How to compare arrays in JavaScript?

Question

I'd like to compare two arrays... ideally, efficiently. Nothing fancy, just true if they are identical, and false if not. Not surprisingly, the comparison operator doesn't seem to work.

var a1 = [1,2,3];
var a2 = [1,2,3];
console.log(a1==a2);    // Returns false
console.log(JSON.stringify(a1)==JSON.stringify(a2));    // Returns true

JSON encoding each array does, but is there a faster or "better" way to simply compare arrays without having to iterate through each value?

Answer 1

If you are using a testing framework like Mocha with the Chai assertion library, you can use deep equality to compare arrays.

expect(a1).to.deep.equal(a2)

This should return true only if the arrays have equal elements at corresponding indices.

Answer 2

If they are two arrays of numbers or strings only, this is a quick one-line one

const array1 = [1, 2, 3];
const array2 = [1, 3, 4];
console.log(array1.join(',') === array2.join(',')) //false

const array3 = [1, 2, 3];
const array4 = [1, 2, 3];
console.log(array3.join(',') === array4.join(',')) //true

Answer 3

this script compares Object, Arrays and multidimensional array

function compare(a,b){
     var primitive=['string','number','boolean'];
     if(primitive.indexOf(typeof a)!==-1 && primitive.indexOf(typeof a)===primitive.indexOf(typeof b))return a===b;
     if(typeof a!==typeof b || a.length!==b.length)return false;
     for(i in a){
          if(!compare(a[i],b[i]))return false;
     }
     return true;
}

first line checks whether it's a primitive type. if so it compares the two parameters.

if they are Objects. it iterates over the Object and check every element recursivly.

Usage:

var a=[1,2,[1,2]];
var b=[1,2,[1,2]];
var isEqual=compare(a,b);  //true

Answer 4

In the spirit of the original question:

I'd like to compare two arrays... ideally, efficiently. Nothing fancy, just true if they are identical, and false if not.

I have been running performance tests on some of the more simple suggestions proposed here with the following results (fast to slow):

while (67%) by Tim Down

var i = a1.length;
while (i--) {
    if (a1[i] !== a2[i]) return false;
}
return true

every (69%) by user2782196

a1.every((v,i)=> v === a2[i]);

reduce (74%) by DEIs

a1.reduce((a, b) => a && a2.includes(b), true);

join & toString (78%) by Gaizka Allende & vivek

a1.join('') === a2.join('');

a1.toString() === a2.toString();

half toString (90%) by Victor Palomo

a1 == a2.toString();

stringify (100%) by radtek

JSON.stringify(a1) === JSON.stringify(a2);

Note the examples below assumes the arrays are sorted, single-dimensional arrays. .length comparison has been removed for a common benchmark (add a1.length === a2.length to any of the suggestions and you will get a ~10% performance boost). Choose whatever solutions that works best for you knowing the speed and limitation of each.

Unrelated note: it is interesting to see people getting all trigger-happy John Waynes on the down vote button on perfectly legitimate answers to this question.

Answer 5

Another approach with very few code (using Array reduce and Array includes):

arr1.length == arr2.length && arr1.reduce((a, b) => a && arr2.includes(b), true)

If you want to compare also the equality of order:

arr1.length == arr2.length && arr1.reduce((a, b, i) => a && arr2[i], true)

• The length check ensures that the set of elements in one array isn't just a subset of the other one.

• The reducer is used to walk through one array and search for each item in other array. If one item isn't found the reduce function returns false.

  1. In the first example it's being tested that an element is included
  2. The second example check for the order too

Answer 6

This I think is the simplest way to do it using JSON stringify, and it may be the best solution in some situations:

JSON.stringify(a1) === JSON.stringify(a2);

This converts the objects a1 and a2 into strings so they can be compared. The order is important in most cases, for that can sort the object using a sort algorithm shown in one of the above answers.

Please do note that you are no longer comparing the object but the string representation of the object. It may not be exactly what you want.