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 Zaremba, 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 Zaremba, 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 Zaremba, 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 Zaremba, 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

81,915
7
76
137

@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 Zaremba 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

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

16,804
3
16
39

answered Jun 23 '19 at 16:53

Ghoul Ahmed

4,012
1
13
22

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