Javascript: How to retrieve the number of decimals of a string number?

Question

I have a set of string numbers having decimals, for example: 23.456, 9.450, 123.01... I need to retrieve the number of decimals for each number, knowing that they have at least 1 decimal.

In other words, the retr_dec() method should return the following:

retr_dec("23.456") -> 3
retr_dec("9.450")  -> 3
retr_dec("123.01") -> 2

Trailing zeros do count as a decimal in this case, unlike in this related question.

Is there an easy/delivered method to achieve this in Javascript or should I compute the decimal point position and compute the difference with the string length? Thanks

Mike Samuel · Accepted Answer · 2012-05-04T18:59:50.983

117

function decimalPlaces(num) {
  var match = (''+num).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
  if (!match) { return 0; }
  return Math.max(
       0,
       // Number of digits right of decimal point.
       (match[1] ? match[1].length : 0)
       // Adjust for scientific notation.
       - (match[2] ? +match[2] : 0));
}

The extra complexity is to handle scientific notation so

decimalPlaces('.05')
2
decimalPlaces('.5')
1
decimalPlaces('1')
0
decimalPlaces('25e-100')
100
decimalPlaces('2.5e-99')
100
decimalPlaces('.5e1')
0
decimalPlaces('.25e1')
1

edited May 04 '12 at 18:59

answered May 04 '12 at 18:53

Mike Samuel

118,113
30
216
245

Note that the OP has already stated that it is not a number, but a string with a number inside. – Phrogz May 04 '12 at 18:55
1

@Phrogz, I don't think that affects the validity of my answer, except that maybe I shouldn't count 0's on the end of the \.(\d+) – Mike Samuel May 04 '12 at 18:57
It doesn't, I just wanted to point out that the `''+` was unnecessary in this case. – Phrogz May 04 '12 at 19:28
1.00 returns 0 :(, instead of 2 – arod Oct 21 '15 at 03:06
2

@arod, `decimalPlaces("1.00") === 2` but `decimalPlaces(1.00) === 0` because `("" + 1.00) === "1"`. The function operates on string representations of numbers, not primitive numeric values because primitive numeric values aren't inherently decimal -- [IEEE 854 didn't make it into EcmaScript 5](http://intertwingly.net/blog/2008/07/11/Decimal-in-ECMAScript). – Mike Samuel Oct 21 '15 at 13:14
I wish I had more upvotes to give this lovely function. – russellmania Jul 13 '16 at 19:10
Isn't this a dupe of [this other question you've answered](https://stackoverflow.com/questions/9539513/is-there-a-reliable-way-in-javascript-to-obtain-the-number-of-decimal-places-of)? – Dan Dascalescu Jun 14 '17 at 08:25
@DanDascalescu, seems like it. I'm always surprised when I find that I disagree with myself. – Mike Samuel Mar 02 '18 at 17:04

score 59 · Answer 2 · edited Dec 09 '12 at 23:13

59

function retr_dec(num) {
  return (num.split('.')[1] || []).length;
}

edited Dec 09 '12 at 23:13

Community

1
1

answered May 04 '12 at 18:51

Jack

9,448
3
29
33

4

@Deleteman Thankfully OP specified `knowing that they have at least 1 decimal` – Andrew May 04 '12 at 18:52
This won't work for really small numbers that are printed in exponential notation. – Mike Samuel May 04 '12 at 18:53
3

Works great. Comments saying doesn't work when no decimal in string refer to this answer before the edit. – Brent Sandstrom Mar 02 '16 at 16:25
If the number is really small like 0.00000001, the num.toString() will return something like 1e-7 and this code will not work. – Orden Feb 26 '17 at 08:02
This also doesn't work for some "international" numbers where the numbers may be written like so `3.000,00` – Hanna Nov 13 '17 at 17:10

score 10 · Answer 3 · answered May 04 '12 at 18:53

10

function retr_dec(numStr) {
    var pieces = numStr.split(".");
    return pieces[1].length;
}

answered May 04 '12 at 18:53

Daniel Bidulock

2,344
1
26
27

1

Or even simplier: numStr.split(".")[1].length; – Saša Feb 02 '15 at 15:45
1

worth pointing out that if your number is not a string this wont work. To get it working `var pieces = numStr.toString().split(".");` – Claire Jun 08 '16 at 12:26
1

This case will throw an error if there is no decimal, e.g: `123`. – lobati Dec 08 '17 at 07:52

Phrogz · Answer 4 · 2012-05-16T12:41:59.673

5

Since there is not already a regex-based answer:

/\d*$/.exec(strNum)[0].length

Note that this "fails" for integers, but per the problem specification they will never occur.

edited May 16 '12 at 12:41

answered May 04 '12 at 18:52

Phrogz

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112
651
745

5

@LarryBattle I appreciate your comment and thoroughness, but the question clearly states, _"knowing that they have at least 1 decimal"_. If you down voted my correct answer for this reason, I ask you to reconsider your vote. – Phrogz May 16 '12 at 12:40

score 3 · Answer 5 · edited May 19 '17 at 09:20

3

You could get the length of the decimal part of your number this way:

var value = 192.123123;
stringValue = value.toString();
length = stringValue.split('.')[1].length;

It makes the number a string, splits the string in two (at the decimal point) and returns the length of the second element of the array returned by the split operation and stores it in the 'length' variable.

edited May 19 '17 at 09:20

Florian Humblot

1,121
11
29

answered May 19 '17 at 07:40

Kamlesh Kumar

87
6

OP stated that the numbers are already strings. Your answer does not provide anything new that hasn't already been suggested by Jack and Daniel Bidulock – Mike Scotty May 19 '17 at 08:51
This is by far the most straightforward answer to OP's question. OP does not ask for other notations to be taken into account (octal? hex?) – PEWColina Jun 30 '17 at 19:06
Look out for the scientific representation of the number before the decimals. It will kick in when the numbers length is 20 and above – Jens Alenius Mar 23 '18 at 14:45

score 2 · Answer 6 · answered Jan 13 '16 at 15:36

2

Try using String.prototype.match() with RegExp /\..*/ , return .length of matched string -1

function retr_decs(args) {
  return /\./.test(args) && args.match(/\..*/)[0].length - 1 || "no decimal found"
}

console.log(
  retr_decs("23.456") // 3
  , retr_decs("9.450") // 3
  , retr_decs("123.01") // 2
  , retr_decs("123") // "no decimal found"
)

answered Jan 13 '16 at 15:36

guest271314

1
15
104
177

1

FYI, this does not count trailing 0s. `retr_decs("9.00") // "no decimal found"` – Gaʀʀʏ Mar 07 '17 at 19:13
@Gaʀʀʏ Returns `2` here https://jsfiddle.net/24rqjr83/. Can you reproduce `"no decimal found"` being returned at jsfiddle? – guest271314 Mar 07 '17 at 21:49
This is very strange. When I was testing it yesterday I was not having trailing zeros counted. I verified your jsfiddle and locally as well that the function is handling them. Weird. You have my upvote. – Gaʀʀʏ Mar 08 '17 at 20:28

score 2 · Answer 7 · answered Feb 26 '17 at 08:30

I had to deal with very small numbers so I created a version that can handle numbers like 1e-7.

Number.prototype.getPrecision = function() {
  var v = this.valueOf();
  if (Math.floor(v) === v) return 0;
  var str = this.toString();
  var ep = str.split("e-");
  if (ep.length > 1) {
    var np = Number(ep[0]);
    return np.getPrecision() + Number(ep[1]);
  }
  var dp = str.split(".");
  if (dp.length > 1) {
    return dp[1].length;
  }
  return 0;
}
document.write("NaN => " + Number("NaN").getPrecision() + "<br>");
document.write("void => " + Number("").getPrecision() + "<br>");
document.write("12.1234 => " + Number("12.1234").getPrecision() + "<br>");
document.write("1212 => " + Number("1212").getPrecision() + "<br>");
document.write("0.0000001 => " + Number("0.0000001").getPrecision() + "<br>");
document.write("1.12e-23 => " + Number("1.12e-23").getPrecision() + "<br>");
document.write("1.12e8 => " + Number("1.12e8").getPrecision() + "<br>");

Gabriel Nahmias · Answer 8 · 2013-09-16T00:39:05.057

A slight modification of the currently accepted answer, this adds to the Number prototype, thereby allowing all number variables to execute this method:

if (!Number.prototype.getDecimals) {
    Number.prototype.getDecimals = function() {
        var num = this,
            match = ('' + num).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
        if (!match)
            return 0;
        return Math.max(0, (match[1] ? match[1].length : 0) - (match[2] ? +match[2] : 0));
    }
}

It can be used like so:

// Get a number's decimals.
var number = 1.235256;
console.debug(number + " has " + number.getDecimals() + " decimal places.");

// Get a number string's decimals.
var number = "634.2384023";
console.debug(number + " has " + parseFloat(number).getDecimals() + " decimal places.");

Utilizing our existing code, the second case could also be easily added to the String prototype like so:

if (!String.prototype.getDecimals) {
    String.prototype.getDecimals = function() {
        return parseFloat(this).getDecimals();
    }
}

Use this like:

console.debug("45.2342".getDecimals());

score 1 · Answer 9 · edited Sep 21 '15 at 13:16

1

A bit of a hybrid of two others on here but this worked for me. Outside cases in my code weren't handled by others here. However, I had removed the scientific decimal place counter. Which I would have loved at uni!

numberOfDecimalPlaces: function (number) {
    var match = ('' + number).match(/(?:\.(\d+))?(?:[eE]([+-]?\d+))?$/);
    if (!match || match[0] == 0) {
        return 0;
    }
     return match[0].length;
}

edited Sep 21 '15 at 13:16

joyBlanks

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1
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answered Feb 02 '15 at 15:19

Liam Middleton

11
2

If you remove the scientific decimal place counter, you can simplify your regex: /(?:\.(\d+))?$/ Plus you use match[0] instead of match[1]. – Fl4v Feb 15 '17 at 16:41

score 0 · Answer 10 · answered Feb 15 '17 at 16:56

Based on Liam Middleton's answer, here's what I did (without scientific notation):

numberOfDecimalPlaces = (number) => {
            let match = (number + "").match(/(?:\.(\d+))?$/);
            if (!match || !match[1]) {
                return 0;
            }

            return match[1].length;
        };
        
alert(numberOfDecimalPlaces(42.21));

score 0 · Answer 11 · answered Apr 02 '18 at 20:37

0

function decimalPlaces(n) {
  if (n === NaN || n === Infinity)
    return 0;
  n = ('' + n).split('.');
  if (n.length == 1) {
    if (Boolean(n[0].match(/e/g)))
      return ~~(n[0].split('e-'))[1];
    return 0;

  }
  n = n[1].split('e-');
  return n[0].length + ~~n[1];

}

answered Apr 02 '18 at 20:37

dke

1

This gonna handle every condition rather be scientific or not – dke Apr 02 '18 at 20:37
Casting to string and checking for scientific notation is not really the scope of this question (which is almost 6 years old). – piksel bitworks Apr 02 '18 at 20:51

score 0 · Answer 12 · answered Apr 02 '23 at 21:24

0

if it's guaranteed to have digits below the radix, already in string format, no trailing non-digit bytes, and no chance of scientific notations, then why not just

  # 1-based indices

  length( decstr ) - index( decstr , "." )

answered Apr 02 '23 at 21:24

RARE Kpop Manifesto

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