Best way to check if values of two arrays are the same/equal in JavaScript

Question

What is the best way to check if two arrays have the same/equal values (in any order) in JavaScript?

These values are just a primary keys of database entities, so they always will be different

const result = [1, 3, 8, 77]
const same = [8, 3, 1, 77]
const diff = [8, 3, 5, 77]

areValuesTheSame(result, same) // true
areValuesTheSame(result, diff) // false

How should areValuesTheSame method look like?

P.S. This question looks like a duplicate but I didn't find anything relative to Javascript.

https://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript — Irfanullah Jan, Jun 23 '19 at 16:50
not sure, they compare array [1, 2, 3] == [1, 2, 3]. I need only values [2, 3, 1] == [3, 2, 1] — Yegor, Jun 23 '19 at 16:53
honestly, I didn't think about it, cause it is needed for database sorting, when I compare `id` of entities, they are always different — Yegor, Jun 23 '19 at 16:58
@YegorZaremba Then why not just sort both arrays and iterate over them in parallel? — melpomene, Jun 23 '19 at 17:00
@melpomene Closing as a dupe of the canonical "compare arrays for equality by their elements", did I overlook something? — Bergi, Jun 23 '19 at 17:00
@Bergi We're doing a bag comparison. Order of the elements doesn't matter. (Apparently they're also unique, so sets would work.) — melpomene, Jun 23 '19 at 17:01
@Bergi, nope, compare only values of array, place is not matter, like [1, 3, 2] === [3, 1, 2]. Your dupe its just [1, 2, 3] === [1, 2, 3] — Yegor, Jun 23 '19 at 17:02
@melpomene Ah, I see now, I only read "array" so I thought "ordered". Would [comparing ECMA6 sets for equality](https://stackoverflow.com/q/31128855/1048572) or [Compare arrays as (multi-) sets](https://stackoverflow.com/q/16913668/1048572) work? — Bergi, Jun 23 '19 at 17:08
last one looks like a dupe, but answers are so old/messy, I really enjoy @ghoul-ahmed answer — Yegor, Jun 23 '19 at 17:11
@Bergi Those look good to me, but OP's actual problem is a bit more constrained, so simpler answers are possible. — melpomene, Jun 23 '19 at 17:12

score 3 · Accepted Answer · answered Jun 23 '19 at 17:18

3

I'm making the following assumptions:

The arrays only contain numbers.
You don't care about the order of the elements; rearranging the arrays is OK.

Under those conditions we can simply convert each array to a canonical string by sorting it and joining the elements with e.g. a space. Then (multi-)set equality boils down to simple string equality.

function areValuesTheSame(a, b) {
    return a.sort().join(' ') === b.sort().join(' ');
}

const result = [1, 3, 8, 77];
const same = [8, 3, 1, 77];
const diff = [8, 3, 5, 77];

console.log(areValuesTheSame(result, same));
console.log(areValuesTheSame(result, diff));

This is probably the laziest / shortest approach.

answered Jun 23 '19 at 17:18

melpomene

84,125
8
85
148

@MisterJojo Yes, that's what I said (assumption #2). – melpomene Jun 23 '19 at 17:27
your answer is so good, cause its more understandable, as someone mentioned in question, order/values place isn't matter(and it's my case), so I'll accept your solution. But @ghoul-ahmed have a nice one, too) – Yegor Jun 23 '19 at 17:36
You could also use `JSON.stringify` instead of join, to make it also work for string or any other primitive arrays. – georg Jun 23 '19 at 18:42
Pretty certain that `sort` modifies the array in place, so if you are relying on the original order of either of your arrays staying the same, this might not be the best option for you. – Alex Currie-Clark Aug 12 '22 at 22:14

score 2 · Answer 2 · answered Jun 23 '19 at 16:55

You could count all elements with a Map (this is type save) up for the one array and down for the other and check if all items have a final count of zero.

function haveSameValues(a, b) {
    const count = d => (m, v) => m.set(v, (m.get(v) || 0) + d)
    return Array
        .from(b.reduce(count(-1), a.reduce(count(1), new Map)).values())
        .every(v => v === 0);
}

const result = [1, 3, 8, 77]
const same = [8, 3, 1, 77]
const diff = [8, 3, 5, 77]

console.log(haveSameValues(result, same)); // true
console.log(haveSameValues(result, diff)); // false

score 1 · Answer 3 · edited Jun 23 '19 at 17:21

1

Try this:

const result = [1, 3, 8, 77]
const same = [8, 3, 1, 77]
const diff = [8, 3, 5, 77]
const areValuesTheSame = (a,b) => (a.length === b.length) && Object.keys(a.sort()).every(i=>a[i] === b.sort()[i])


console.log(areValuesTheSame(result, same)) // true
console.log(areValuesTheSame(result, diff)) // false

edited Jun 23 '19 at 17:21

Mister Jojo

20,093
6
21
40

answered Jun 23 '19 at 16:53

Ghoul Ahmed

4,446
1
14
23

1

That doesn't work at all. If you add `9` to `same` (as in `[8, 3, 1, 77, 9]`), it still thinks it's equal to `result`. – melpomene Jun 23 '19 at 16:57
Instead of `Object.keys()`, just use `a.sort(), b.sort().every((x, i) => x == a[i])`. Especially don't call `b.sort()` during every comparison! – Bergi Jun 23 '19 at 17:53
@MisterJojo If order doesn't matter, it probably doesn't matter to change it. If it matters, it's trivial to clone the arrays beforehand. – Bergi Jun 23 '19 at 17:53

Best way to check if values of two arrays are the same/equal in JavaScript

3 Answers3