: round(num * p) / p
Naive implementation
Using the following function with halfway numbers, you will get either the upper rounded value as expected, or the lower rounded value sometimes depending on the input.
This inconsistency in rounding may introduce hard to detect bugs in the client code.
function naiveRound(num, decimalPlaces = 0) {
var p = Math.pow(10, decimalPlaces);
return Math.round(num * p) / p;
}
console.log( naiveRound(1.245, 2) ); // 1.25 correct (rounded as expected)
console.log( naiveRound(1.255, 2) ); // 1.25 incorrect (should be 1.26)
// testing edge cases
console.log( naiveRound(1.005, 2) ); // 1 incorrect (should be 1.01)
console.log( naiveRound(2.175, 2) ); // 2.17 incorrect (should be 2.18)
console.log( naiveRound(5.015, 2) ); // 5.01 incorrect (should be 5.02)
In order to determine whether a rounding operation involves a midpoint value, the Round function multiplies the original value to be rounded by 10 ** n, where n is the desired number of fractional digits in the return value, and then determines whether the remaining fractional portion of the value is greater than or equal to .5. This "Exact Testing for Equality" with floating-point values are problematic because of the floating-point format's issues with binary representation and precision. This means that any fractional portion of a number that is slightly less than .5 (because of a loss of precision) will not be rounded upward.
In the previous example, 5.015 is a midpoint value if it is to be rounded to two decimal places, the value of 5.015 * 100 is actually 501.49999999999994. Because .49999999999994 is less than .5, it is rounded down to 501 and finally the result is 5.01.
Better implementations
Exponential notation
By converting the number to a string in the exponential notation, positive numbers are rounded as expected.
But, be aware that negative numbers round differently than positive numbers.
In fact, it performs what is basically equivalent to "round half up" as the rule, you will see that round(-1.005, 2) evaluates to -1 even though round(1.005, 2) evaluates to 1.01. The lodash _.round method uses this technique.
/**
* Round half up ('round half towards positive infinity')
* Negative numbers round differently than positive numbers.
*/
function round(num, decimalPlaces = 0) {
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
// test rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // 0
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1
console.log( round(-2.175, 2) ); // -2.17
console.log( round(-5.015, 2) ); // -5.01
If you want the usual behavior when rounding negative numbers, you would need to convert negative numbers to positive before calling Math.round(), and then convert them back to negative numbers before returning.
// Round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
Approximate rounding
To correct the rounding problem shown in the previous naiveRound example, we can define a custom rounding function that performs a "nearly equal" test to determine whether a fractional value is sufficiently close to a midpoint value to be subject to midpoint rounding.
// round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
var p = Math.pow(10, decimalPlaces);
var n = num * p;
var f = n - Math.floor(n);
var e = Number.EPSILON * n;
// Determine whether this fraction is a midpoint value.
return (f >= .5 - e) ? Math.ceil(n) / p : Math.floor(n) / p;
}
// test rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
Number.EPSILON
There is a different purely mathematical technique to perform round-to-nearest (using "round half away from zero"), in which epsilon correction is applied before calling the rounding function.
Simply, we add the smallest possible float value (= 1.0 ulp; unit in the last place) to the product before rounding. This moves to the next representable float value, away from zero, thus it will offset the binary round-off error that may occur during the multiplication by 10 ** n.
/**
* Round half away from zero ('commercial' rounding)
* Uses correction to offset floating-point inaccuracies.
* Works symmetrically for positive and negative numbers.
*/
function round(num, decimalPlaces = 0) {
var p = Math.pow(10, decimalPlaces);
var n = (num * p) * (1 + Number.EPSILON);
return Math.round(n) / p;
}
// rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
After adding 1 ulp, the value of 5.015 * 100 which is 501.49999999999994 will be corrected to 501.50000000000006, this will rounded up to 502 and finally the result is 5.02.
Note that the size of a unit in last place ("ulp") is determined by (1) the magnitude of the number and (2) the relative machine epsilon (2^-52). Ulps are relatively larger at numbers with bigger magnitudes than they are at numbers with smaller magnitudes.
Double rounding
Here, we use the toPrecision() method to strip the floating-point round-off errors in the intermediate calculations. Simply, we round to 15 significant figures to strip the round-off error at the 16th significant digit. This technique to preround the result to significant digits is also used by PHP 7 round function.
The value of 5.015 * 100 which is 501.49999999999994 will be rounded first to 15 significant digits as 501.500000000000, then it will rounded up again to 502 and finally the result is 5.02.
// Round half away from zero
function round(num, decimalPlaces = 0) {
if (num < 0)
return -round(-num, decimalPlaces);
var p = Math.pow(10, decimalPlaces);
var n = (num * p).toPrecision(15);
return Math.round(n) / p;
}
// rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
Arbitrary-precision JavaScript library - decimal.js
// Round half away from zero
function round(num, decimalPlaces = 0) {
return new Decimal(num).toDecimalPlaces(decimalPlaces).toNumber();
}
// rounding of half
console.log( round(0.5) ); // 1
console.log( round(-0.5) ); // -1
// testing edge cases
console.log( round(1.005, 2) ); // 1.01
console.log( round(2.175, 2) ); // 2.18
console.log( round(5.015, 2) ); // 5.02
console.log( round(-1.005, 2) ); // -1.01
console.log( round(-2.175, 2) ); // -2.18
console.log( round(-5.015, 2) ); // -5.02
<script src="https://cdnjs.cloudflare.com/ajax/libs/decimal.js/10.2.1/decimal.js" integrity="sha512-GKse2KVGCCMVBn4riigHjXE8j5hCxYLPXDw8AvcjUtrt+a9TbZFtIKGdArXwYOlZvdmkhQLWQ46ZE3Q1RIa7uQ==" crossorigin="anonymous"></script>
Solution 1: string in exponential notation
Inspired by the solution provided by KFish here: https://stackoverflow.com/a/55521592/4208440
A simple drop in solution that provides accurate decimal rounding, flooring, and ceiling to a specific number of decimal places without adding a whole library. It treats floats more like decimals by fixing the binary rounding issues to avoid unexpected results: for example, floor((0.1+0.7)*10) will return the expected result 8.
Numbers are rounded to a specific number of fractional digits. Specifying a negative precision will round to any number of places to the left of the decimal point.
// Solution 1
var DecimalPrecision = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var shift = function(value, exponent) {
value = (value + 'e').split('e');
return +(value[0] + 'e' + (+value[1] + (exponent || 0)));
};
var n = shift(num, +decimalPlaces);
return shift(Math[type](n), -decimalPlaces);
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces);
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
return decimalAdjust('ceil', num, decimalPlaces);
},
// Decimal floor
floor: function(num, decimalPlaces) {
return decimalAdjust('floor', num, decimalPlaces);
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return decimalAdjust('trunc', num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision.round(0.5)); // 1
console.log(DecimalPrecision.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision.round(5.12, 1) === 5.1);
console.log(DecimalPrecision.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision.round(1.005, 2) === 1.01);
console.log(DecimalPrecision.round(39.425, 2) === 39.43);
console.log(DecimalPrecision.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision.round(1262.48, -1) === 1260);
console.log(DecimalPrecision.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision.toFixed(1.005, 2) === "1.01");
Solution 2: purely mathematical (Number.EPSILON)
This solution avoids any string conversion / manipulation of any kind for performance reasons.
// Solution 2
var DecimalPrecision2 = (function() {
if (Number.EPSILON === undefined) {
Number.EPSILON = Math.pow(2, -52);
}
if (Math.sign === undefined) {
Math.sign = function(x) {
return ((x > 0) - (x < 0)) || +x;
};
}
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces || 0);
var n = (num * p) * (1 + Number.EPSILON);
return Math.round(n) / p;
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces || 0);
var n = (num * p) * (1 - Math.sign(num) * Number.EPSILON);
return Math.ceil(n) / p;
},
// Decimal floor
floor: function(num, decimalPlaces) {
var p = Math.pow(10, decimalPlaces || 0);
var n = (num * p) * (1 + Math.sign(num) * Number.EPSILON);
return Math.floor(n) / p;
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return (num < 0 ? this.ceil : this.floor)(num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return this.round(num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision2.round(0.5)); // 1
console.log(DecimalPrecision2.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision2.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision2.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision2.round(5.12, 1) === 5.1);
console.log(DecimalPrecision2.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision2.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision2.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision2.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision2.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision2.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision2.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision2.round(1.005, 2) === 1.01);
console.log(DecimalPrecision2.round(39.425, 2) === 39.43);
console.log(DecimalPrecision2.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision2.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision2.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision2.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision2.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision2.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision2.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision2.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision2.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision2.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision2.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision2.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision2.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision2.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision2.round(1262.48, -1) === 1260);
console.log(DecimalPrecision2.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision2.toFixed(1.005, 2) === "1.01");
Solution 3: double rounding
This solution uses the toPrecision() method to strip the floating-point round-off errors.
// Solution 3
var DecimalPrecision3 = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7,
1e8, 1e9, 1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
/* Not in lookup table */
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
// Eliminate binary floating-point inaccuracies.
var stripError = function(num) {
if (Number.isInteger(num))
return num;
return parseFloat(num.toPrecision(15));
};
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var p = intpow10(decimalPlaces || 0);
var n = stripError(num * p);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces);
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
return decimalAdjust('ceil', num, decimalPlaces);
},
// Decimal floor
floor: function(num, decimalPlaces) {
return decimalAdjust('floor', num, decimalPlaces);
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return decimalAdjust('trunc', num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision3.round(0.5)); // 1
console.log(DecimalPrecision3.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision3.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision3.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision3.round(5.12, 1) === 5.1);
console.log(DecimalPrecision3.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision3.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision3.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision3.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision3.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision3.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision3.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision3.round(1.005, 2) === 1.01);
console.log(DecimalPrecision3.round(39.425, 2) === 39.43);
console.log(DecimalPrecision3.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision3.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision3.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision3.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision3.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision3.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision3.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision3.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision3.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision3.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision3.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision3.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision3.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision3.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision3.round(1262.48, -1) === 1260);
console.log(DecimalPrecision3.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision3.toFixed(1.005, 2) === "1.01");
Solution 4: double rounding v2
This solution is just like Solution 3, however it uses a custom toPrecision() function.
// Solution 4
var DecimalPrecision4 = (function() {
if (Math.trunc === undefined) {
Math.trunc = function(v) {
return v < 0 ? Math.ceil(v) : Math.floor(v);
};
}
var powers = [
1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7,
1e8, 1e9, 1e10, 1e11, 1e12, 1e13, 1e14, 1e15,
1e16, 1e17, 1e18, 1e19, 1e20, 1e21, 1e22
];
var intpow10 = function(power) {
/* Not in lookup table */
if (power < 0 || power > 22) {
return Math.pow(10, power);
}
return powers[power];
};
var toPrecision = function(num, significantDigits) {
// Return early for ±0, NaN and Infinity.
if (!num || !Number.isFinite(num))
return num;
// Compute shift of the decimal point (sf - leftSidedDigits).
var shift = significantDigits - 1 - Math.floor(Math.log10(Math.abs(num)));
// Return if rounding to the same or higher precision.
var decimalPlaces = 0;
for (var p = 1; num != Math.round(num * p) / p; p *= 10) decimalPlaces++;
if (shift >= decimalPlaces)
return num;
// Round to "shift" fractional digits
var scale = intpow10(Math.abs(shift));
return shift > 0 ?
Math.round(num * scale) / scale :
Math.round(num / scale) * scale;
};
// Eliminate binary floating-point inaccuracies.
var stripError = function(num) {
if (Number.isInteger(num))
return num;
return toPrecision(num, 15);
};
var decimalAdjust = function myself(type, num, decimalPlaces) {
if (type === 'round' && num < 0)
return -myself(type, -num, decimalPlaces);
var p = intpow10(decimalPlaces || 0);
var n = stripError(num * p);
return Math[type](n) / p;
};
return {
// Decimal round (half away from zero)
round: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces);
},
// Decimal ceil
ceil: function(num, decimalPlaces) {
return decimalAdjust('ceil', num, decimalPlaces);
},
// Decimal floor
floor: function(num, decimalPlaces) {
return decimalAdjust('floor', num, decimalPlaces);
},
// Decimal trunc
trunc: function(num, decimalPlaces) {
return decimalAdjust('trunc', num, decimalPlaces);
},
// Format using fixed-point notation
toFixed: function(num, decimalPlaces) {
return decimalAdjust('round', num, decimalPlaces).toFixed(decimalPlaces);
}
};
})();
// test rounding of half
console.log(DecimalPrecision4.round(0.5)); // 1
console.log(DecimalPrecision4.round(-0.5)); // -1
// testing very small numbers
console.log(DecimalPrecision4.ceil(1e-8, 2) === 0.01);
console.log(DecimalPrecision4.floor(1e-8, 2) === 0);
// testing simple cases
console.log(DecimalPrecision4.round(5.12, 1) === 5.1);
console.log(DecimalPrecision4.round(-5.12, 1) === -5.1);
console.log(DecimalPrecision4.ceil(5.12, 1) === 5.2);
console.log(DecimalPrecision4.ceil(-5.12, 1) === -5.1);
console.log(DecimalPrecision4.floor(5.12, 1) === 5.1);
console.log(DecimalPrecision4.floor(-5.12, 1) === -5.2);
console.log(DecimalPrecision4.trunc(5.12, 1) === 5.1);
console.log(DecimalPrecision4.trunc(-5.12, 1) === -5.1);
// testing edge cases for round
console.log(DecimalPrecision4.round(1.005, 2) === 1.01);
console.log(DecimalPrecision4.round(39.425, 2) === 39.43);
console.log(DecimalPrecision4.round(-1.005, 2) === -1.01);
console.log(DecimalPrecision4.round(-39.425, 2) === -39.43);
// testing edge cases for ceil
console.log(DecimalPrecision4.ceil(9.13, 2) === 9.13);
console.log(DecimalPrecision4.ceil(65.18, 2) === 65.18);
console.log(DecimalPrecision4.ceil(-2.26, 2) === -2.26);
console.log(DecimalPrecision4.ceil(-18.15, 2) === -18.15);
// testing edge cases for floor
console.log(DecimalPrecision4.floor(2.26, 2) === 2.26);
console.log(DecimalPrecision4.floor(18.15, 2) === 18.15);
console.log(DecimalPrecision4.floor(-9.13, 2) === -9.13);
console.log(DecimalPrecision4.floor(-65.18, 2) === -65.18);
// testing edge cases for trunc
console.log(DecimalPrecision4.trunc(2.26, 2) === 2.26);
console.log(DecimalPrecision4.trunc(18.15, 2) === 18.15);
console.log(DecimalPrecision4.trunc(-2.26, 2) === -2.26);
console.log(DecimalPrecision4.trunc(-18.15, 2) === -18.15);
// testing round to tens and hundreds
console.log(DecimalPrecision4.round(1262.48, -1) === 1260);
console.log(DecimalPrecision4.round(1262.48, -2) === 1300);
// testing toFixed()
console.log(DecimalPrecision4.toFixed(1.005, 2) === "1.01");
Benchmarks
http://jsbench.github.io/#31ec3a8b3d22bd840f8e6822e681a3ac
Here is a benchmark comparing the operations per second in the solutions above on Chrome 109.0.0.0. Rounding functions using the Number.EPSILON is at least 10x-20x faster. Obviously all browsers differ, so your mileage may vary.
(Note: More is better)
Thanks #Mike for adding a screenshot of the benchmark.

This question is complicated.
Suppose we have a function, roundTo2DP(num), that takes a float as an argument and returns a value rounded to 2 decimal places. What should each of these expressions evaluate to?
roundTo2DP(0.014999999999999999)
roundTo2DP(0.0150000000000000001)
roundTo2DP(0.015)
The 'obvious' answer is that the first example should round to 0.01 (because it's closer to 0.01 than to 0.02) while the other two should round to 0.02 (because 0.0150000000000000001 is closer to 0.02 than to 0.01, and because 0.015 is exactly halfway between them and there is a mathematical convention that such numbers get rounded up).
The catch, which you may have guessed, is that roundTo2DP cannot possibly be implemented to give those obvious answers, because all three numbers passed to it are the same number. IEEE 754 binary floating point numbers (the kind used by JavaScript) can't exactly represent most non-integer numbers, and so all three numeric literals above get rounded to a nearby valid floating point number. This number, as it happens, is exactly
0.01499999999999999944488848768742172978818416595458984375
which is closer to 0.01 than to 0.02.
You can see that all three numbers are the same at your browser console, Node shell, or other JavaScript interpreter. Just compare them:
> 0.014999999999999999 === 0.0150000000000000001
true
So when I write m = 0.0150000000000000001, the exact value of m that I end up with is closer to 0.01 than it is to 0.02. And yet, if I convert m to a String...
> var m = 0.0150000000000000001;
> console.log(String(m));
0.015
> var m = 0.014999999999999999;
> console.log(String(m));
0.015
... I get 0.015, which should round to 0.02, and which is noticeably not the 56-decimal-place number I earlier said that all of these numbers were exactly equal to. So what dark magic is this?
The answer can be found in the ECMAScript specification, in section 7.1.12.1: ToString applied to the Number type. Here the rules for converting some Number m to a String are laid down. The key part is point 5, in which an integer s is generated whose digits will be used in the String representation of m:
let n, k, and s be integers such that k ≥ 1, 10k-1 ≤ s < 10k, the Number value for s × 10n-k is m, and k is as small as possible. Note that k is the number of digits in the decimal representation of s, that s is not divisible by 10, and that the least significant digit of s is not necessarily uniquely determined by these criteria.
The key part here is the requirement that "k is as small as possible". What that requirement amounts to is a requirement that, given a Number m, the value of String(m) must have the least possible number of digits while still satisfying the requirement that Number(String(m)) === m. Since we already know that 0.015 === 0.0150000000000000001, it's now clear why String(0.0150000000000000001) === '0.015' must be true.
Of course, none of this discussion has directly answered what roundTo2DP(m) should return. If m's exact value is 0.01499999999999999944488848768742172978818416595458984375, but its String representation is '0.015', then what is the correct answer - mathematically, practically, philosophically, or whatever - when we round it to two decimal places?
There is no single correct answer to this. It depends upon your use case. You probably want to respect the String representation and round upwards when:
The value being represented is inherently discrete, e.g. an amount of currency in a 3-decimal-place currency like dinars. In this case, the true value of a Number like 0.015 is 0.015, and the 0.0149999999... representation that it gets in binary floating point is a rounding error. (Of course, many will argue, reasonably, that you should use a decimal library for handling such values and never represent them as binary floating point Numbers in the first place.)
The value was typed by a user. In this case, again, the exact decimal number entered is more 'true' than the nearest binary floating point representation.
On the other hand, you probably want to respect the binary floating point value and round downwards when your value is from an inherently continuous scale - for instance, if it's a reading from a sensor.
These two approaches require different code. To respect the String representation of the Number, we can (with quite a bit of reasonably subtle code) implement our own rounding that acts directly on the String representation, digit by digit, using the same algorithm you would've used in school when you were taught how to round numbers. Below is an example which respects the OP's requirement of representing the number to 2 decimal places "only when necessary" by stripping trailing zeroes after the decimal point; you may, of course, need to tweak it to your precise needs.
/**
* Converts num to a decimal string (if it isn't one already) and then rounds it
* to at most dp decimal places.
*
* For explanation of why you'd want to perform rounding operations on a String
* rather than a Number, see http://stackoverflow.com/a/38676273/1709587
*
* #param {(number|string)} num
* #param {number} dp
* #return {string}
*/
function roundStringNumberWithoutTrailingZeroes (num, dp) {
if (arguments.length != 2) throw new Error("2 arguments required");
num = String(num);
if (num.indexOf('e+') != -1) {
// Can't round numbers this large because their string representation
// contains an exponent, like 9.99e+37
throw new Error("num too large");
}
if (num.indexOf('.') == -1) {
// Nothing to do
return num;
}
if (num[0] == '-') {
return "-" + roundStringNumberWithoutTrailingZeroes(num.slice(1), dp)
}
var parts = num.split('.'),
beforePoint = parts[0],
afterPoint = parts[1],
shouldRoundUp = afterPoint[dp] >= 5,
finalNumber;
afterPoint = afterPoint.slice(0, dp);
if (!shouldRoundUp) {
finalNumber = beforePoint + '.' + afterPoint;
} else if (/^9+$/.test(afterPoint)) {
// If we need to round up a number like 1.9999, increment the integer
// before the decimal point and discard the fractional part.
// We want to do this while still avoiding converting the whole
// beforePart to a Number (since that could cause loss of precision if
// beforePart is bigger than Number.MAX_SAFE_INTEGER), so the logic for
// this is once again kinda complicated.
// Note we can (and want to) use early returns here because the
// zero-stripping logic at the end of
// roundStringNumberWithoutTrailingZeroes does NOT apply here, since
// the result is a whole number.
if (/^9+$/.test(beforePoint)) {
return "1" + beforePoint.replaceAll("9", "0")
}
// Starting from the last digit, increment digits until we find one
// that is not 9, then stop
var i = beforePoint.length - 1;
while (true) {
if (beforePoint[i] == '9') {
beforePoint = beforePoint.substr(0, i) +
'0' +
beforePoint.substr(i+1);
i--;
} else {
beforePoint = beforePoint.substr(0, i) +
(Number(beforePoint[i]) + 1) +
beforePoint.substr(i+1);
break;
}
}
return beforePoint
} else {
// Starting from the last digit, increment digits until we find one
// that is not 9, then stop
var i = dp-1;
while (true) {
if (afterPoint[i] == '9') {
afterPoint = afterPoint.substr(0, i) +
'0' +
afterPoint.substr(i+1);
i--;
} else {
afterPoint = afterPoint.substr(0, i) +
(Number(afterPoint[i]) + 1) +
afterPoint.substr(i+1);
break;
}
}
finalNumber = beforePoint + '.' + afterPoint;
}
// Remove trailing zeroes from fractional part before returning
return finalNumber.replace(/0+$/, '')
}
Example usage:
> roundStringNumberWithoutTrailingZeroes(1.6, 2)
'1.6'
> roundStringNumberWithoutTrailingZeroes(10000, 2)
'10000'
> roundStringNumberWithoutTrailingZeroes(0.015, 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes('0.015000', 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes(1, 1)
'1'
> roundStringNumberWithoutTrailingZeroes('0.015', 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes(0.01499999999999999944488848768742172978818416595458984375, 2)
'0.02'
> roundStringNumberWithoutTrailingZeroes('0.01499999999999999944488848768742172978818416595458984375', 2)
'0.01'
> roundStringNumberWithoutTrailingZeroes('16.996', 2)
'17'
The function above is probably what you want to use to avoid users ever witnessing numbers that they have entered being rounded wrongly.
(As an alternative, you could also try the round10 library which provides a similarly-behaving function with a wildly different implementation.)
But what if you have the second kind of Number - a value taken from a continuous scale, where there's no reason to think that approximate decimal representations with fewer decimal places are more accurate than those with more? In that case, we don't want to respect the String representation, because that representation (as explained in the spec) is already sort-of-rounded; we don't want to make the mistake of saying "0.014999999...375 rounds up to 0.015, which rounds up to 0.02, so 0.014999999...375 rounds up to 0.02".
Here we can simply use the built-in toFixed method. Note that by calling Number() on the String returned by toFixed, we get a Number whose String representation has no trailing zeroes (thanks to the way JavaScript computes the String representation of a Number, discussed earlier in this answer).
/**
* Takes a float and rounds it to at most dp decimal places. For example
*
* roundFloatNumberWithoutTrailingZeroes(1.2345, 3)
*
* returns 1.234
*
* Note that since this treats the value passed to it as a floating point
* number, it will have counterintuitive results in some cases. For instance,
*
* roundFloatNumberWithoutTrailingZeroes(0.015, 2)
*
* gives 0.01 where 0.02 might be expected. For an explanation of why, see
* http://stackoverflow.com/a/38676273/1709587. You may want to consider using the
* roundStringNumberWithoutTrailingZeroes function there instead.
*
* #param {number} num
* #param {number} dp
* #return {number}
*/
function roundFloatNumberWithoutTrailingZeroes (num, dp) {
var numToFixedDp = Number(num).toFixed(dp);
return Number(numToFixedDp);
}

Consider .toFixed() and .toPrecision():
http://www.javascriptkit.com/javatutors/formatnumber.shtml

One can use .toFixed(NumberOfDecimalPlaces).
var str = 10.234.toFixed(2); // => '10.23'
var number = Number(str); // => 10.23

Here is a simple way to do it:
Math.round(value * 100) / 100
You might want to go ahead and make a separate function to do it for you though:
function roundToTwo(value) {
return(Math.round(value * 100) / 100);
}
Then you would simply pass in the value.
You could enhance it to round to any arbitrary number of decimals by adding a second parameter.
function myRound(value, places) {
var multiplier = Math.pow(10, places);
return (Math.round(value * multiplier) / multiplier);
}

A precise rounding method. Source: Mozilla
(function(){
/**
* Decimal adjustment of a number.
*
* #param {String} type The type of adjustment.
* #param {Number} value The number.
* #param {Integer} exp The exponent (the 10 logarithm of the adjustment base).
* #returns {Number} The adjusted value.
*/
function decimalAdjust(type, value, exp) {
// If the exp is undefined or zero...
if (typeof exp === 'undefined' || +exp === 0) {
return Math[type](value);
}
value = +value;
exp = +exp;
// If the value is not a number or the exp is not an integer...
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0)) {
return NaN;
}
// Shift
value = value.toString().split('e');
value = Math[type](+(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp));
}
// Decimal round
if (!Math.round10) {
Math.round10 = function(value, exp) {
return decimalAdjust('round', value, exp);
};
}
// Decimal floor
if (!Math.floor10) {
Math.floor10 = function(value, exp) {
return decimalAdjust('floor', value, exp);
};
}
// Decimal ceil
if (!Math.ceil10) {
Math.ceil10 = function(value, exp) {
return decimalAdjust('ceil', value, exp);
};
}
})();
Examples:
// Round
Math.round10(55.55, -1); // 55.6
Math.round10(55.549, -1); // 55.5
Math.round10(55, 1); // 60
Math.round10(54.9, 1); // 50
Math.round10(-55.55, -1); // -55.5
Math.round10(-55.551, -1); // -55.6
Math.round10(-55, 1); // -50
Math.round10(-55.1, 1); // -60
Math.round10(1.005, -2); // 1.01 -- compare this with Math.round(1.005*100)/100 above
// Floor
Math.floor10(55.59, -1); // 55.5
Math.floor10(59, 1); // 50
Math.floor10(-55.51, -1); // -55.6
Math.floor10(-51, 1); // -60
// Ceil
Math.ceil10(55.51, -1); // 55.6
Math.ceil10(51, 1); // 60
Math.ceil10(-55.59, -1); // -55.5
Math.ceil10(-59, 1); // -50

None of the answers found here is correct. stinkycheeseman asked to round up, but you all rounded the number.
To round up, use this:
Math.ceil(num * 100)/100;

This may help you:
var result = Math.round(input*100)/100;
For more information, you can have a look at Math.round(num) vs num.toFixed(0) and browser inconsistencies

For me Math.round() was not giving correct answer. I found toFixed(2) works better.
Below are examples of both:
console.log(Math.round(43000 / 80000) * 100); // wrong answer
console.log(((43000 / 80000) * 100).toFixed(2)); // correct answer

Use this function Number(x).toFixed(2);

If you are using the Lodash library, you can use the round method of Lodash like following.
_.round(number, precision)
For example:
_.round(1.7777777, 2) = 1.78

+(10).toFixed(2); // = 10
+(10.12345).toFixed(2); // = 10.12
(10).toFixed(2); // = 10.00
(10.12345).toFixed(2); // = 10.12

Try this lightweight solution:
function round(x, digits){
return parseFloat(x.toFixed(digits))
}
round(1.222, 2);
// 1.22
round(1.222, 10);
// 1.222

There are a couple of ways to do that. For people like me, Lodash's variant
function round(number, precision) {
var pair = (number + 'e').split('e')
var value = Math.round(pair[0] + 'e' + (+pair[1] + precision))
pair = (value + 'e').split('e')
return +(pair[0] + 'e' + (+pair[1] - precision))
}
Usage:
round(0.015, 2) // 0.02
round(1.005, 2) // 1.01
If your project uses jQuery or Lodash, you can also find the proper round method in the libraries.

2017
Just use native code .toFixed()
number = 1.2345;
number.toFixed(2) // "1.23"
If you need to be strict and add digits just if needed it can use replace
number = 1; // "1"
number.toFixed(5).replace(/\.?0*$/g,'');

Another simple solution (without writing any function) may to use toFixed() and then convert to float again:
For example:
var objNumber = 1201203.1256546456;
objNumber = parseFloat(objNumber.toFixed(2))

Since ES6 there is a 'proper' way (without overriding statics and creating workarounds) to do this by using toPrecision
var x = 1.49999999999;
console.log(x.toPrecision(4));
console.log(x.toPrecision(3));
console.log(x.toPrecision(2));
var y = Math.PI;
console.log(y.toPrecision(6));
console.log(y.toPrecision(5));
console.log(y.toPrecision(4));
var z = 222.987654
console.log(z.toPrecision(6));
console.log(z.toPrecision(5));
console.log(z.toPrecision(4));
then you can just parseFloat and zeroes will 'go away'.
console.log(parseFloat((1.4999).toPrecision(3)));
console.log(parseFloat((1.005).toPrecision(3)));
console.log(parseFloat((1.0051).toPrecision(3)));
It doesn't solve the '1.005 rounding problem' though - since it is intrinsic to how float fractions are being processed.
console.log(1.005 - 0.005);
If you are open to libraries you can use bignumber.js
console.log(1.005 - 0.005);
console.log(new BigNumber(1.005).minus(0.005));
console.log(new BigNumber(1.005).round(4));
console.log(new BigNumber(1.005).round(3));
console.log(new BigNumber(1.005).round(2));
console.log(new BigNumber(1.005).round(1));
<script src="https://cdnjs.cloudflare.com/ajax/libs/bignumber.js/2.3.0/bignumber.min.js"></script>

The easiest approach would be to use toFixed and then strip trailing zeros using the Number function:
const number = 15.5;
Number(number.toFixed(2)); // 15.5
const number = 1.7777777;
Number(number.toFixed(2)); // 1.78

One way to achieve such a rounding only if necessary is to use Number.prototype.toLocaleString():
myNumber.toLocaleString('en', {maximumFractionDigits:2, useGrouping:false})
This will provide exactly the output you expect, but as strings. You can still convert those back to numbers if that's not the data type you expect.

MarkG and Lavamantis offered a much better solution than the one that has been accepted. It's a shame they don't get more upvotes!
Here is the function I use to solve the floating point decimals issues also based on MDN. It is even more generic (but less concise) than Lavamantis's solution:
function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);
value = +value;
exp = +exp;
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
Use it with:
round(10.8034, 2); // Returns 10.8
round(1.275, 2); // Returns 1.28
round(1.27499, 2); // Returns 1.27
round(1.2345678e+2, 2); // Returns 123.46
Compared to Lavamantis's solution, we can do...
round(1234.5678, -2); // Returns 1200
round("123.45"); // Returns 123

It may work for you,
Math.round(num * 100)/100;
to know the difference between toFixed and round. You can have a look at Math.round(num) vs num.toFixed(0) and browser inconsistencies.

Keep the type as integer for later sorting or other arithmetic operations:
Math.round(1.7777777 * 100)/100
1.78
// Round up!
Math.ceil(1.7777777 * 100)/100
1.78
// Round down!
Math.floor(1.7777777 * 100)/100
1.77
Or convert to a string:
(1.7777777).toFixed(2)
"1.77"

This is the simplest, more elegant solution (and I am the best of the world;):
function roundToX(num, X) {
return +(Math.round(num + "e+"+X) + "e-"+X);
}
//roundToX(66.66666666,2) => 66.67
//roundToX(10,2) => 10
//roundToX(10.904,2) => 10.9
Modern syntax alternative with fallback values
const roundToX = (num = 0, X = 20) => +(Math.round(num + `e${X}`) + `e-${X}`)

var roundUpto = function(number, upto){
return Number(number.toFixed(upto));
}
roundUpto(0.1464676, 2);
toFixed(2): Here 2 is the number of digits up to which we want to round this number.

See AmrAli's answer for a more thorough run through and performance breakdown of all the various adaptations of this solution.
var DecimalPrecision = (function(){
if (Number.EPSILON === undefined) {
Number.EPSILON = Math.pow(2, -52);
}
if(Number.isInteger === undefined){
Number.isInteger = function(value) {
return typeof value === 'number' &&
isFinite(value) &&
Math.floor(value) === value;
};
}
this.isRound = function(n,p){
let l = n.toString().split('.')[1].length;
return (p >= l);
}
this.round = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
if(n<0)
o *= -1;
return Math.round((n + r) * o) / o;
}
this.ceil = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
return Math.ceil((n + r) * o) / o;
}
this.floor = function(n, p=2){
if(Number.isInteger(n) || this.isRound(n,p))
return n;
let r = 0.5 * Number.EPSILON * n;
let o = 1; while(p-- > 0) o *= 10;
return Math.floor((n + r) * o) / o;
}
return this;
})();
console.log(DecimalPrecision.round(1.005));
console.log(DecimalPrecision.ceil(1.005));
console.log(DecimalPrecision.floor(1.005));
console.log(DecimalPrecision.round(1.0049999));
console.log(DecimalPrecision.ceil(1.0049999));
console.log(DecimalPrecision.floor(1.0049999));
console.log(DecimalPrecision.round(2.175495134384,7));
console.log(DecimalPrecision.round(2.1753543549,8));
console.log(DecimalPrecision.round(2.1755465135353,4));
console.log(DecimalPrecision.ceil(17,4));
console.log(DecimalPrecision.ceil(17.1,4));
console.log(DecimalPrecision.ceil(17.1,15));

which are alternative of tofixed() in javascript [duplicate]

Suppose I have a value of 15.7784514, I want to display it 15.77 with no rounding.
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
Results in -
15.8
15.78
15.778
15.7784514000
How do I display 15.77?

Convert the number into a string, match the number up to the second decimal place:
function calc(theform) {
var num = theform.original.value, rounded = theform.rounded
var with2Decimals = num.toString().match(/^-?\d+(?:\.\d{0,2})?/)[0]
rounded.value = with2Decimals
}
<form onsubmit="return calc(this)">
Original number: <input name="original" type="text" onkeyup="calc(form)" onchange="calc(form)" />
<br />"Rounded" number: <input name="rounded" type="text" placeholder="readonly" readonly>
</form>
The toFixed method fails in some cases unlike toString, so be very careful with it.

Update 5 Nov 2016
New answer, always accurate
function toFixed(num, fixed) {
var re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fixed || -1) + '})?');
return num.toString().match(re)[0];
}
As floating point math in javascript will always have edge cases, the previous solution will be accurate most of the time which is not good enough.
There are some solutions to this like num.toPrecision, BigDecimal.js, and accounting.js.
Yet, I believe that merely parsing the string will be the simplest and always accurate.
Basing the update on the well written regex from the accepted answer by #Gumbo, this new toFixed function will always work as expected.
Old answer, not always accurate.
Roll your own toFixed function:
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}

Another single-line solution :
number = Math.trunc(number*100)/100
I used 100 because you want to truncate to the second digit, but a more flexible solution would be :
number = Math.trunc(number*Math.pow(10, digits))/Math.pow(10, digits)
where digits is the amount of decimal digits to keep.
See Math.trunc specs for details and browser compatibility.

I opted to write this instead to manually remove the remainder with strings so I don't have to deal with the math issues that come with numbers:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
This will give you "15.77" with num = 15.7784514;

Update (Jan 2021)
Depending on its range, a number in javascript may be shown in scientific notation. For example, if you type 0.0000001 in the console, you may see it as 1e-7, whereas 0.000001 appears unchanged (0.000001).
If your application works on a range of numbers for which scientific notation is not involved, you can just ignore this update and use the original answer below.
This update is about adding a function that checks if the number is in scientific format and, if so, converts it into decimal format. Here I'm proposing this one, but you can use any other function that achieves the same goal, according to your application's needs:
function toFixed(x) {
if (Math.abs(x) < 1.0) {
let e = parseInt(x.toString().split('e-')[1]);
if (e) {
x *= Math.pow(10,e-1);
x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
}
} else {
let e = parseInt(x.toString().split('+')[1]);
if (e > 20) {
e -= 20;
x /= Math.pow(10,e);
x += (new Array(e+1)).join('0');
}
}
return x;
}
Now just apply that function to the parameter (that's the only change with respect to the original answer):
function toFixedTrunc(x, n) {
x = toFixed(x)
// From here on the code is the same than the original answer
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
This updated version addresses also a case mentioned in a comment:
toFixedTrunc(0.000000199, 2) => "0.00"
Again, choose what fits your application needs at best.
Original answer (October 2017)
General solution to truncate (no rounding) a number to the n-th decimal digit and convert it to a string with exactly n decimal digits, for any n≥0.
function toFixedTrunc(x, n) {
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
where x can be either a number (which gets converted into a string) or a string.
Here are some tests for n=2 (including the one requested by OP):
0 => 0.00
0.01 => 0.01
0.5839 => 0.58
0.999 => 0.99
1.01 => 1.01
2 => 2.00
2.551 => 2.55
2.99999 => 2.99
4.27 => 4.27
15.7784514 => 15.77
123.5999 => 123.59
And for some other values of n:
15.001097 => 15.0010 (n=4)
0.000003298 => 0.0000032 (n=7)
0.000003298257899 => 0.000003298257 (n=12)

parseInt is faster then Math.floor
function floorFigure(figure, decimals){
if (!decimals) decimals = 2;
var d = Math.pow(10,decimals);
return (parseInt(figure*d)/d).toFixed(decimals);
};
floorFigure(123.5999) => "123.59"
floorFigure(123.5999, 3) => "123.599"

num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)
This will return 19.66

Simple do this
number = parseInt(number * 100)/100;

Just truncate the digits:
function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}

This is not a safe alternative, as many others commented examples with numbers that turn into exponential notation, that scenery is not covered by this function
// typescript
// function formatLimitDecimals(value: number, decimals: number): number {
function formatLimitDecimals(value, decimals) {
const stringValue = value.toString();
if(stringValue.includes('e')) {
// TODO: remove exponential notation
throw 'invald number';
} else {
const [integerPart, decimalPart] = stringValue.split('.');
if(decimalPart) {
return +[integerPart, decimalPart.slice(0, decimals)].join('.')
} else {
return integerPart;
}
}
}
console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4
console.log(formatLimitDecimals(0, 4)); // 0
// not covered
console.log(formatLimitDecimals(0.000000199, 2)); // 0.00

These solutions do work, but to me seem unnecessarily complicated. I personally like to use the modulus operator to obtain the remainder of a division operation, and remove that. Assuming that num = 15.7784514:
num-=num%.01;
This is equivalent to saying num = num - (num % .01).

I fixed using following simple way-
var num = 15.7784514;
Math.floor(num*100)/100;
Results will be 15.77

My version for positive numbers:
function toFixed_norounding(n,p)
{
var result = n.toFixed(p);
return result <= n ? result: (result - Math.pow(0.1,p)).toFixed(p);
}
Fast, pretty, obvious. (version for positive numbers)

The answers here didn't help me, it kept rounding up or giving me the wrong decimal.
my solution converts your decimal to a string, extracts the characters and then returns the whole thing as a number.
function Dec2(num) {
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length == 1) {
return Number(num);
}
else {
return Number(numarr[0]+"."+numarr[1].charAt(0)+numarr[1].charAt(1));
}
}
else {
return Number(num);
}
}
Dec2(99); // 99
Dec2(99.9999999); // 99.99
Dec2(99.35154); // 99.35
Dec2(99.8); // 99.8
Dec2(10265.985475); // 10265.98

The following code works very good for me:
num.toString().match(/.\*\\..{0,2}|.\*/)[0];

This worked well for me. I hope it will fix your issues too.
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
// 5.56789 ----> $5.56
// 0.342 ----> $0.34
// -10.3484534 ----> $-10.34
// 600 ----> $600.00
function convertNumber(){
var result = toFixedNumber(document.getElementById("valueText").value);
document.getElementById("resultText").value = result;
}
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
<div>
<input type="text" id="valueText" placeholder="Input value here..">
<br>
<button onclick="convertNumber()" >Convert</button>
<br><hr>
<input type="text" id="resultText" placeholder="result" readonly="true">
</div>

An Easy way to do it is the next but is necessary ensure that the amount parameter is given as a string.
function truncate(amountAsString, decimals = 2){
var dotIndex = amountAsString.indexOf('.');
var toTruncate = dotIndex !== -1 && ( amountAsString.length > dotIndex + decimals + 1);
var approach = Math.pow(10, decimals);
var amountToTruncate = toTruncate ? amountAsString.slice(0, dotIndex + decimals +1) : amountAsString;
return toTruncate
? Math.floor(parseFloat(amountToTruncate) * approach ) / approach
: parseFloat(amountAsString);
}
console.log(truncate("7.99999")); //OUTPUT ==> 7.99
console.log(truncate("7.99999", 3)); //OUTPUT ==> 7.999
console.log(truncate("12.799999999999999")); //OUTPUT ==> 7.99

Here you are. An answer that shows yet another way to solve the problem:
// For the sake of simplicity, here is a complete function:
function truncate(numToBeTruncated, numOfDecimals) {
var theNumber = numToBeTruncated.toString();
var pointIndex = theNumber.indexOf('.');
return +(theNumber.slice(0, pointIndex > -1 ? ++numOfDecimals + pointIndex : undefined));
}
Note the use of + before the final expression. That is to convert our truncated, sliced string back to number type.
Hope it helps!

truncate without zeroes
function toTrunc(value,n){
return Math.floor(value*Math.pow(10,n))/(Math.pow(10,n));
}
or
function toTrunc(value,n){
x=(value.toString()+".0").split(".");
return parseFloat(x[0]+"."+x[1].substr(0,n));
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.4
truncate with zeroes
function toTruncFixed(value,n){
return toTrunc(value,n).toFixed(n);
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.40

If you exactly wanted to truncate to 2 digits of precision, you can go with a simple logic:
function myFunction(number) {
var roundedNumber = number.toFixed(2);
if (roundedNumber > number)
{
roundedNumber = roundedNumber - 0.01;
}
return roundedNumber;
}

I used (num-0.05).toFixed(1) to get the second decimal floored.

It's more reliable to get two floating points without rounding.
Reference Answer
var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);
Thank You !

My solution in typescript (can easily be ported to JS):
/**
* Returns the price with correct precision as a string
*
* #param price The price in decimal to be formatted.
* #param decimalPlaces The number of decimal places to use
* #return string The price in Decimal formatting.
*/
type toDecimal = (price: number, decimalPlaces?: number) => string;
const toDecimalOdds: toDecimal = (
price: number,
decimalPlaces: number = 2,
): string => {
const priceString: string = price.toString();
const pointIndex: number = priceString.indexOf('.');
// Return the integer part if decimalPlaces is 0
if (decimalPlaces === 0) {
return priceString.substr(0, pointIndex);
}
// Return value with 0s appended after decimal if the price is an integer
if (pointIndex === -1) {
const padZeroString: string = '0'.repeat(decimalPlaces);
return `${priceString}.${padZeroString}`;
}
// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {
const padZeroString: string = '0'.repeat(padZeroLen);
return `${priceString}${padZeroString}`;
}
return priceString.substr(0, pointIndex + decimalPlaces + 1);
};
Test cases:
expect(filters.toDecimalOdds(3.14159)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 2)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 0)).toBe('3');
expect(filters.toDecimalOdds(3.14159, 10)).toBe('3.1415900000');
expect(filters.toDecimalOdds(8.2)).toBe('8.20');
Any improvements?

Another solution, that truncates and round:
function round (number, decimals, truncate) {
if (truncate) {
number = number.toFixed(decimals + 1);
return parseFloat(number.slice(0, -1));
}
var n = Math.pow(10.0, decimals);
return Math.round(number * n) / n;
};

function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);

Roll your own toFixed function: for positive values Math.floor works fine.
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
For negative values Math.floor is round of the values. So you can use Math.ceil instead.
Example,
Math.ceil(-15.778665 * 10000) / 10000 = -15.7786
Math.floor(-15.778665 * 10000) / 10000 = -15.7787 // wrong.

Gumbo's second solution, with the regular expression, does work but is slow because of the regular expression. Gumbo's first solution fails in certain situations due to imprecision in floating points numbers. See the JSFiddle for a demonstration and a benchmark. The second solution takes about 1636 nanoseconds per call on my current system, Intel Core i5-2500 CPU at 3.30 GHz.
The solution I've written involves adding a small compensation to take care of floating point imprecision. It is basically instantaneous, i.e. on the order of nanoseconds. I clocked 2 nanoseconds per call but the JavaScript timers are not very precise or granular. Here is the JS Fiddle and the code.
function toFixedWithoutRounding (value, precision)
{
var factorError = Math.pow(10, 14);
var factorTruncate = Math.pow(10, 14 - precision);
var factorDecimal = Math.pow(10, precision);
return Math.floor(Math.floor(value * factorError + 1) / factorTruncate) / factorDecimal;
}
var values = [1.1299999999, 1.13, 1.139999999, 1.14, 1.14000000001, 1.13 * 100];
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 2));
}
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 4));
}
console.log("type of result is " + typeof toFixedWithoutRounding(1.13 * 100 / 100, 2));
// Benchmark
var value = 1.13 * 100;
var startTime = new Date();
var numRun = 1000000;
var nanosecondsPerMilliseconds = 1000000;
for (var run = 0; run < numRun; run++)
toFixedWithoutRounding(value, 2);
var endTime = new Date();
var timeDiffNs = nanosecondsPerMilliseconds * (endTime - startTime);
var timePerCallNs = timeDiffNs / numRun;
console.log("Time per call (nanoseconds): " + timePerCallNs);

Building on David D's answer:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156

Already there are some suitable answer with regular expression and arithmetic calculation, you can also try this
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23

Here is what is did it with string
export function withoutRange(number) {
const str = String(number);
const dotPosition = str.indexOf('.');
if (dotPosition > 0) {
const length = str.substring().length;
const end = length > 3 ? 3 : length;
return str.substring(0, dotPosition + end);
}
return str;
}

Convert Fraction String to Decimal?

I'm trying to create a javascript function that can take a fraction input string such as '3/2' and convert it to decimal—either as a string '1.5' or number 1.5
function ratio(fraction) {
var fraction = (fraction !== undefined) ? fraction : '1/1',
decimal = ??????????;
return decimal;
});
Is there a way to do this?

Since no one has mentioned it yet there is a quick and dirty solution:
var decimal = eval(fraction);
Which has the perks of correctly evaluating all sorts of mathematical strings.
eval("3/2") // 1.5
eval("6") // 6
eval("6.5/.5") // 13, works with decimals (floats)
eval("12 + 3") // 15, you can add subtract and multiply too
People here will be quick to mention the dangers of using a raw eval but I submit this as the lazy mans answer.

Here is the bare bones minimal code needed to do this:
var a = "3/2";
var split = a.split('/');
var result = parseInt(split[0], 10) / parseInt(split[1], 10);
alert(result); // alerts 1.5
JsFiddle: http://jsfiddle.net/XS4VE/
Things to consider:
division by zero
if the user gives you an integer instead of a fraction, or any other invalid input
rounding issues (like 1/3 for example)

Something like this:
bits = fraction.split("/");
return parseInt(bits[0],10)/parseInt(bits[1],10);

I have a function I use to handle integers, mixed fractions (including unicode vulgar fraction characters), and decimals. Probably needs some polishing but it works for my purpose (recipe ingredient list parsing).
NPM
GitHub
Inputs "2 1/2", "2½", "2 ½", and "2.5" will all return 2.5. Examples:
var numQty = require("numeric-quantity");
numQty("1 1/4") === 1.25; // true
numQty("3 / 4") === 0.75; // true
numQty("¼" ) === 0.25; // true
numQty("2½") === 2.5; // true
numQty("¾") === 0.75; // true
numQty("⅓") === 0.333; // true
numQty("⅔") === 0.667; // true
One thing it doesn't handle is decimals within the fraction, e.g. "2.5 / 5".

I created a nice function to do just that, everything was based off of this question and answers but it will take the string and output the decimal value but will also output whole numbers as well with out errors
https://gist.github.com/drifterz28/6971440
function toDeci(fraction) {
fraction = fraction.toString();
var result,wholeNum=0, frac, deci=0;
if(fraction.search('/') >=0){
if(fraction.search('-') >=0){
wholeNum = fraction.split('-');
frac = wholeNum[1];
wholeNum = parseInt(wholeNum,10);
}else{
frac = fraction;
}
if(fraction.search('/') >=0){
frac = frac.split('/');
deci = parseInt(frac[0], 10) / parseInt(frac[1], 10);
}
result = wholeNum+deci;
}else{
result = fraction
}
return result;
}
/* Testing values / examples */
console.log('1 ',toDeci("1-7/16"));
console.log('2 ',toDeci("5/8"));
console.log('3 ',toDeci("3-3/16"));
console.log('4 ',toDeci("12"));
console.log('5 ',toDeci("12.2"));

Too late, but can be helpful:
You can use Array.prototype.reduce instead of eval
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
ES6
const fractionStrToDecimal = str => str.split('/').reduce((p, c) => p / c);
console.log(fractionStrToDecimal('1/4/2')); // Logs 0.125
console.log(fractionStrToDecimal('3/2')); // Logs 1.5
CJS
function fractionStrToDecimal(str) {
return str.split('/').reduce((p, c) => p / c);
}
console.log(fractionStrToDecimal('1/4')); // Logs 0.25
[EDIT] Removed reducer initial value and now the function works for numerators greater than 1. Thanks, James Furey.

Function (ES6):
function fractionToDecimal(fraction) {
return fraction
.split('/')
.reduce((numerator, denominator, i) =>
numerator / (i ? denominator : 1)
);
}
Function (ES6, condensed):
function fractionToDecimal(f) {
return f.split('/').reduce((n, d, i) => n / (i ? d : 1));
}
Examples:
fractionToDecimal('1/2'); // 0.5
fractionToDecimal('5/2'); // 2.5
fractionToDecimal('1/2/2'); // 0.25
fractionToDecimal('10/5/10'); // 0.2
fractionToDecimal('0/1'); // 0
fractionToDecimal('1/0'); // Infinity
fractionToDecimal('cat/dog'); // NaN

With modern destructuring syntax, the best/safest answer can be simplified to:
const parseFraction = fraction => {
const [numerator, denominator] = fraction.split('/').map(Number);
return numerator / denominator;
}
// example
parseFraction('3/2'); // 1.5
In other words, split the faction by its / symbol, turn both resulting strings into numbers, then return the first number divided by the second ...
... all with only two (very readable) lines of code.

If you don't mind using an external library, math.js offers some useful functions to convert fractions to decimals as well as perform fractional number arithmetic.
console.log(math.number(math.fraction("1/3"))); //returns 0.3333333333333333
console.log(math.fraction("1/3") * 9) //returns 3
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjs/3.20.1/math.js"></script>

const fractionStringToNumber = s => s.split("/").map(s => Number(s)).reduce((a, b) => a / b);
console.log(fractionStringToNumber("1/2"));
console.log(fractionStringToNumber("1/3"));
console.log(fractionStringToNumber("3/2"));
console.log(fractionStringToNumber("3/1"));
console.log(fractionStringToNumber("22/7"));
console.log(fractionStringToNumber("355 / 113"));
console.log(fractionStringToNumber("8/4/2"));
console.log(fractionStringToNumber("3")); // => 3, not "3"

From a readability, step through debugging perspective, this may be easier to follow:
// i.e. '1/2' -> .5
// Invalid input returns 0 so impact on upstream callers are less likely to be impacted
function fractionToNumber(fraction = '') {
const fractionParts = fraction.split('/');
const numerator = fractionParts[0] || '0';
const denominator = fractionParts[1] || '1';
const radix = 10;
const number = parseInt(numerator, radix) / parseInt(denominator, radix);
const result = number || 0;
return result;
}

To convert a fraction to a decimal, just divide the top number by the bottom number. 5 divided by 3 would be 5/3 or 1.67. Much like:
function decimal(top,bottom) {
return (top/bottom)
}
Hope this helps, haha

It works with eval() method but you can use parseFloat method. I think it is better!
Unfortunately it will work only with that kind of values - "12.2" not with "5/8", but since you can handle with calculation I think this is good approach!

If you want to use the result as a fraction and not just get the answer from the string, a library like https://github.com/infusion/Fraction.js would do the job quite well.
var f = new Fraction("3/2");
console.log(f.toString()); // Returns string "1.5"
console.log(f.valueOf()); // Returns number 1.5
var g = new Fraction(6.5).div(.5);
console.log(f.toString()); // Returns string "13"

Also a bit late to the party, but an alternative to eval() with less security issues (according to MDN at least) is the Function() factory.
var fraction = "3/2";
console.log( Function("return (" + fraction + ");")() );
This would output the result "1.5" in the console.
Also as a side note: Mixed fractions like 1 1/2 will not work with neither eval() nor the solution with Function() as written as they both stumble on the space.

safer eval() according to MDN
const safeEval = (str) => {
return Function('"use strict";return (' + str + ")")();
}
safeEval("1 1/2") // 1.5
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/eval#Do_not_ever_use_eval!

This too will work:
let y = "2.9/59"
let a = y.split('')
let b = a.splice(a.indexOf("/"))
console.log(parseFloat(a.join('')))
a = parseFloat(a.join(''))
console.log(b)
let c = parseFloat(b.slice(1).join(''))
let d = a/c
console.log(d) // Answer for y fraction

I developed a function to convert a value using a factor that may be passed as a fraction of integers or decimals. The user input and conversion factor might not be in the correct format, so it checks for the original value to be a number, as well as that the conversion can be converted to a fraction assuming that /number means 1/number, or there are a numerator and a denominator in the format number/number.
/**
* Convert value using conversion factor
* #param {float} value - number to convert
* #param {string} conversion - factor
* #return {float} converted value
*/
function convertNumber(value, conversion) {
try {
let numberValue = eval(value);
if (isNaN(numberValue)) {
throw value + " is not a number.";
}
let fraction = conversion.toString();
let divider = fraction.indexOf("/");
let upper = 1;
let denominator = 1;
if (divider == -1) {
upper = eval(fraction);
} else {
let split = fraction.split("/");
if (split.length > 2) {
throw fraction + " cannot be evaluated to a fraction.";
} else {
denominator = eval(split[1]);
if (divider > 0) {
upper = eval(split[0]);
}
}
}
let factor = upper/denominator;
if (isNaN(factor)) {
throw fraction + " cannot be converted to a factor.";
}
let result = numberValue * factor;
if (isNaN(result)) {
throw numberValue + " * " + factor + " is not a number.";
}
return result
} catch (err) {
let message = "Unable to convert '" + value + "' using '" + conversion + "'. " + err;
throw message;
}
}

You can use eval() with regex to implement a secure method to calculate fraction
var input = "1/2";
return input.match(/^[0-9\/\.]+$/) != null ? eval(input) : "invalid number";

How to format a float in javascript?

In JavaScript, when converting from a float to a string, how can I get just 2 digits after the decimal point? For example, 0.34 instead of 0.3445434.

There are functions to round numbers. For example:
var x = 5.0364342423;
print(x.toFixed(2));
will print 5.04.
EDIT:
Fiddle

var result = Math.round(original*100)/100;
The specifics, in case the code isn't self-explanatory.
edit: ...or just use toFixed, as proposed by Tim Büthe. Forgot that one, thanks (and an upvote) for reminder :)

Be careful when using toFixed():
First, rounding the number is done using the binary representation of the number, which might lead to unexpected behaviour. For example
(0.595).toFixed(2) === '0.59'
instead of '0.6'.
Second, there's an IE bug with toFixed(). In IE (at least up to version 7, didn't check IE8), the following holds true:
(0.9).toFixed(0) === '0'
It might be a good idea to follow kkyy's suggestion or to use a custom toFixed() function, eg
function toFixed(value, precision) {
var power = Math.pow(10, precision || 0);
return String(Math.round(value * power) / power);
}

One more problem to be aware of, is that toFixed() can produce unnecessary zeros at the end of the number.
For example:
var x=(23-7.37)
x
15.629999999999999
x.toFixed(6)
"15.630000"
The idea is to clean up the output using a RegExp:
function humanize(x){
return x.toFixed(6).replace(/\.?0*$/,'');
}
The RegExp matches the trailing zeros (and optionally the decimal point) to make sure it looks good for integers as well.
humanize(23-7.37)
"15.63"
humanize(1200)
"1200"
humanize(1200.03)
"1200.03"
humanize(3/4)
"0.75"
humanize(4/3)
"1.333333"

var x = 0.3445434
x = Math.round (x*100) / 100 // this will make nice rounding

The key here I guess is to round up correctly first, then you can convert it to String.
function roundOf(n, p) {
const n1 = n * Math.pow(10, p + 1);
const n2 = Math.floor(n1 / 10);
if (n1 >= (n2 * 10 + 5)) {
return (n2 + 1) / Math.pow(10, p);
}
return n2 / Math.pow(10, p);
}
// All edge cases listed in this thread
roundOf(95.345, 2); // 95.35
roundOf(95.344, 2); // 95.34
roundOf(5.0364342423, 2); // 5.04
roundOf(0.595, 2); // 0.60
roundOf(0.335, 2); // 0.34
roundOf(0.345, 2); // 0.35
roundOf(551.175, 2); // 551.18
roundOf(0.3445434, 2); // 0.34
Now you can safely format this value with toFixed(p).
So with your specific case:
roundOf(0.3445434, 2).toFixed(2); // 0.34

There is a problem with all those solutions floating around using multipliers. Both kkyy and Christoph's solutions are wrong unfortunately.
Please test your code for number 551.175 with 2 decimal places - it will round to 551.17 while it should be 551.18 ! But if you test for ex. 451.175 it will be ok - 451.18. So it's difficult to spot this error at a first glance.
The problem is with multiplying: try 551.175 * 100 = 55117.49999999999 (ups!)
So my idea is to treat it with toFixed() before using Math.round();
function roundFix(number, precision)
{
var multi = Math.pow(10, precision);
return Math.round( (number * multi).toFixed(precision + 1) ) / multi;
}

If you want the string without round you can use this RegEx (maybe is not the most efficient way... but is really easy)
(2.34567778).toString().match(/\d+\.\d{2}/)[0]
// '2.34'

function trimNumber(num, len) {
const modulu_one = 1;
const start_numbers_float=2;
var int_part = Math.trunc(num);
var float_part = String(num % modulu_one);
float_part = float_part.slice(start_numbers_float, start_numbers_float+len);
return int_part+'.'+float_part;
}

There is no way to avoid inconsistent rounding for prices with x.xx5 as actual value using either multiplication or division. If you need to calculate correct prices client-side you should keep all amounts in cents. This is due to the nature of the internal representation of numeric values in JavaScript. Notice that Excel suffers from the same problems so most people wouldn't notice the small errors caused by this phenomen. However errors may accumulate whenever you add up a lot of calculated values, there is a whole theory around this involving the order of calculations and other methods to minimize the error in the final result. To emphasize on the problems with decimal values, please note that 0.1 + 0.2 is not exactly equal to 0.3 in JavaScript, while 1 + 2 is equal to 3.

Maybe you'll also want decimal separator? Here is a function I just made:
function formatFloat(num,casasDec,sepDecimal,sepMilhar) {
if (num < 0)
{
num = -num;
sinal = -1;
} else
sinal = 1;
var resposta = "";
var part = "";
if (num != Math.floor(num)) // decimal values present
{
part = Math.round((num-Math.floor(num))*Math.pow(10,casasDec)).toString(); // transforms decimal part into integer (rounded)
while (part.length < casasDec)
part = '0'+part;
if (casasDec > 0)
{
resposta = sepDecimal+part;
num = Math.floor(num);
} else
num = Math.round(num);
} // end of decimal part
while (num > 0) // integer part
{
part = (num - Math.floor(num/1000)*1000).toString(); // part = three less significant digits
num = Math.floor(num/1000);
if (num > 0)
while (part.length < 3) // 123.023.123 if sepMilhar = '.'
part = '0'+part; // 023
resposta = part+resposta;
if (num > 0)
resposta = sepMilhar+resposta;
}
if (sinal < 0)
resposta = '-'+resposta;
return resposta;
}

/** don't spend 5 minutes, use my code **/
function prettyFloat(x,nbDec) {
if (!nbDec) nbDec = 100;
var a = Math.abs(x);
var e = Math.floor(a);
var d = Math.round((a-e)*nbDec); if (d == nbDec) { d=0; e++; }
var signStr = (x<0) ? "-" : " ";
var decStr = d.toString(); var tmp = 10; while(tmp<nbDec && d*tmp < nbDec) {decStr = "0"+decStr; tmp*=10;}
var eStr = e.toString();
return signStr+eStr+"."+decStr;
}
prettyFloat(0); // "0.00"
prettyFloat(-1); // "-1.00"
prettyFloat(-0.999); // "-1.00"
prettyFloat(0.5); // "0.50"

I use this code to format floats. It is based on toPrecision() but it strips unnecessary zeros. I would welcome suggestions for how to simplify the regex.
function round(x, n) {
var exp = Math.pow(10, n);
return Math.floor(x*exp + 0.5)/exp;
}
Usage example:
function test(x, n, d) {
var rounded = rnd(x, d);
var result = rounded.toPrecision(n);
result = result.replace(/\.?0*$/, '');
result = result.replace(/\.?0*e/, 'e');
result = result.replace('e+', 'e');
return result;
}
document.write(test(1.2000e45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2000e+45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2340e45, 3, 2) + '=' + '1.23e45' + '<br>');
document.write(test(1.2350e45, 3, 2) + '=' + '1.24e45' + '<br>');
document.write(test(1.0000, 3, 2) + '=' + '1' + '<br>');
document.write(test(1.0100, 3, 2) + '=' + '1.01' + '<br>');
document.write(test(1.2340, 4, 2) + '=' + '1.23' + '<br>');
document.write(test(1.2350, 4, 2) + '=' + '1.24' + '<br>');

countDecimals = value => {
if (Math.floor(value) === value) return 0;
let stringValue = value.toString().split(".")[1];
if (stringValue) {
return value.toString().split(".")[1].length
? value.toString().split(".")[1].length
: 0;
} else {
return 0;
}
};
formatNumber=(ans)=>{
let decimalPlaces = this.countDecimals(ans);
ans = 1 * ans;
if (decimalPlaces !== 0) {
let onePlusAns = ans + 1;
let decimalOnePlus = this.countDecimals(onePlusAns);
if (decimalOnePlus < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
} else {
let tenMulAns = ans * 10;
let decimalTenMul = this.countDecimals(tenMulAns);
if (decimalTenMul + 1 < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
}
}
}
}
I just add 1 to the value and count the decimal digits present in the original value and the added value. If I find the decimal digits after adding one less than the original decimal digits, I just call the toFixed() with (original decimals - 1). I also check by multiplying the original value by 10 and follow the same logic in case adding one doesn't reduce redundant decimal places.
A simple workaround to handle floating-point number rounding in JS. Works in most cases I tried.