A Loosely-typed Indexof()?

Let's suppose I have the array: [1, 2, 3, 4] Now let's suppose I want to know the first index of a value that ==s (not ===s) a given value, like '3'. For this case I would want to

Solution 1:

Update 2021-04-13: ES6 introduced findIndex() and arrow functions, so you can now do...

> [1,2,3,4].findIndex(v => v == '3')
2

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[Original answer]

Well, there's the naive implementation ...

functionlooseIndex(needle, haystack) {
  for (var i = 0, len = haystack.length; i < len; i++) {
    if (haystack[i] == needle) return i;
  }
  return-1;
}

E.g.

> [0,1,2].indexOf('1')
-1// Native implementation doesn't match
> looseIndex('1', [0, 1, 2])
1// Our implementation detects `'1' == 1`
> looseIndex('1', [0, true, 2])
1// ... and also that `true == 1`

(One bonus of this approach is that it works with objects like NodeLists and arguments that may not support the full Array API.)

If your array is sorted, you can do a binary search for better performance. See Underscore.js' sortedIndex code for an example

Solution 2:

indexOf does strict equals, so do a conversion on either of the sides, You can try this way:

Either:

arr.map(String).indexOf("3");

or

var str = "3";
arr.indexOf(+str);

Fiddle

Solution 3:

Perhaps:

functiongetTypelessIndex(a, v) {
  return a.indexOf(String(v)) == -1)? a.indexOf(Number(v)) : array.indexOf(String(v));
}

But that may do indexOf(String(v)) more than is desirable. To only do the index each way once at most:

functiongetTypelessIndex(a, v) {
  var index = a.indexOf(String(v));

  if (index == -1) {
    index = a.indexOf(Number(v));
  }
  return index;
}

Note that Array.prototype.indexOf is not supported in IE 9 and lower.