Truncate (not round off) decimal numbers in javascript

Question

I am trying to truncate decimal numbers to decimal places. Something like this:

5.467   -> 5.46  
985.943 -> 985.94

toFixed(2) does just about the right thing but it rounds off the value. I don't need the value rounded off. Hope this is possible in javascript.

Solution 1

upd:

So, after all it turned out, rounding bugs will always haunt you, no matter how hard you try to compensate them. Hence the problem should be attacked by representing numbers exactly in decimal notation.

Number.prototype.toFixedDown = function(digits) {
    var re = new RegExp("(\\d+\\.\\d{" + digits + "})(\\d)"),
        m = this.toString().match(re);
    return m ? parseFloat(m[1]) : this.valueOf();
};

[   5.467.toFixedDown(2),
    985.943.toFixedDown(2),
    17.56.toFixedDown(2),
    (0).toFixedDown(1),
    1.11.toFixedDown(1) + 22];

// [5.46, 985.94, 17.56, 0, 23.1]

Old error-prone solution based on compilation of others':

Number.prototype.toFixedDown = function(digits) {
  var n = this - Math.pow(10, -digits)/2;
  n += n / Math.pow(2, 53); // added 1360765523: 17.56.toFixedDown(2) === "17.56"
  return n.toFixed(digits);
}

Solution 2

Dogbert's answer is good, but if your code might have to deal with negative numbers, Math.floor by itself may give unexpected results.

E.g. Math.floor(4.3) = 4, but Math.floor(-4.3) = -5

Use a helper function like this one instead to get consistent results:

truncateDecimals = function (number) {
    return Math[number < 0 ? 'ceil' : 'floor'](number);
};

// Applied to Dogbert's answer:
var a = 5.467;
var truncated = truncateDecimals(a * 100) / 100; // = 5.46

Here's a more convenient version of this function:

truncateDecimals = function (number, digits) {
    var multiplier = Math.pow(10, digits),
        adjustedNum = number * multiplier,
        truncatedNum = Math[adjustedNum < 0 ? 'ceil' : 'floor'](adjustedNum);

    return truncatedNum / multiplier;
};

// Usage:
var a = 5.467;
var truncated = truncateDecimals(a, 2); // = 5.46

// Negative digits:
var b = 4235.24;
var truncated = truncateDecimals(b, -2); // = 4200

If that isn't desired behaviour, insert a call to Math.abs on the first line:

var multiplier = Math.pow(10, Math.abs(digits)),

EDIT: shendz correctly points out that using this solution with a = 17.56 will incorrectly produce 17.55. For more about why this happens, read What Every Computer Scientist Should Know About Floating-Point Arithmetic. Unfortunately, writing a solution that eliminates all sources of floating-point error is pretty tricky with javascript. In another language you'd use integers or maybe a Decimal type, but with javascript...

This solution should be 100% accurate, but it will also be slower:

function truncateDecimals (num, digits) {
    var numS = num.toString(),
        decPos = numS.indexOf('.'),
        substrLength = decPos == -1 ? numS.length : 1 + decPos + digits,
        trimmedResult = numS.substr(0, substrLength),
        finalResult = isNaN(trimmedResult) ? 0 : trimmedResult;

    return parseFloat(finalResult);
}

For those who need speed but also want to avoid floating-point errors, try something like BigDecimal.js. You can find other javascript BigDecimal libraries in this SO question: "Is there a good Javascript BigDecimal library?" and here's a good blog post about math libraries for Javascript

Solution 3

var a = 5.467;
var truncated = Math.floor(a * 100) / 100; // = 5.46

Solution 4

You can fix the rounding by subtracting 0.5 for toFixed, e.g.

(f - 0.005).toFixed(2)

Solution 5

Consider taking advantage of the double tilde: ~~.

Take in the number. Multiply by significant digits after the decimal so that you can truncate to zero places with ~~. Divide that multiplier back out. Profit.

function truncator(numToTruncate, intDecimalPlaces) {    
    var numPower = Math.pow(10, intDecimalPlaces); // "numPowerConverter" might be better
    return ~~(numToTruncate * numPower)/numPower;
}

I'm trying to resist wrapping the ~~ call in parens; order of operations should make that work correctly, I believe.

alert(truncator(5.1231231, 1)); // is 5.1

alert(truncator(-5.73, 1)); // is -5.7

alert(truncator(-5.73, 0)); // is -5

JSFiddle link.

EDIT: Looking back over, I've unintentionally also handled cases to round off left of the decimal as well.

alert(truncator(4343.123, -2)); // gives 4300.

The logic's a little wacky looking for that usage, and may benefit from a quick refactor. But it still works. Better lucky than good.

View more solutions

Comments

[Felix Kling]

jQuery is just a framework and your problem is not jQuery related. It is more about doing some basic computation in JavaScript. I hope you are also satisfied with a non-jQuery solution.

[MsTapp]

I found it to be too much work to get my calculations to return just 2 decimals using Javascript. I was able to do it easily in my database view instead. I realize that this method won't fit every situation, but I want to put it out here because it might save somebody a lot of time.

[Nick Knowlson]

This works well but will give results that are probably undesirable if he (or someone else looking at this answer later) has to deal with negative numbers. See stackoverflow.com/a/9232092/224354

[Thomas W]

Yeah, Prototypes don't work reliably cross-browser. Instead of defining this (limited purpose) function through the type system, in a way that doesn't work reliably, why not just put it in a library.

[Thomas W]

Why undesirable? Changing the direction of rounding when you go below 0, causes all sorts of arithmetical artefacts.

[Thomas W]

Why unexpected? Changing the direction of rounding when you go below 0, causes all sorts of arithmetical artefacts & crappy math. For example, twice as many numbers will round to 0, as any other integer. For graphics, accounting & many other uses, you'll get crappy results. To tell you the truth, it'd be harder to say what your suggestion is good for as to say what it's not.

[Nick Knowlson]

It's good for exactly what it says - when you want to truncate decimals rather than rounding.

[shendz]

This does not work as excepted. Try the number 17.56 and digits = 2. It should be 17.56, but this function returns 17.55.

[shendz]

Not going to work with 17.56 because browser gives 17.56 * 100 = 1755.9999999999998 not 1756

[kirilloid]

Thanks for report, this is an important notice. Unfortunately 17.56 != 17.55 - 0.01. I updated the code.

[Nick Knowlson]

Good point shendz. I've updated my answer with a solution that eliminates all floating-point error for those who need that.

[Jeremy Witmer]

This won't work for numbers that are less than 1 if you want no decimals - truncateDecimals(.12345, 0) results in NaN unless you add a check: if(isNAN(result) result = 0; Depends on the behavior you want.

[Nick Knowlson]

I didn't know what you were talking about at first, until I realized that ".1".toString doesn't always return "0.1", maybe only in Firefox's developer tools. I'll fix it, thanks.

[Nick Knowlson]

There is a difference between rounding and truncating. Truncating is clearly the behaviour this question is seeking. If I call truncate(-3.14) and receive -4 back, I would definitely call that undesirable.

[lukas.pukenis]

if(isNan( is missing a closing bracket

[Nick Knowlson]

Heads up: as it is this doesn't work for very small numbers, numbers with more than 3 decimal places, or negative numbers. Try .0045, 5.4678 and -5.467

[Hari Pachuveetil]

+1. This does not handle integers (with no decimal point in them). Checking for decPos to be -1 will fix it. Like this: result = decPos === -1 ? num : numS.substr(0, 1 + decPos + digits);

[Nick Knowlson]

Aha, so it doesn't. Small integers end up working fine, that's probably why I didn't notice back when I first wrote it. Fixed now, thanks!

[Nick Knowlson]

Two inconsistencies with this function: This function returns a string so 1.11.toFixedDown(1) + 22 ends up as 1.122 instead of 23.1. Also 0.toFixedDown(1) should produce 0 but instead it produces -0.1.

[rgajrawala]

Note that this function removes the negative sign. Ex: (-10.2131).toFixedDown(2) // ==> 10.21.

[rgajrawala]

Also, (1e-7).toFixedDown(0) // ==> 1e-7. Does that for 1e-(>=7) (ex: 1e-8, 1e-9, ...).

[Liglo App]

This is the best answer. If you extend the Math prototype with this and check for NaN-s before executing, it would be just perfect.

[Alex Pi]

This is solution is good enough in my case, I only changed the regExp a little: "(-*\\d+\\.\\d{" + digits + "})(\\d)"

[kirilloid]

I'd write "([-+]?\\d+\\.\\d{" + digits + "})(\\d)" then.

[Admin]

I agree with Thomas. The difference in perspective may come with whether you are usually truncating for display, or for computation. From a computational perspective, this avoids "arithmetical artifacts"

[ruffin]

Good call. Using a bitwise operator coerces the value into an int, and oring with 0 means "just keep what I've already got". Does what my ~~ answer does, but with a single bitwise operation. Though it has the same limitation as written too: We can't go over 2^31.

[Gene Parcellano]

I like this solution, but just keep in mind that it's not fully supported by all browsers. (developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…‌​)

[jperelli]

How does this work reference: stackoverflow.com/questions/7487977/…

[Sanyam Jain]

var a = 65.1 var truncated = Math.floor(a * 100) / 100; // = 65.09 Hence this is not a correct solution

[Salman A]

How does it truncate something to, say, 2 decimal places?

[dmarra]

This will work so long as you match the value you are subtracting with the length you wish to have. whatever you pass to toFixed() needs to be the number of 0's after the decimal.

[Mike Makuch]

That doesn't work in all cases i.e. console.log(Math.trunc(9.28 * 100) / 100); // 9.27

[zurfyx]

@MikeMakuch that's not a problem with Math.trunc, but rather than 9.28 * 100 is 927.9999 rather than 928. You might want to read over The Perils of Floating Point

[giles123]

Two years after this was posted but stumbled across this when I was trying to work out the best way using Math.trunc, regex, etc. I really like this solution. Dead simple but works perfectly (for my use case anyway).

[giles123]

Don't forget to account for n=0 though.

[jStaff]

This answer is obsurd. You shouldn't be editting the javascript Number class to remove decimal points. It complicates the issue a lot more than it has to be.

[Ling Loeng]

x = 0.0000 test truncateDecimals (x, 2) fails. returns 0. not as expected 0.00

[Konstantin Smolyanin]

It seems that function works correctly but author's statement: "rounding bugs will always haunt you" is incorrect. There are also correct and more simple (and fast) solutions are provided in some other answers here.

[Alex]

not correct when truncate((10 * 2.9) / 100); this code return 0.28 instead of 0.29 jsfiddle.net/9pf0732d

[Alex]

truncator((10 * 2.9) / 100, 2) return 0.28 instead of 0.29 ... jsfiddle.net/25tgrzq1

[ruffin]

@Alex As I'm guessing you realize... welcome to JavaScript!. There are fixes. Perhaps you'd like to share one? :D

[Alex]

@ruffin I know about this issue =) I thought this answer was the solution to this problem. Unfortunately, I haven’t found an exact solution yet, everywhere there is such a problem.

[Esteban]

I'd write "([-+]?\\d*\\.\\d{" + digits + "})(\\d)" => Ex: .00 , .00009, .999999

[iCode]

I like that this works with strings, thus eliminating the nuances of floating point numbers. Thanks!

[taxilian]

this is the method that I would have expected to use

[HelpfulPanda]

Not working for me, truncator(1000.12345678, 7) returns 141.1299975

[HelpfulPanda]

Would love if that worked 100%, but Math.trunc(0.29 * Math.pow(10, 2)) / Math.pow(10, 2) gives you 0.28. To be honest the only reliable way I've seen so far is by splitting/cutting strings.

[HelpfulPanda]

Same issue as others pointed out, truncate(0.29, 2) gives you 0.28.

[HelpfulPanda]

console.log((-5).trim(2)); throws Uncaught TypeError: d[1] is undefined

[ruffin]

@Alex If you want to avoid floating point errors, use a decimal type. Or I guess we could rewrite to use string manipulation, but that seems insane. You'll notice similar answers have the same issue.

[ruffin]

@HelpfulPanda That's because JavaScript uses 32-bit ints for bitwise operators. The max 32-bit int is 2,147,483,647, and 100012345678 is significantly bigger than 2147483647. If you really have numbers larger than 32-bit (rather, if you need that many significant digits), this isn't the droid you're looking for. This is a quick and dirty (and fast) answer for, say, 4-5 sig digits. For potentially overkill-ish perfection, try the accepted answer. ;^D

[Ankur144]

This code will add padding "0" is places after decimal is than what is required.

[Ankur144]

eg- for 3 decimal places (with padding"0") 5.3 --->5.300 0.01 ---->0.010 0.00001 --->0.000 5.32195--->5.321 -3.66696---> -3.666

[stackoverflow]

@HelpfulPanda pff yeah, wtf :| I'm currently having hard times solving this issue, so I think I'll fallback as you proposed to strings :)