JavaScript comparing two arrays

Question

Is there i way in JavaScript to compare two arrays;

This is my example.

array1 = ['jpg','png'];

array 2 = ['jpg','pdf','png','bmp'];

I my case i need if are all elements of array1 part of array2 return TRUE, or something. In case only one element of array 1 match, element in array 2 return FALSE. Order its not important. Basciclly this is validation of uploaded files, i try to removing button, if two file are not with right extension.

I try function inarray, but i think it only works with string not array

`array1` compared to `array2` return `true` and `array2` compared to `array1` return `false`? — Bharadwaj, Dec 29 '14 at 13:18
I heard somewhere that you can just compare the `toString` of each array. — Mr. Polywhirl, Dec 29 '14 at 13:20
I don't think this is a duplicate of [Comparing two arrays in Javascript](http://stackoverflow.com/questions/7837456/comparing-two-arrays-in-javascript). This question asks how to test if `array1` is a subset of `array2` where order doesn't matter, which seems quite different from testing the equality of two arrays. I'm not 100% sure if this is a *good* question (maybe it's too specific to be helpful to any likely future readers? maybe not?), but it's not a duplicate. — apsillers, Dec 29 '14 at 13:24
This is not duplicated at all. But it's already closed, so whatever. @Vladimir Štus here is an answer with proper validation http://jsfiddle.net/wxfnr2f2/. Also check loxxy's answer for another working solution for your problem. — dfsq, Dec 29 '14 at 13:26

score 2 · Accepted Answer · answered Dec 29 '14 at 13:19

2

If legacy is not a problem, Something like this would do:

var array1 = ['jpg','png','d'];
var array2 = ['jpg','pdf','png','bmp'];

var result = !array1.filter(function(a) { return array2.indexOf(a)==-1; }).length;   

// result is False

answered Dec 29 '14 at 13:19

loxxy

12,990
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OP wants to check **all elements** are there, `intersection.length === array1.length` – Paul S. Dec 29 '14 at 13:24
@PaulS. The code in this answer appears to be behave correctly. It *eliminates* items that *are* a match and tests for any remaining non-match items in `array1` (thus a non-zero length is a failure condition, implying that something wasn't eliminated as a match). – apsillers Dec 29 '14 at 13:28
This is correct although not optimal answer. No need to check further is something is already returned `-1`. – dfsq Dec 29 '14 at 13:32
1

Per @dfsq's suggestion, a better solution might be `array1.every(function(a) { return array2.indexOf(a)!=-1; })`, which will return `false` early on the first failure (and the use of `every` here seems more idiomatic to me than `filter`, but that's a bit nitpicky). – apsillers Dec 29 '14 at 13:33
1

@apsillers No, my suggestion was `!array1.some(function(el) { return array2.indexOf(el) == -1; });`, so it never checks further is something is already `-1`. – dfsq Dec 29 '14 at 13:35
@dfsq True, Valid Point.. But something tells me, that OP has a pretty small set, for this to matter at all. – loxxy Dec 29 '14 at 13:36
@dfsq Our solutions appear to be behaviorally identical `:)` I decided to put my `!` inside the callback; you prefer it outside. Both `some` and `every` perform early termination. – apsillers Dec 29 '14 at 13:44
so is it loxxy answer valid – Vladimir Štus Dec 29 '14 at 13:48
@VladimirŠtus Yup, it is. However, Close your eyes & pick one from the comments of apsillers or dsfq, if you are still worried about performance. – loxxy Dec 29 '14 at 13:57

score 0 · Answer 2 · answered Dec 29 '14 at 13:30

if are all elements of array1 part of array2 return TRUE
only one element of array1 match, element in array2 return FALSE

You can think about this in two ways,

take the intersection and check the intersection is the same as the input
loop over the input and return false on the first mismatch

Here is an example of using a loop

var a = [1, 3],
    b = [1, 5],
    c = [0, 1, 2, 3, 4];

function test(needles, haystack) {
    var i;
    for (i = 0; i < needles.length; ++i) {
        if (haystack.indexOf(needles[i]) === -1) {
            return false;
        }
    }
    return true;
}

test(a, c); // true
test(b, c); // false

KooiInc · Answer 3 · 2014-12-29T13:40:12.560

If the result of filtering the second array with the values of the first is an array with length equal to the length of the first array, the first array is a subset of the second. You can write a function for that, or assing an extra method to Array.prototype:

var array1 = ['jpg', 'png'];
var array2 = ['jpg', 'pdf', 'png', 'bmp'];
var array3 = ['jpg', 'bmp', 'pdf'];

Helpers.log2Screen('array1 subset of array2? ', 
                    isSubset(array1, array2) ? 'yes' : 'no' );

// assign and use Array.prototype.subsetOf
Array.prototype.subsetOf = isSubsetProto;
Helpers.log2Screen('array1 subset of array3? ', 
                    array1.subsetOf(array3) ? 'yes' : 'no' );

function isSubset(arr1, arr2) {
  return arr2.filter(
          function (v) {return arr1.indexOf(v) > -1; }
         ).length == arr1.length;
}

function isSubsetProto(arr) {
    return arr.filter(
          function (v) {return this.indexOf(v) > -1; }, this
         ).length == this.length;
}

<script src="http://kooiinc.github.io/JSHelpers/Helpers-min.js"></script>

JavaScript comparing two arrays

3 Answers3