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I am trying to convert a string to a big integer to perform some arithmetic calculations. However, when I try this:
Number("9007199254740993")
...I am getting this unexpected result:
9007199254740992
I suspect that this is probably because of the limit on the size of integers that Number is capable of working with.
Basically, I want to check if two strings are consecutive numbers or not. Since Number is not returning the correct value, I am getting the incorrect difference for "9007199254740993" and "9007199254740992". Specifically, I am expecting 1, but getting 0.
One possibility I considered is dividing each number by a factor to make each of them smaller. Is there any other solution?
Javascript's Number type is a numeric data type in the double-precision 64-bit floating point format (IEEE 754).
If you are dealing with large integers, use a BigInt or a corresponding library.
I you don't want to rely on BigInt and only have positive integers in mind, you can also write the successor test yourself. Full code in the snippet below.
Notes
A string representation of a positive integer is easily convertible to a decimal array where the index represents the exponent to the base 10. For example "42" ~> [2, 4] (since 42 = 2*10^0 + 4*10^1). You can also just as easily convert it back.
Now for the successor test you just need to define the increment operation (which is just adding 1 with carry). With that you can just compare if the increment of one number is equal to the unincremented other number (and vice versa).
Code
// Convert a string representation of positive decimal integer to an array of decimals.
const toArray = numberString => Array.from(numberString, c => parseInt(c))
.reverse();
// Convert the array representation of a positive decimal integer string back to the corresponding string representation (this is the inverse of `toArray`).
const fromArray = numberArray => numberArray.map(String)
.reverse()
.join('');
console.log(fromArray(toArray("9007199254740993")) === "9007199254740993"); // true
// Perform the increment operation on the array representation of the positive decimal integer.
const increment = numberArray => {
let carry = 1;
const incrementedNumberArray = [];
numberArray.forEach(i => {
let j;
if (carry === 0) {
j = i;
} else if (carry === 1) {
if (i === 9) {
j = 0;
} else {
j = i + 1;
carry = 0;
}
}
incrementedNumberArray.push(j);
});
if (carry === 1) {
incrementedNumberArray.push(1);
}
return incrementedNumberArray;
};
console.log(fromArray(increment(toArray("9007199254740993"))) === "9007199254740994"); // true
console.log(fromArray(increment(toArray("9999999999999999"))) === "10000000000000000"); // true
// Test if two strings represent positive integers where one is the other's successor.
const isSuccessor = (a, b) => {
const a_ = increment(toArray(a));
const b_ = increment(toArray(b));
return fromArray(a_) === b || fromArray(b_) === a;
};
console.log(isSuccessor("9007199254740993", "9007199254740994")); // true
console.log(isSuccessor("9007199254740994", "9007199254740993")); // true
console.log(isSuccessor("9999999999999999", "10000000000000000")); // true
console.log(isSuccessor("10000000000000000", "9999999999999999")); // true
console.log(isSuccessor("10000000000000000", "10000000000000002")); // false
You can use BIG integer library like one in JAVA.
check here
npm install big-integer
var bigInt = require("big-integer");
var largeNumber1 = bigInt("9007199254740993");
var largeNumber2 = bigInt("9007199254740994"); // any other number
var ans = largeNumber1.minus(largeNumber2);
if(ans == 1 || ans == -1){
console.log('consecutive ')
}else{
console.log('not consecutive ')
}
Note: I recommend you to use BigInt(as suggested by #Andreas in comment), if you are dealing with Big Numbers.
UPDATED
Use this code to compare big positive integers(The arguments should be in string format)
function compareBigNumber(num1, num2) {
if (num1 > Number.MAX_SAFE_INTEGER && num2 > Number.MAX_SAFE_INTEGER) {
var newNum1 = num1.split('').reverse();
var newNum2 = num2.split('').reverse();
do {
newNum1.pop();
newNum2.pop();
} while (newNum1[newNum1.length-1] === '0' || newNum2[newNum2.length-1] === '0')
return compareBigNumber(newNum1.reverse().join(''), newNum2.reverse().join(''));
} else if(num1 > Number.MAX_SAFE_INTEGER){
return 'num1 is greater'
} else if (num2 > Number.MAX_SAFE_INTEGER) {
return 'num2 is greater'
}
else {
var num1Int = parseInt(num1);
var num2Int = parseInt(num2);
if (num1Int > num2Int) {
return 'Num1 is greater';
} else if (num2Int > num1Int){
return 'Num2 is greater'
} else {
return 'Num1 is equal to Num2';
}
}
}
console.log(compareBigNumber("9007199254740992", "9007199254740993"))
console.log(compareBigNumber("100000000000000000000", "0"))
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;
}
Test cases:
var num1 = 10.66;
var num2 = 10.7898
The function I found on stackOverFlow:
function formatUserCurrencyValue(fieldValue){
var num = parseFloat(fieldValue);
var str = num.toFixed(10);
str = str.substring(0, str.length-7);
return str;
};
I would like the result to be like this: if 10.66 then 10.670 and if 10.78998 then 10.789. Basically if the value has 2 decimal places then the result should round up the first and then format as 3 decimals. If more than 2 decimals (eg. 10.78998) then 10.789, trimming out values after 3 decimals.
Could somebody please tell me how I can achieve this? I tried with the above function as well as some others I found but the result is not what I expected for the 10.66 scenario. I am getting 10.660 instead of 10.670.
It looks like a similar question has already been asked here: Formatting a number with exactly two decimals in JavaScript
I liked the answer that #Miguel had. Using a function to do the conversion.
function precise_round(num, decimals) {
var t=Math.pow(10, decimals);
return (Math.round((num * t) + (decimals>0?1:0)*(Math.sign(num) * (10 / Math.pow(100, decimals)))) / t).toFixed(decimals);
}
Then use the function.
precise_round(num,3)
Setting aside the fact that rounding num1 will produce 10.66 and not 10.67 what you want can be achieved with the below function which will print the appropriate results to the console.
var num1 = 10.66;
var num2 = 10.7898;
var roundIt = function(num) {
console.log(parseFloat(Math.round(num * 1000) / 1000).toFixed(3));
};
roundIt(num2); //10.790
roundIt(num1); //10.660
Since using the toFixed() method returns a string we wrap the result in parseFloat() so that we get a floating point number.
Here you have what you ask, it seems weird to me. Round all like this is something "special" at least... but, is exactly what you ask.
var weirdRound = function(n) {
var splited = n.toString().split(".");
var res = 0;
if(splited[1]) {
var len = splited[1].length;
if(len > 3) {
splited[1] = splited[1].substr(0, 3);
res = (splited.join(".") *1).toFixed(3);
} else if(len == 2) {
splited[1] = splited[1].substr(0, 1) + ((splited[1].substr(len -1, len) *1) +1);
res = (splited.join(".") *1).toFixed(3);
} else if(len == 1) {
res = n.toFixed(2)
} else {
res = n.toFixed(3);
}
}
return res.toFixed(3);
}
console.log(weirdRound(10.66));
console.log(weirdRound(10.9));
console.log(weirdRound(10.7898));
console.log(weirdRound(1.12));
console.log(weirdRound(1.12565));
console.log(weirdRound(1.125));
I have this line of code which rounds my numbers to two decimal places. But I get numbers like this: 10.8, 2.4, etc. These are not my idea of two decimal places so how I can improve the following?
Math.round(price*Math.pow(10,2))/Math.pow(10,2);
I want numbers like 10.80, 2.40, etc. Use of jQuery is fine with me.
To format a number using fixed-point notation, you can simply use the toFixed method:
(10.8).toFixed(2); // "10.80"
var num = 2.4;
alert(num.toFixed(2)); // "2.40"
Note that toFixed() returns a string.
IMPORTANT: Note that toFixed does not round 90% of the time, it will return the rounded value, but for many cases, it doesn't work.
For instance:
2.005.toFixed(2) === "2.00"
UPDATE:
Nowadays, you can use the Intl.NumberFormat constructor. It's part of the ECMAScript Internationalization API Specification (ECMA402). It has pretty good browser support, including even IE11, and it is fully supported in Node.js.
const formatter = new Intl.NumberFormat('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
});
console.log(formatter.format(2.005)); // "2.01"
console.log(formatter.format(1.345)); // "1.35"
You can alternatively use the toLocaleString method, which internally will use the Intl API:
const format = (num, decimals) => num.toLocaleString('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
});
console.log(format(2.005)); // "2.01"
console.log(format(1.345)); // "1.35"
This API also provides you a wide variety of options to format, like thousand separators, currency symbols, etc.
This is an old topic but still top-ranked Google results and the solutions offered share the same floating point decimals issue. Here is the (very generic) function I use, thanks to MDN:
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));
}
As we can see, we don't get these issues:
round(1.275, 2); // Returns 1.28
round(1.27499, 2); // Returns 1.27
This genericity also provides some cool stuff:
round(1234.5678, -2); // Returns 1200
round(1.2345678e+2, 2); // Returns 123.46
round("123.45"); // Returns 123
Now, to answer the OP's question, one has to type:
round(10.8034, 2).toFixed(2); // Returns "10.80"
round(10.8, 2).toFixed(2); // Returns "10.80"
Or, for a more concise, less generic function:
function round2Fixed(value) {
value = +value;
if (isNaN(value))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + 2) : 2)));
// Shift back
value = value.toString().split('e');
return (+(value[0] + 'e' + (value[1] ? (+value[1] - 2) : -2))).toFixed(2);
}
You can call it with:
round2Fixed(10.8034); // Returns "10.80"
round2Fixed(10.8); // Returns "10.80"
Various examples and tests (thanks to #t-j-crowder!):
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));
}
function naive(value, exp) {
if (!exp) {
return Math.round(value);
}
var pow = Math.pow(10, exp);
return Math.round(value * pow) / pow;
}
function test(val, places) {
subtest(val, places);
val = typeof val === "string" ? "-" + val : -val;
subtest(val, places);
}
function subtest(val, places) {
var placesOrZero = places || 0;
var naiveResult = naive(val, places);
var roundResult = round(val, places);
if (placesOrZero >= 0) {
naiveResult = naiveResult.toFixed(placesOrZero);
roundResult = roundResult.toFixed(placesOrZero);
} else {
naiveResult = naiveResult.toString();
roundResult = roundResult.toString();
}
$("<tr>")
.append($("<td>").text(JSON.stringify(val)))
.append($("<td>").text(placesOrZero))
.append($("<td>").text(naiveResult))
.append($("<td>").text(roundResult))
.appendTo("#results");
}
test(0.565, 2);
test(0.575, 2);
test(0.585, 2);
test(1.275, 2);
test(1.27499, 2);
test(1234.5678, -2);
test(1.2345678e+2, 2);
test("123.45");
test(10.8034, 2);
test(10.8, 2);
test(1.005, 2);
test(1.0005, 2);
table {
border-collapse: collapse;
}
table, td, th {
border: 1px solid #ddd;
}
td, th {
padding: 4px;
}
th {
font-weight: normal;
font-family: sans-serif;
}
td {
font-family: monospace;
}
<table>
<thead>
<tr>
<th>Input</th>
<th>Places</th>
<th>Naive</th>
<th>Thorough</th>
</tr>
</thead>
<tbody id="results">
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
I usually add this to my personal library, and after some suggestions and using the #TIMINeutron solution too, and making it adaptable for decimal length then, this one fits best:
function precise_round(num, decimals) {
var t = Math.pow(10, decimals);
return (Math.round((num * t) + (decimals>0?1:0)*(Math.sign(num) * (10 / Math.pow(100, decimals)))) / t).toFixed(decimals);
}
will work for the exceptions reported.
FAST AND EASY
parseFloat(number.toFixed(2))
Example
let number = 2.55435930
let roundedString = number.toFixed(2) // "2.55"
let twoDecimalsNumber = parseFloat(roundedString) // 2.55
let directly = parseFloat(number.toFixed(2)) // 2.55
One way to be 100% sure that you get a number with 2 decimals:
(Math.round(num*100)/100).toFixed(2)
If this causes rounding errors, you can use the following as James has explained in his comment:
(Math.round((num * 1000)/10)/100).toFixed(2)
I don't know why can't I add a comment to a previous answer (maybe I'm hopelessly blind, I don't know), but I came up with a solution using #Miguel's answer:
function precise_round(num,decimals) {
return Math.round(num*Math.pow(10, decimals)) / Math.pow(10, decimals);
}
And its two comments (from #bighostkim and #Imre):
Problem with precise_round(1.275,2) not returning 1.28
Problem with precise_round(6,2) not returning 6.00 (as he wanted).
My final solution is as follows:
function precise_round(num,decimals) {
var sign = num >= 0 ? 1 : -1;
return (Math.round((num*Math.pow(10,decimals)) + (sign*0.001)) / Math.pow(10,decimals)).toFixed(decimals);
}
As you can see I had to add a little bit of "correction" (it's not what it is, but since Math.round is lossy - you can check it on jsfiddle.net - this is the only way I knew how to "fix" it). It adds 0.001 to the already padded number, so it is adding a 1 three 0s to the right of the decimal value. So it should be safe to use.
After that I added .toFixed(decimal) to always output the number in the correct format (with the right amount of decimals).
So that's pretty much it. Use it well ;)
EDIT: added functionality to the "correction" of negative numbers.
toFixed(n) provides n length after the decimal point; toPrecision(x)
provides x total length.
Use this method below
// Example: toPrecision(4) when the number has 7 digits (3 before, 4 after)
// It will round to the tenths place
num = 500.2349;
result = num.toPrecision(4); // result will equal 500.2
AND if you want the number to be fixed use
result = num.toFixed(2);
I didn't find an accurate solution for this problem, so I created my own:
function inprecise_round(value, decPlaces) {
return Math.round(value*Math.pow(10,decPlaces))/Math.pow(10,decPlaces);
}
function precise_round(value, decPlaces){
var val = value * Math.pow(10, decPlaces);
var fraction = (Math.round((val-parseInt(val))*10)/10);
//this line is for consistency with .NET Decimal.Round behavior
// -342.055 => -342.06
if(fraction == -0.5) fraction = -0.6;
val = Math.round(parseInt(val) + fraction) / Math.pow(10, decPlaces);
return val;
}
Examples:
function inprecise_round(value, decPlaces) {
return Math.round(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
function precise_round(value, decPlaces) {
var val = value * Math.pow(10, decPlaces);
var fraction = (Math.round((val - parseInt(val)) * 10) / 10);
//this line is for consistency with .NET Decimal.Round behavior
// -342.055 => -342.06
if (fraction == -0.5) fraction = -0.6;
val = Math.round(parseInt(val) + fraction) / Math.pow(10, decPlaces);
return val;
}
// This may produce different results depending on the browser environment
console.log("342.055.toFixed(2) :", 342.055.toFixed(2)); // 342.06 on Chrome & IE10
console.log("inprecise_round(342.055, 2):", inprecise_round(342.055, 2)); // 342.05
console.log("precise_round(342.055, 2) :", precise_round(342.055, 2)); // 342.06
console.log("precise_round(-342.055, 2) :", precise_round(-342.055, 2)); // -342.06
console.log("inprecise_round(0.565, 2) :", inprecise_round(0.565, 2)); // 0.56
console.log("precise_round(0.565, 2) :", precise_round(0.565, 2)); // 0.57
Here's a simple one
function roundFloat(num,dec){
var d = 1;
for (var i=0; i<dec; i++){
d += "0";
}
return Math.round(num * d) / d;
}
Use like alert(roundFloat(1.79209243929,4));
Jsfiddle
Round down
function round_down(value, decPlaces) {
return Math.floor(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Round up
function round_up(value, decPlaces) {
return Math.ceil(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Round nearest
function round_nearest(value, decPlaces) {
return Math.round(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Merged https://stackoverflow.com/a/7641824/1889449 and
https://www.kirupa.com/html5/rounding_numbers_in_javascript.htm Thanks
them.
Building on top of Christian C. Salvadó's answer, doing the following will output a Number type, and also seems to be dealing with rounding well:
const roundNumberToTwoDecimalPlaces = (num) => Number(new Intl.NumberFormat('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
}).format(num));
roundNumberToTwoDecimalPlaces(1.344); // => 1.34
roundNumberToTwoDecimalPlaces(1.345); // => 1.35
The difference between the above and what has already been mentioned is that you don't need the .format() chaining when you're using it[, and that it outputs a Number type].
#heridev and I created a small function in jQuery.
You can try next:
HTML
<input type="text" name="one" class="two-digits"><br>
<input type="text" name="two" class="two-digits">
jQuery
// apply the two-digits behaviour to elements with 'two-digits' as their class
$( function() {
$('.two-digits').keyup(function(){
if($(this).val().indexOf('.')!=-1){
if($(this).val().split(".")[1].length > 2){
if( isNaN( parseFloat( this.value ) ) ) return;
this.value = parseFloat(this.value).toFixed(2);
}
}
return this; //for chaining
});
});
DEMO ONLINE:
http://jsfiddle.net/c4Wqn/
The trouble with floating point values is that they are trying to represent an infinite amount of (continuous) values with a fixed amount of bits. So naturally, there must be some loss in play, and you're going to be bitten with some values.
When a computer stores 1.275 as a floating point value, it won't actually remember whether it was 1.275 or 1.27499999999999993, or even 1.27500000000000002. These values should give different results after rounding to two decimals, but they won't, since for computer they look exactly the same after storing as floating point values, and there's no way to restore the lost data. Any further calculations will only accumulate such imprecision.
So, if precision matters, you have to avoid floating point values from the start. The simplest options are to
use a devoted library
use strings for storing and passing around the values (accompanied by string operations)
use integers (e.g. you could be passing around the amount of hundredths of your actual value, e.g. amount in cents instead of amount in dollars)
For example, when using integers to store the number of hundredths, the function for finding the actual value is quite simple:
function descale(num, decimals) {
var hasMinus = num < 0;
var numString = Math.abs(num).toString();
var precedingZeroes = '';
for (var i = numString.length; i <= decimals; i++) {
precedingZeroes += '0';
}
numString = precedingZeroes + numString;
return (hasMinus ? '-' : '')
+ numString.substr(0, numString.length-decimals)
+ '.'
+ numString.substr(numString.length-decimals);
}
alert(descale(127, 2));
With strings, you'll need rounding, but it's still manageable:
function precise_round(num, decimals) {
var parts = num.split('.');
var hasMinus = parts.length > 0 && parts[0].length > 0 && parts[0].charAt(0) == '-';
var integralPart = parts.length == 0 ? '0' : (hasMinus ? parts[0].substr(1) : parts[0]);
var decimalPart = parts.length > 1 ? parts[1] : '';
if (decimalPart.length > decimals) {
var roundOffNumber = decimalPart.charAt(decimals);
decimalPart = decimalPart.substr(0, decimals);
if ('56789'.indexOf(roundOffNumber) > -1) {
var numbers = integralPart + decimalPart;
var i = numbers.length;
var trailingZeroes = '';
var justOneAndTrailingZeroes = true;
do {
i--;
var roundedNumber = '1234567890'.charAt(parseInt(numbers.charAt(i)));
if (roundedNumber === '0') {
trailingZeroes += '0';
} else {
numbers = numbers.substr(0, i) + roundedNumber + trailingZeroes;
justOneAndTrailingZeroes = false;
break;
}
} while (i > 0);
if (justOneAndTrailingZeroes) {
numbers = '1' + trailingZeroes;
}
integralPart = numbers.substr(0, numbers.length - decimals);
decimalPart = numbers.substr(numbers.length - decimals);
}
} else {
for (var i = decimalPart.length; i < decimals; i++) {
decimalPart += '0';
}
}
return (hasMinus ? '-' : '') + integralPart + (decimals > 0 ? '.' + decimalPart : '');
}
alert(precise_round('1.275', 2));
alert(precise_round('1.27499999999999993', 2));
Note that this function rounds to nearest, ties away from zero, while IEEE 754 recommends rounding to nearest, ties to even as the default behavior for floating point operations. Such modifications are left as an exercise for the reader :)
Round your decimal value, then use toFixed(x) for your expected digit(s).
function parseDecimalRoundAndFixed(num,dec){
var d = Math.pow(10,dec);
return (Math.round(num * d) / d).toFixed(dec);
}
Call
parseDecimalRoundAndFixed(10.800243929,4) => 10.80
parseDecimalRoundAndFixed(10.807243929,2) => 10.81
Number(Math.round(1.005+'e2')+'e-2'); // 1.01
This worked for me: Rounding Decimals in JavaScript
With these examples you will still get an error when trying to round the number 1.005 the solution is to either use a library like Math.js or this function:
function round(value: number, decimals: number) {
return Number(Math.round(value + 'e' + decimals) + 'e-' + decimals);
}
Here is my 1-line solution: Number((yourNumericValueHere).toFixed(2));
Here's what happens:
1) First, you apply .toFixed(2) onto the number that you want to round off the decimal places of. Note that this will convert the value to a string from number. So if you are using Typescript, it will throw an error like this:
"Type 'string' is not assignable to type 'number'"
2) To get back the numeric value or to convert the string to numeric value, simply apply the Number() function on that so-called 'string' value.
For clarification, look at the example below:
EXAMPLE:
I have an amount that has upto 5 digits in the decimal places and I would like to shorten it to upto 2 decimal places. I do it like so:
var price = 0.26453;
var priceRounded = Number((price).toFixed(2));
console.log('Original Price: ' + price);
console.log('Price Rounded: ' + priceRounded);
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) {
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)
Better implementations
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')
* Uses exponential notation to avoid floating-point issues.
* Negative numbers round differently than positive numbers.
*/
function round(num, decimalPlaces) {
num = Math.round(num + "e" + decimalPlaces);
return Number(num + "e" + -decimalPlaces);
}
// test rounding of half
console.log( round(0.5, 0) ); // 1
console.log( round(-0.5, 0) ); // 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) {
num = Math.round(Math.abs(num) + "e" + decimalPlaces) * Math.sign(num);
return Number(num + "e" + -decimalPlaces);
}
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 number before rounding. This moves to the next representable value after the number, away from zero.
/**
* 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) {
var p = Math.pow(10, decimalPlaces);
var e = Number.EPSILON * num * p;
return Math.round((num * p) + e) / p;
}
// test rounding of half
console.log( round(0.5, 0) ); // 1
console.log( round(-0.5, 0) ); // -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
This is needed to offset the implicit round-off error that may occur during encoding of decimal numbers, particularly those having "5" in the last decimal position, like 1.005, 2.675 and 16.235. Actually, 1.005 in decimal system is encoded to 1.0049999999999999 in 64-bit binary float; while, 1234567.005 in decimal system is encoded to 1234567.0049999998882413 in 64-bit binary float.
It is worth noting that the maximum binary round-off error is dependent upon (1) the magnitude of the number and (2) the relative machine epsilon (2^-52).
Put the following in some global scope:
Number.prototype.getDecimals = function ( decDigCount ) {
return this.toFixed(decDigCount);
}
and then try:
var a = 56.23232323;
a.getDecimals(2); // will return 56.23
Update
Note that toFixed() can only work for the number of decimals between 0-20 i.e. a.getDecimals(25) may generate a javascript error, so to accomodate that you may add some additional check i.e.
Number.prototype.getDecimals = function ( decDigCount ) {
return ( decDigCount > 20 ) ? this : this.toFixed(decDigCount);
}
Number(((Math.random() * 100) + 1).toFixed(2))
this will return a random number from 1 to 100 rounded to 2 decimal places.
Using this response by reference: https://stackoverflow.com/a/21029698/454827
I build a function to get dynamic numbers of decimals:
function toDec(num, dec)
{
if(typeof dec=='undefined' || dec<0)
dec = 2;
var tmp = dec + 1;
for(var i=1; i<=tmp; i++)
num = num * 10;
num = num / 10;
num = Math.round(num);
for(var i=1; i<=dec; i++)
num = num / 10;
num = num.toFixed(dec);
return num;
}
here working example: https://jsfiddle.net/wpxLduLc/
parse = function (data) {
data = Math.round(data*Math.pow(10,2))/Math.pow(10,2);
if (data != null) {
var lastone = data.toString().split('').pop();
if (lastone != '.') {
data = parseFloat(data);
}
}
return data;
};
$('#result').html(parse(200)); // output 200
$('#result1').html(parse(200.1)); // output 200.1
$('#result2').html(parse(200.10)); // output 200.1
$('#result3').html(parse(200.109)); // output 200.11
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<div id="result"></div>
<div id="result1"></div>
<div id="result2"></div>
<div id="result3"></div>
I got some ideas from this post a few months back, but none of the answers here, nor answers from other posts/blogs could handle all the scenarios (e.g. negative numbers and some "lucky numbers" our tester found). In the end, our tester did not find any problem with this method below. Pasting a snippet of my code:
fixPrecision: function (value) {
var me = this,
nan = isNaN(value),
precision = me.decimalPrecision;
if (nan || !value) {
return nan ? '' : value;
} else if (!me.allowDecimals || precision <= 0) {
precision = 0;
}
//[1]
//return parseFloat(Ext.Number.toFixed(parseFloat(value), precision));
precision = precision || 0;
var negMultiplier = value < 0 ? -1 : 1;
//[2]
var numWithExp = parseFloat(value + "e" + precision);
var roundedNum = parseFloat(Math.round(Math.abs(numWithExp)) + 'e-' + precision) * negMultiplier;
return parseFloat(roundedNum.toFixed(precision));
},
I also have code comments (sorry i forgot all the details already)...I'm posting my answer here for future reference:
9.995 * 100 = 999.4999999999999
Whereas 9.995e2 = 999.5
This discrepancy causes Math.round(9.995 * 100) = 999 instead of 1000.
Use e notation instead of multiplying /dividing by Math.Pow(10,precision).
I'm fix the problem the modifier.
Support 2 decimal only.
$(function(){
//input number only.
convertNumberFloatZero(22); // output : 22.00
convertNumberFloatZero(22.5); // output : 22.50
convertNumberFloatZero(22.55); // output : 22.55
convertNumberFloatZero(22.556); // output : 22.56
convertNumberFloatZero(22.555); // output : 22.55
convertNumberFloatZero(22.5541); // output : 22.54
convertNumberFloatZero(22222.5541); // output : 22,222.54
function convertNumberFloatZero(number){
if(!$.isNumeric(number)){
return 'NaN';
}
var numberFloat = number.toFixed(3);
var splitNumber = numberFloat.split(".");
var cNumberFloat = number.toFixed(2);
var cNsplitNumber = cNumberFloat.split(".");
var lastChar = splitNumber[1].substr(splitNumber[1].length - 1);
if(lastChar > 0 && lastChar < 5){
cNsplitNumber[1]--;
}
return Number(splitNumber[0]).toLocaleString('en').concat('.').concat(cNsplitNumber[1]);
};
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
(Math.round((10.2)*100)/100).toFixed(2)
That should yield: 10.20
(Math.round((.05)*100)/100).toFixed(2)
That should yield: 0.05
(Math.round((4.04)*100)/100).toFixed(2)
That should yield: 4.04
etc.
/*Due to all told stuff. You may do 2 things for different purposes:
When showing/printing stuff use this in your alert/innerHtml= contents:
YourRebelNumber.toFixed(2)*/
var aNumber=9242.16;
var YourRebelNumber=aNumber-9000;
alert(YourRebelNumber);
alert(YourRebelNumber.toFixed(2));
/*and when comparing use:
Number(YourRebelNumber.toFixed(2))*/
if(YourRebelNumber==242.16)alert("Not Rounded");
if(Number(YourRebelNumber.toFixed(2))==242.16)alert("Rounded");
/*Number will behave as you want in that moment. After that, it'll return to its defiance.
*/
This is very simple and works just as well as any of the others:
function parseNumber(val, decimalPlaces) {
if (decimalPlaces == null) decimalPlaces = 0
var ret = Number(val).toFixed(decimalPlaces)
return Number(ret)
}
Since toFixed() can only be called on numbers, and unfortunately returns a string, this does all the parsing for you in both directions. You can pass a string or a number, and you get a number back every time! Calling parseNumber(1.49) will give you 1, and parseNumber(1.49,2) will give you 1.50. Just like the best of 'em!
You could also use the .toPrecision() method and some custom code, and always round up to the nth decimal digit regardless the length of int part.
function glbfrmt (number, decimals, seperator) {
return typeof number !== 'number' ? number : number.toPrecision( number.toString().split(seperator)[0].length + decimals);
}
You could also make it a plugin for a better use.
Here's a TypeScript implementation of https://stackoverflow.com/a/21323330/916734. It also dries things up with functions, and allows for a optional digit offset.
export function round(rawValue: number | string, precision = 0, fractionDigitOffset = 0): number | string {
const value = Number(rawValue);
if (isNaN(value)) return rawValue;
precision = Number(precision);
if (precision % 1 !== 0) return NaN;
let [ stringValue, exponent ] = scientificNotationToParts(value);
let shiftExponent = exponentForPrecision(exponent, precision, Shift.Right);
const enlargedValue = toScientificNotation(stringValue, shiftExponent);
const roundedValue = Math.round(enlargedValue);
[ stringValue, exponent ] = scientificNotationToParts(roundedValue);
const precisionWithOffset = precision + fractionDigitOffset;
shiftExponent = exponentForPrecision(exponent, precisionWithOffset, Shift.Left);
return toScientificNotation(stringValue, shiftExponent);
}
enum Shift {
Left = -1,
Right = 1,
}
function scientificNotationToParts(value: number): Array<string> {
const [ stringValue, exponent ] = value.toString().split('e');
return [ stringValue, exponent ];
}
function exponentForPrecision(exponent: string, precision: number, shift: Shift): number {
precision = shift * precision;
return exponent ? (Number(exponent) + precision) : precision;
}
function toScientificNotation(value: string, exponent: number): number {
return Number(`${value}e${exponent}`);
}
fun Any.twoDecimalPlaces(numInDouble: Double): String {
return "%.2f".format(numInDouble)
}
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";