Remove "unnecessary" spaces

Question

I would like to remove all the "unnecessary" spaces from a string. Specifically:

"a b c d" => "a b c d" // spaces between two words are left in
" a b c d " => "a b c d" // spaces not between two words are removed
"a  b c   d" => "a b c d" // as are duplicate spaces

Is there any regex out there that I can put into String.replace() to do this for me?

Please post what you have tried. – Wiktor Stribiżew Sep 02 '15 at 21:36 — Wiktor Stribiżew, Sep 02 '15 at 21:36

score 5 · Accepted Answer · answered Sep 02 '15 at 21:37

5

Use trim() to remove spaces at the ends of the strings, and regexp replacement to replace multiple spaces in the middle.

str = str.trim().replace(/\s{2,}/g, ' ');

answered Sep 02 '15 at 21:37

Barmar

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asdf · Answer 2 · 2015-09-02T21:37:59.097

3

Use split, filter and join. This will split the string, filter out the extra empty array entries then rejoin them with a space:

variable.split(" ").filter(Boolean).join(" ");

edited Sep 02 '15 at 21:37

answered Sep 02 '15 at 21:36

asdf

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Ah, of course. Thank you for the answer, but I should have thought of that myself :) – Bluefire Sep 02 '15 at 21:37
1

@Qix Performance was not among the requirements that OP stated. It it obviously important to **you**, but to show how the problem can be solved this is still the most *human readable* way to do it. As long as performance does not become an issue, I'd prefer this alot over your favored solution. – connexo Sep 02 '15 at 22:40
@connexo [Performance is a feature.](http://blog.codinghorror.com/performance-is-a-feature/) It should be a consideration in everything you do coding-wise. Always. It might be a cute one-liner, but the fact is is that it's non-performant and incredibly obtuse. – Qix - MONICA WAS MISTREATED Sep 02 '15 at 22:41
@Qix Performance is a feature, but if the gains from performance are marginal (as they most likely are in this case) readability is favorable. [Readability vs Performance](http://stackoverflow.com/questions/19709334/readability-vs-performance) – asdf Sep 02 '15 at 22:43
1

@asdf `String.replace` is incredibly readable - it's one function call vs. three. As well, this answer is **5x slower** than the other answers. I understand the necessity of readability, but in this case it's a no brainer. – Qix - MONICA WAS MISTREATED Sep 02 '15 at 22:45
1

@connexo "this is still the most human readable way to do it" --- well, it's far from readability and semantic: the question is about removing (replacing) substrings from a string, not about splitting a string into chunks. I honestly don't believe someone would find `"foo".replace` code less readable than this. – zerkms Sep 02 '15 at 22:46
1

So readability in JS code does in your opinion involve knowledge about regex syntax? I do recognize the power in regular expressions - and that was actually what OP asked for - but I wouldn't exactly consider regexes as human readable. – connexo Sep 02 '15 at 22:51
1

@connexo it's more about `String.prototype.replace` semantics. It clearly addresses the operation to be performed: to replace a substring with another substring. And yes, I believe understanding `/\s{2,}/g` is a strong "must" requirement for a developer. And yes, I find this regex trivial and 101 of regexes. – zerkms Sep 02 '15 at 22:52
@connexo regexes are neither readable nor unreadable. The official Email regex is unreadable; `\s+` is readable. If you don't know regex, you're going to have a hard time with Javascript since it's heavily used. – Qix - MONICA WAS MISTREATED Sep 02 '15 at 22:54

connexo · Answer 3 · 2015-09-02T22:34:45.480

3

This should work in ES 3 enabled browsers as well.

function deleteUnwantedWhitespace (searchstring) {
   searchstring = searchstring.trim().split(" ");
   var tempArray = [];
   for (var i = 0; i < searchstring.length; i++) {
      if (searchstring[i]) {
          tempArray.push(searchstring[i]);
      }
   }
   return tempArray.join(" ");
}

Edit: I just learned that String.prototype.trim is part of ES 5.1, so here's the polyfill:

if (!String.prototype.trim) {
  String.prototype.trim = function () {
    return this.replace(/^[\s\uFEFF\xA0]+|[\s\uFEFF\xA0]+$/g, '');
  };
}

edited Sep 02 '15 at 22:34

answered Sep 02 '15 at 21:42

connexo

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This is the fastest ES3-compatible solution. – Qix - MONICA WAS MISTREATED Sep 02 '15 at 22:43
What clients does the ES3-compliance address? Curious what makes one deliberately limit themselves without (?) a good reason. – zerkms Sep 02 '15 at 22:44
2

It addresses IE8, which is still important in many company environments, as sad as it is. – connexo Sep 02 '15 at 22:46

score 1 · Answer 4 · edited Sep 02 '15 at 22:28

1

Do it manually

var strArr = ["a b c d", " a b c d ", "a  b c   d"];

for (var i = 0; i < strArr.length; i++) {
  console.log(removeSpaces(strArr[i]));
}

function removeSpaces(str) {
  str = str.replace(/ +/g, ' ');
  str = str.trim();

  return str;
}

edited Sep 02 '15 at 22:28

Qix - MONICA WAS MISTREATED

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answered Sep 02 '15 at 21:43

AmmarCSE

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Just a note, you don't need code snippets just to have a code block. – Qix - MONICA WAS MISTREATED Sep 02 '15 at 22:29

Qix - MONICA WAS MISTREATED · Answer 5 · 2015-09-02T22:22:13.533

0

Perfectly possible. The first regex replaces 1 or more \s characters with a single space. This includes spaces, tabs, and new-lines.

str.replace(/\s+/g, ' ');

If you just want spaces, use

str.replace(/ +/g, ' ');

This is also more than 5x faster than using a Boolean filter.

edited Sep 02 '15 at 22:22

answered Sep 02 '15 at 22:16

Qix - MONICA WAS MISTREATED

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Remove "unnecessary" spaces

5 Answers5