Finding all indexes of a specified character within a string

Question

For example, if I had "scissors" in variable and wanted to know the position of all occurrences of the letter "s", it should print out 1, 4, 5, 8.

How can I do this in JavaScript in most efficient way? I don't think looping through the whole is terribly efficient

Answer 1

A simple loop works well:

var str = "scissors"; var indices = []; for(var i=0; i<str.length;i++) {     if (str[i] === "s") indices.push(i); } 

Now, you indicate that you want 1,4,5,8. This will give you 0, 3, 4, 7 since indexes are zero-based. So you could add one:

if (str[i] === "s") indices.push(i+1); 

and now it will give you your expected result.

A fiddle can be see here.

I don't think looping through the whole is terribly efficient

As far as performance goes, I don't think this is something that you need to be gravely worried about until you start hitting problems.

Here is a jsPerf test comparing various answers. In Safari 5.1, the IndexOf performs the best. In Chrome 19, the for loop is the fastest.

Answer 2

Using the native String.prototype.indexOf method to most efficiently find each offset.

function locations(substring,string){   var a=[],i=-1;   while((i=string.indexOf(substring,i+1)) >= 0) a.push(i);   return a; }  console.log(locations("s","scissors")); //-> [0, 3, 4, 7] 

This is a micro-optimization, however. For a simple and terse loop that will be fast enough:

// Produces the indices in reverse order; throw on a .reverse() if you want for (var a=[],i=str.length;i--;) if (str[i]=="s") a.push(i);     

In fact, a native loop is faster on chrome that using indexOf!