I am having an issue with the way Javascript is rounding numbers when hitting 0.5.
I am writing levies calculators, and am noticing a 0.1c discrepancy in the results.
The problem is that the result for them is 21480.705 which my application translates into 21480.71, whereas the tariff says 21480.70.
This is what I am seeing with Javascript:
(21480.105).toFixed(2)
"21480.10"
(21480.205).toFixed(2)
"21480.21"
(21480.305).toFixed(2)
"21480.31"
(21480.405).toFixed(2)
"21480.40"
(21480.505).toFixed(2)
"21480.51"
(21480.605).toFixed(2)
"21480.60"
(21480.705).toFixed(2)
"21480.71"
(21480.805).toFixed(2)
"21480.81"
(21480.905).toFixed(2)
"21480.90"
Questions:
What the hell is going on with this erratic rouding?
What's the quickest easiest way to get a "rounded up" result (when hitting 0.5)?
So as some of the others already explained the reason for the 'erratic' rounding is a floating point precision problem. You can investigate this by using the toExponential() method of a JavaScript number.
(21480.905).toExponential(20)
#>"2.14809049999999988358e+4"
(21480.805).toExponential(20)
#>"2.14808050000000002910e+4"
As you can see here 21480.905, gets a double representation that is slightly smaller than 21480.905, while 21480.805 gets a double representation slightly larger than the original value. Since the toFixed() method works with the double representation and has no idea of your original intended value, it does all it can and should do with the information it has.
One way to work around this, is to shift the decimal point to the number of decimals you require by multiplication, then use the standard Math.round(), then shift the decimal point back again, either by division or multiplication by the inverse. Then finally we call toFixed() method to make sure the output value gets correctly zero-padded.
var x1 = 21480.905;
var x2 = -21480.705;
function round_up(x,nd)
{
var rup=Math.pow(10,nd);
var rdwn=Math.pow(10,-nd); // Or you can just use 1/rup
return (Math.round(x*rup)*rdwn).toFixed(nd)
}
function round_down(x,nd)
{
var rup=Math.pow(10,nd);
var rdwn=Math.pow(10,-nd);
return (Math.round(x*-rup)*-rdwn).toFixed(nd)
}
function round_tozero(x,nd)
{
return x>0?round_down(x,nd):round_up(x,nd)
}
console.log(x1,'up',round_up(x1,2));
console.log(x1,'down',round_down(x1,2));
console.log(x1,'to0',round_tozero(x1,2));
console.log(x2,'up',round_up(x2,2));
console.log(x2,'down',round_down(x2,2));
console.log(x2,'to0',round_tozero(x2,2));
Finally:
Encountering a problem like this is usually a good time to sit down and have a long think about wether you are actually using the correct data type for your problem. Since floating point errors can accumulate with iterative calculation, and since people are sometimes strangely sensitive with regards to money magically disappearing/appearing in the CPU, maybe you would be better off keeping monetary counters in integer 'cents' (or some other well thought out structure) rather than floating point 'dollar'.
The why -
You may have heard that in some languages, such as JavaScript, numbers with a fractional part are calling floating-point numbers, and floating-point numbers are about dealing with approximations of numeric operations. Not exact calculations, approximations. Because how exactly would you expect to compute and store 1/3 or square root of 2, with exact calculations?
If you had not, then now you've heard of it.
That means that when you type in the number literal 21480.105, the actual value that ends up stored in computer memory is not actually 21480.105, but an approximation of it. The value closest to 21480.105 that can be represented as a floating-point number.
And since this value is not exactly 21480.105, that means it is either slightly more than that, or slightly less than that. More will be rounded up, and less will be rounded down, as expected.
The solution -
Your problem comes from approximations, that it seems you cannot afford. The solution is to work with exact numbers, not approximate.
Use whole numbers. Those are exact. Add in a fractional dot when you convert your numbers to string.
This works in most cases. (See note below.)
The rounding problem can be avoided by using numbers represented in
exponential notation:
function round(value, decimals) {
return Number(Math.round(value+'e'+decimals)+'e-'+decimals);
}
console.log(round(21480.105, 2).toFixed(2));
Found at http://www.jacklmoore.com/notes/rounding-in-javascript/
NOTE: As pointed out by Mark Dickinson, this is not a general solution because it returns NaN in certain cases, such as round(0.0000001, 2) and with large inputs.
Edits to make this more robust are welcome.
You could round to an Integer, then shift in a comma while displaying:
function round(n, digits = 2) {
// rounding to an integer is accurate in more cases, shift left by "digits" to get the number of digits behind the comma
const str = "" + Math.round(n * 10 ** digits);
return str
.padStart(digits + 1, "0") // ensure there are enough digits, 0 -> 000 -> 0.00
.slice(0, -digits) + "." + str.slice(-digits); // add a comma at "digits" counted from the end
}
What the hell is going on with this erratic rouding?
Please reference the cautionary Mozilla Doc, which identifies the cause for these discrepancies. "Floating point numbers cannot represent all decimals precisely in binary which can lead to unexpected results..."
Also, please reference Is floating point math broken? (Thank you Robby Cornelissen for the reference)
What's the quickest easiest way to get a "rounded up" result (when hitting 0.5)?
Use a JS library like accounting.js to round, format, and present currency.
For example...
function roundToNearestCent(rawValue) {
return accounting.toFixed(rawValue, 2);
}
const roundedValue = roundToNearestCent(21480.105);
console.log(roundedValue);
<script src="https://combinatronics.com/openexchangerates/accounting.js/master/accounting.js"></script>
Also, consider checking out BigDecimal in JavaScript.
Hope that helps!
Related
I am aware of the fact that floating point calculation is likely to be inaccurate. For example:
var x = 100*0.333;
Here x gets set to 33.300000000000004 rather than 33. This might seem to be a minor problem, but it becomes serious if rounding is involved. For example, Math.ceil(x) will be incorrect - will yield 34 rather than 33.
The problem is that I was told that JS doesn't make much difference between integers and floating points. So I'm becoming worried if there's any way to ensure that integral division is correct.
The usual formula for integer division in JS seems to be Math.floor(n/m), where n, m are integers. So now let's say that m divides n evenly. Since JS treats every number as if it was a floating point number, n/m will return a floating point number. If this floating point number is not exact, but is even a very little bit smaller than the actual calculation result, then Math.floor(n/m) will get rounded down to one integer below it should.
To illustrate: if, for example, 9/3 yields 2.999999999999999 rather than 3, then Math.floor(9/3) will incorrectly yield 2 rather than 3. Well, actually, Math.floor(9/3) does give out correct results, but this is of course only an example; substitute 9 and 3 with any unfortunate choice of integers.
Can such scenario ever happen? If so, is there any way to ensure correct results?
Can such scenario ever happen?
No. If they are integers (i.e. floating numbers with no fractional part) and evenly divide, then the result will be exact and an integer again.
If the don't evenly divide, the result will always be greater than the exact result from integer division. You'll never get a 2.9999999… number that was supposed to be a 3. The floating point results are as precise as representable, rounding does not destroy the total preorder.
if you want an exact integral division result. Then do this
Math.round((m - m%n)/n)
If the value of dividing two numbers is supposed to be 3 then it will be. If the value is supposed to be 2.999999999999, it will be. Java doesn't do some kind of strange hidden rounding, up or down. If you ALWAYS need to have the value round down then Use Floor(x/y) Even if the value is 2.99999999999999999999999999999999999 it will still come back as 2. That's what Floor does. Same thing with celling. if the value is 3.0000000000000000000001 then it will be set to 4. I teach this in college, this is how it's supposed to and does work.
I was making a calculator (something like excel in javascript) and I have found a strange behavior with ParseFloat.
parseFloat(999999999999999) //999999999999999
parseFloat(9999999999999999) //10000000000000000
parseFloat(9999999999999899) //9999999999999900
Is there a limit with parseFloat function in javascript? Following ECMA Standard there is no issue about this.
Float is not an endless container. Consider this example:
console.log(0.1 + 0.2 == 0.3) // Prints... FALSE!
Or, another case:
console.log(99999999999999999999999999999999999) // Prints 1e+35
...while 1e+35 is just 1 with 35 zeroes. Original number (9999...) is so large and precise that JS starts cutting lower digits to store at least something - the source is too big to save in float.
This actually happens because of internal float conversions made by JavaScript engine and the philosophy of float type is that higher digits are more important that lower.
Your case is somewhat similar. This is because floating point type accuracy depends on its value length. So, If your value is too big or too small, you will lose precision for lower digits.
Thus you should never trust float and never compare it with other values using '==' of '===' - it may be anything.
I was bored, so I started fidlling around in the console, and stumbled onto this (ignore the syntax error):
Some variable "test" has a value, which I multiply by 10K, it suddenly changes into different number (you could call it a rounding error, but that depends on how much accuracy you need). I then multiply that number by 10, and it changes back/again.
That raises a few questions for me:
How in accurate is Javascript? Has this been determined? I.e. a number that can be taken into account?
Is there a way to fix this? I.e. to do math in Javascript with complete accuracy (within the limitations of its datatype).
Should the changed number after the second operation be interpreted as 'changing back to the original number' or 'changing again, because of the inaccuracy'?
I'm not sure whether this should be a separate question, but I was actually trying to round numbers to a certain amount after the decimal point. I've researched it a bit, and have found two methods:
> Method A
function roundNumber(number, digits) {
var multiple = Math.pow(10, digits);
return Math.floor(number * multiple) / multiple;
}
> Method B
function roundNumber(number, digits) {
return Number(number.toFixed(digits));
}
Intuitively I like method B more (looks more efficient), but I don't know what going on behind the scenes so I can't really judge. Anyone have an idea on that? Or a way to benchmark this? And why is there no native round_to_this_many_decimals function? (one that returns an integer, not a string)
How in accurate is Javascript?
Javascript uses standard double precision floating point numbers, so the precision limitations are the same as for any other language that uses them, which is most languages. It's the native format used by the processor to handle floating point numbers.
Is there a way to fix this? I.e. to do math in Javascript with complete accuracy (within the limitations of its datatype).
No. The precision limitations lies in the way that the number is stored. Floating point numbers doesn't have complete accuracy, so no matter how you do the calculations you can't achieve absolute accuracy as the result goes back into a floating point number.
If you want complete accuracy then you need to use a different data type.
Should the changed number after the second operation be interpreted as
'changing back to the original number' or 'changing again, because of
the inaccuracy'?
It's changing again.
When a number is converted to text to be displayed, it's rounded to a certain number of digits. The numbers that look like they are exact aren't, it's just that the limitations in precision doesn't show up.
When the number "changes back" it's just because the rounding again hides the limitations in the precision. Each calculation adds or subtracts a small inaccuracy in the number, and sometimes it just happens to take the number closer to the number that you had originally. Eventhough it looks like it's more accurate, it's actually less accurate as each calculation adds a bit of uncertainty.
Internally, JavaScript uses 64-bit IEEE 754 floating-point numbers, which are a widely used standard and usually guarantee about 16 digits of accuracy. The error you witnessesed was on the 17th significant digit of the number and was reeeally tiny.
Is there a way to [...] do math in Javascript with complete accuracy (within the limitations of its datatype).
I would say that JavaScript's math is completely accurate within the limitations of its datatype. The error you witnessed was outside of those limitations.
Are you working with calculations that require a higher degree of precision than that?
Should the changed number after the second operation be interpreted as 'changing back to the original number' or 'changing again, because of the inaccuracy'?
The number never really became more or less accurate than the original value. It was only when the value was converted into a decimal value that a rounding error became apparent. But this was not a case of the value "changing back" to an accurate number. The rounding error was just too small to display.
And why is there no native round_to_this_many_decimals function? (one that returns an integer, not a string)
"Why is the language this way" questions are not considered very productive here, but it is easy to get around this limitation (assuming you mean numbers and not integers). This answer has 337 upvotes: +numb.toFixed(digits);, but note that if you try to display a number produced with that expression, there's no guarantee that it will actually display with only six digits. That's probably one of the reasons why JavaScript's "round to N places" function produces a string and not a number.
I came across the same few times and with further research I was able solve the little issues by using the library below
Math.js Library
Sample
import {
atan2, chain, derivative, e, evaluate, log, pi, pow, round, sqrt
} from 'mathjs'
// functions and constants
round(e, 3) // 2.718
atan2(3, -3) / pi // 0.75
log(10000, 10) // 4
sqrt(-4) // 2i
pow([[-1, 2], [3, 1]], 2) // [[7, 0], [0, 7]]
derivative('x^2 + x', 'x') // 2 * x + 1
// expressions
evaluate('12 / (2.3 + 0.7)') // 4
evaluate('12.7 cm to inch') // 5 inch
evaluate('sin(45 deg) ^ 2') // 0.5
evaluate('9 / 3 + 2i') // 3 + 2i
evaluate('det([-1, 2; 3, 1])') // -7
// chaining
chain(3)
.add(4)
.multiply(2)
.done() // 14
I started having problems with decimals which made me learn about the whole floating point math. My question is, what's a viable solution for it?
x = 0.1;
y = 0.2;
num = x + y;
num = Math.round(num * 100) / 100;
or
x = 0.1;
y = 0.2;
num = x + y;
num = num.toFixed(2);
num = Number(num);
Are these both 100% viable options? As in, never have to worry about having the same problem anymore? Which one would you recommend? Or would you recommend a different solution? Any reason to use one solution over the other? Thanks in advance for any help.
EDIT:
Sorry I wasn't more specific. I'm fine with it always being 2 decimals, since that won't be a problem for my project. Obviously if you want more decimals you would use 1000 instead of 100 and toFixed(3), and so on. My main concern is, are the above 2 solutions 100% viable, as in, I won't have to worry about any of the same problems? And also, would you recommend the first solution or the second? Or another one altogether? Since I will be using a method quite a lot for many calculations. Thanks again for your help.
This is not a problem with JavaScript's floating point implementation, or something that will go away if you use a string formatting function like toFixed (the MDN docs for it here make clear that it is a string representation returned, not some other format of number). Rather, this is an inherent property of floating point arithmetic as a concept - it has a variable accuracy designed to closely approximate values within a certain range.
If you want your values to always be entirely accurate, the only solution is not to use floating point numbers. Generally, this is done by using integers representing some fraction of the "whole" numbers you're dealing with (e.g. pence/cents instead of pounds/euros/dollars, or milliseconds instead of seconds). Alternatively, you may be able to find a precision maths library which performs fixed-point arithmetic, so avoids the inaccuracies but will have worse performance.
If you don't mind the risk of the inaccuracies slowly building up, you can simply use formatting functions to only display to a certain precision when you output the result of a calculation. There is little point in converting to a string with a fixed precision and then back to a number, as the floating point implementation may still be unable to represent that number with complete precision.
When I pull the values I want to multiply, they're strings. So I pull them, parse them as floats (to preserve the decimal places), and multiply them together.
LineTaxRate = parseFloat(myRate) * parseFloat(myQuantity) * parseFloat(myTaxRateRound);
This has worked for 99% of my invoices but I discovered one very odd problem.
When it multiplied: 78 * 7 * 0.0725
Javascript is returning: 39.584999999999994
When you normally do the math in a calculator its: 39.585
When all is said and done, I take that number and round it using .toFixed(2)
Because Javascript is returning that number, it's not rounding to the desired value of: $39.59
I tried Math.round() the total but I still get the same number.
I have thought of rounding the number to 3 decimals then two, but that seems hacky to me.
I have searched everywhere and all I see is people mention parseFloat loses its precision, and to use .toFixed, however in the example above, that doesn't help.
Here is my test script i made to recreate the issue:
<script>
var num1 = parseFloat("78");
var num2 = parseFloat("7");
var num3 = parseFloat("0.0725");
var myTotal = num1 * num2 * num3;
var result = Math.round(myTotal*100)/100
alert(myTotal);
alert(myTotal.toFixed(2));
alert(result);
</script>
Floating points are represented in binary, not decimal. Some decimal numbers will not be represented precisely. And unfortunately, since Javascript only has one Number class, it's not a very good tool for this job. Other languages have decent decimal libraries designed to avoid precisely this kind of error. You're going to have to either accept one-cent errors, implement a solution server-side, or work very hard to fix this.
edit: ooh! you can do 78 * 7 * 725 and then divide by 10000, or to be even more precise just put the decimal point in the right place. Basically represent the tax rate as something other than a tiny fraction. Less convenient but it'll probably take care of your multiplication errors.
You might find the Accounting.js library useful for this. It has an "improved" toFixed() method.
JavaScript/TypeScript have only one Number class which is not that good. I have the same problem as I am using TypeScript. I solved my problem by using decimal.js-light library.
new Decimal(78).mul(7).mul(0.0725) returns as expected 39.585