using lodash to compare arrays (items existence without order)

I know I can do it using loops, but I am trying to find an elegant way of doing this:

I have two arrays:

var array1 = [['a', 'b'], ['b', 'c']];
var array2 = [['b', 'c'], ['a', 'b']];

I want to use lodash to confirm that the two are the same. By 'the same' I mean that there is no item in array1 that is not contained in array2.

In terms of checking equality between these items:

['a', 'b'] == ['b', 'a'] 


['a', 'b'] == ['a', 'b'] 

both work since the letters will always be in order.

Thanks in advance.



If you sort the outer array, you can use _.isEqual() since the inner array is already sorted.

var array1 = [['a', 'b'], ['b', 'c']];
var array2 = [['b', 'c'], ['a', 'b']];
_.isEqual(array1.sort(), array2.sort()); //true

Note that .sort() will mutate the arrays. If that's a problem for you, make a copy first using (for example) .slice() or the spread operator (...).

Or, do as Daniel Budick recommends in a comment below:

_.isEqual(_.sortBy(array1), _.sortBy(array2))

Lodash's sortBy() will not mutate the array.


You can use lodashs xor for this

doArraysContainSameElements = _.xor(arr1, arr2).length === 0

If you consider array [1, 1] to be different than array [1] then you may improve performance a bit like so:

doArraysContainSameElements = arr1.length === arr2.length === 0 && _.xor(arr1, arr2).length === 0

By 'the same' I mean that there are is no item in array1 that is not contained in array2.

You could use flatten() and difference() for this, which works well if you don't care if there are items in array2 that aren't in array1. It sounds like you're asking is array1 a subset of array2?

var array1 = [['a', 'b'], ['b', 'c']];
var array2 = [['b', 'c'], ['a', 'b']];

function isSubset(source, target) {
    return !_.difference(_.flatten(source), _.flatten(target)).length;

isSubset(array1, array2); // ? true
isSubset(array1, array2); // ? false
isSubset(array2, array1); // ? true

There are already answers here, but here's my pure JS implementation. I'm not sure if it's optimal, but it sure is transparent, readable, and simple.

// Does array a contain elements of array b?
const contains = (a, b) => new Set([...a, ...b]).size === a.length
const isEqualSet = (a, b) => contains(a, b) && contains(b, a)

The rationale in contains() is that if a does contain all the elements of b, then putting them into the same set would not change the size.

For example, if const a = [1,2,3,4] and const b = [1,2], then new Set([...a, ...b]) === {1,2,3,4}. As you can see, the resulting set has the same elements as a.

From there, to make it more concise, we can boil it down to the following:

const isEqualSet = (a, b) => {
  const unionSize = new Set([...a, ...b])
  return unionSize === a.length && unionSize === b.length

PURE JS (works also when arrays and subarrays has more than 2 elements with arbitrary order). If strings contains , use as join('-') parametr character (can be utf) which is not used in strings>x.sort()).sort().join() ===>x.sort()).sort().join()

var array1 = [['a', 'b'], ['b', 'c']];
var array2 = [['b', 'c'], ['b', 'a']];

var r =>x.sort()).sort().join() ===>x.sort()).sort().join();



We can use _.difference function to see if there is any difference or not.

function isSame(arrayOne, arrayTwo) {
   var a = _.unique(arrayOne),
       b = _.unique(arrayTwo);

   if (a.length <= b.length) {
      a = arrayTwo;
      b = arrayOne;
      return _.isEmpty(_.difference(a.sort(), b.sort()));
   } else {
      return false;


// examples
console.log(isSame([1, 2, 3], [1, 2, 3])); // true
console.log(isSame([1, 2, 4], [1, 2, 3])); // false
console.log(isSame([1, 2], [2, 3, 1])); // false
console.log(isSame([2, 3, 1], [1, 2])); // false

// Test cases pointed by Mariano Desanze, Thanks.
console.log(isSame([1, 2, 3], [1, 2, 2])); // false
console.log(isSame([1, 2, 2], [1, 2, 2])); // true
console.log(isSame([1, 2, 2], [1, 2, 3])); // false

I hope this will help you.


Edit: I missed the multi-dimensional aspect of this question, so I'm leaving this here in case it helps people compare one-dimensional arrays

It's an old question, but I was having issues with the speed of using .sort() or sortBy(), so I used this instead:

function arraysContainSameStrings(array1: string[], array2: string[]): boolean {
  return (
    array1.length === array2.length &&
    array1.every((str) => array2.includes(str)) &&
    array2.every((str) => array1.includes(str))

It was intended to fail fast, and for my purposes works fine.


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