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Javascript – Regex to group numbers in threes excluding decimals

How can I define a regex that groups a number into threes excluding decimals?

Example

1123456789 -> ["1","123","456","789"]

1123456789.999 -> ["1","123","456","789"]

I’ve tried this Regex:

/d{1,3}(?=(d{3})*$)/g

But it outputs the following:

1123456789 -> ["1","123","456","789"]

1123456789.999 -> ["999"]

Edit: removed _ from the example strings.

Tags:

4

Answers


  1. 0

    Update to handle updated question

    You could use any script that adds thousand separators, here I assume the number is smaller than MAXINT

    const en_formatter = new Intl.NumberFormat('en-US', { maximumFractionDigits: 0 });
let value = 1123456789.999;
let formatted = en_formatter.format(value);
console.log(formatted.split(','))

let formatted1 = parseInt(value).toString().replace(/B(?=(d{3})+(?!d))/g, ",").split(',');
console.log(formatted1)

    Answers based on original question which was about 1_123_456_789

    Just add the underscore to the regex (assuming there will always be an underscore) and use a negative lookbehind for the decimals
    We then map to get the second part of the match, which is the group

    const str = '1_123_456_789.999';
const regex = /(?<!.d{0,3})(d{1,3})_?/g;
const matches = [...str.matchAll(regex)];
const digitGroups = matches.map(match => match[1]); // Extracting the first captured group from each match
console.log(digitGroups);

    Alternative

    const groups = '1_123_456_789.999'.split('_').map(str => parseInt(str));
console.log(groups);
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  2. 0

    What happens is that the regex is set to find the end of the string, marked with the $. That is why it will return the numbers after the decimal only. If you want the oposite, make it start on the start of the string instead.

    /^d{1,3}(?=(d{3})*)/
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  3. 0

    You should be able to use something like this –

    const str = '1_123_456_789.999';
const regex = /(?<![.d])d{1,3}/gm;
const matches = [...str.matchAll(regex)];
const digitGroups = matches.map(match => match[0]);
console.log(digitGroups);

    to get your matches. It adds a digit search to the negative look-behind to get around matching digit strings that may occur AFTER the decimal

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  4. 0

    If supported in your environment, you can make use of an infinite quantifier in a lookbehind assertion:

    d{1,3}(?<!.d*)(?=(?:d{3})*b)

    The pattern in parts matches:

    • d{1,3} Match 1-3 digits
    • (?<!.d*) Negative lookbehind to assert that the current match is not preceded by a dot and optional digits
    • (?=(?:d{3})*b) Positive lookahead, assert optional repetitions of 3 digits followed by a word boundary

    See a regex demo

    const regex = /d{1,3}(?<!.d*)(?=(?:d{3})*b)/g;
[
  "1123456789",
  "1123456789.999"
].forEach(s =>
  console.log(s.match(regex))
)
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