Fastest method to replace all instances of a character in a string [duplicate]

Answer 1

I tried a number of these suggestions after realizing that an implementation I had written of this probably close to 10 years ago actually didn't work completely (nasty production bug in an long-forgotten system, isn't that always the way?!)... what I noticed is that the ones I tried (I didn't try them all) had the same problem as mine, that is, they wouldn't replace EVERY occurrence, only the first, at least for my test case of getting "test....txt" down to "test.txt" by replacing ".." with "."... maybe I missed so regex situation? But I digress...

So, I rewrote my implementation as follows. It's pretty darned simple, although I suspect not the fastest but I also don't think the difference will matter with modern JS engines, unless you're doing this inside a tight loop of course, but that's always the case for anything...

function replaceSubstring(inSource, inToReplace, inReplaceWith) {

  var outString = inSource;
  while (true) {
    var idx = outString.indexOf(inToReplace);
    if (idx == -1) {
      break;
    }
    outString = outString.substring(0, idx) + inReplaceWith +
      outString.substring(idx + inToReplace.length);
  }
  return outString;

}

Hope that helps someone!

Answer 2

Use the replace() method of the String object.

As mentioned in the selected answer, the /g flag should be used in the regex, in order to replace all instances of the substring in the string.

Answer 3

Use Regex object like this

var regex = new RegExp('"', 'g'); str = str.replace(regex, '\'');

It will replace all occurrence of " into '.

Answer 4

I think the real answer is that it completely depends on what your inputs look like. I created a JsFiddle to try a bunch of these and a couple of my own against various inputs. No matter how I look at the results, I see no clear winner.

• RegExp wasn't the fastest in any of the test cases, but it wasn't bad either.
• Split/Join approach seems fastest for sparse replacements.

• This one I wrote seems fastest for small inputs and dense replacements:

  function replaceAllOneCharAtATime(inSource, inToReplace, inReplaceWith) {
      var output="";
      var firstReplaceCompareCharacter = inToReplace.charAt(0);
      var sourceLength = inSource.length;
      var replaceLengthMinusOne = inToReplace.length - 1;
      for(var i = 0; i < sourceLength; i++){
          var currentCharacter = inSource.charAt(i);
          var compareIndex = i;
          var replaceIndex = 0;
          var sourceCompareCharacter = currentCharacter;
          var replaceCompareCharacter = firstReplaceCompareCharacter;
          while(true){
              if(sourceCompareCharacter != replaceCompareCharacter){
              output += currentCharacter;
              break;
          }
          if(replaceIndex >= replaceLengthMinusOne) {
              i+=replaceLengthMinusOne;
              output += inReplaceWith;
              //was a match
              break;
          }
          compareIndex++; replaceIndex++;
          if(i >= sourceLength){
              // not a match
              break;
          }
          sourceCompareCharacter = inSource.charAt(compareIndex)
              replaceCompareCharacter = inToReplace.charAt(replaceIndex);
          }   
          replaceCompareCharacter += currentCharacter;
      }
      return output;
  }
  

Answer 5

What's the fastest I don't know, but I know what's the most readable - that what's shortest and simplest. Even if it's a little bit slower than other solution it's worth to use.

So use:

 "string".replace("a", "b");
 "string".replace(/abc?/g, "def");

And enjoy good code instead of faster (well... 1/100000 sec. is not a difference) and ugly one. ;)

Answer 6

The easiest would be to use a regular expression with g flag to replace all instances:

str.replace(/foo/g, "bar")

This will replace all occurrences of foo with bar in the string str. If you just have a string, you can convert it to a RegExp object like this:

var pattern = "foobar",
    re = new RegExp(pattern, "g");