Using lodash to compare jagged arrays (items existence without order)

Question

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

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 union = new Set([...a, ...b]);
const contains = (a, b) => union.size === a.length && union.size === b.length;
// Since order is not important, just data validity.
const isEqualSet = (a, b) => union.contains(a, b) || union.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: string[], b: sting[]): boolean => {
  const union = new Set([...a, ...b])
  return union.size === a.length && union.size === b.length;
}

Edit: This will not work with obj[{a: true}, true, 3] but does compare array contents probably as long as they are primitive elements. Method fixed and tested against strings two arrays using the same values in different orders. Does not work with object types. I recommend making a universal helper which calls a helper function depending on the type which needs to be compared. Try _.isEqual(a. b); from the very fantastic lodash library.

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
array1.push('d');
isSubset(array1, array2); // → false
isSubset(array2, array1); // → true