How to check if two arrays are equal with JavaScript? [duplicate]

How to check if two arrays are equal with JavaScript? [duplicate]

This question already has an answer here:

How to compare arrays in JavaScript?

52 answers

var a = [1, 2, 3];
var b = [3, 2, 1];
var c = new Array(1, 2, 3);

alert(a == b + “|” + b == c);

How can I check these array for equality and get a method which returns true if they are equal?
Does jQuery offer any method for this?


Solution 1:

This is what you should do. Please don’t use stringify nor < >.

function arraysEqual(a, b) {
  if (a === b) return true;
  if (a == null || b == null) return false;
  if (a.length != b.length) return false;

  // If you don't care about the order of the elements inside
  // the array, you should sort both arrays here.
  // Please note that calling sort on an array will modify that array.
  // you might want to clone your array first.

  for (var i = 0; i < a.length; ++i) {
    if (a[i] !== b[i]) return false;
  return true;

Solution 2:

Option 1

Easiest option, works in almost all cases, except that null!==undefined but they both are converted to JSON representation null and considered equal:

function arraysEqual(a1,a2) {
    /* WARNING: arrays must not contain {objects} or behavior may be undefined */
    return JSON.stringify(a1)==JSON.stringify(a2);

(This might not work if your array contains objects. Whether this still works with objects depends on whether the JSON implementation sorts keys. For example, the JSON of {1:2,3:4} may or may not be equal to {3:4,1:2}; this depends on the implementation, and the spec makes no guarantee whatsoever. [2017 update: Actually the ES6 specification now guarantees object keys will be iterated in order of 1) integer properties, 2) properties in the order they were defined, then 3) symbol properties in the order they were defined. Thus IF the JSON.stringify implementation follows this, equal objects (in the === sense but NOT NECESSARILY in the == sense) will stringify to equal values. More research needed. So I guess you could make an evil clone of an object with properties in the reverse order, but I cannot imagine it ever happening by accident…] At least on Chrome, the JSON.stringify function tends to return keys in the order they were defined (at least that I’ve noticed), but this behavior is very much subject to change at any point and should not be relied upon. If you choose not to use objects in your lists, this should work fine. If you do have objects in your list that all have a unique id, you can do}{return {id:x.uniqueId}}). If you have arbitrary objects in your list, you can read on for option #2.)

This works for nested arrays as well.

It is, however, slightly inefficient because of the overhead of creating these strings and garbage-collecting them.

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Option 2

More “proper” option, which you can override to deal with special cases (like regular objects and null/undefined and custom objects, if you so desire):

// generally useful functions
function type(x) { // does not work in general, but works on JSONable objects we care about... modify as you see fit
    // e.g.  type(/asdf/g) --> "[object RegExp]"
function zip(arrays) {
    // e.g. zip([[1,2,3],[4,5,6]]) --> [[1,4],[2,5],[3,6]]
    return arrays[0].map(function(_,i){
        return{return array[i]})


// helper functions
function allCompareEqual(array) {
    // e.g.  allCompareEqual([2,2,2,2]) --> true
    // does not work with nested arrays or objects
    return array.every(function(x){return x==array[0]});

function isArray(x){ return type(x)==type([]) }
function getLength(x){ return x.length }
function allTrue(array){ return array.reduce(function(a,b){return a&&b},true) }
    // e.g. allTrue([true,true,true,true]) --> true
    // or just array.every(function(x){return x});


function allDeepEqual(things) {
    // works with nested arrays
    if( things.every(isArray) )
        return allCompareEqual(     // all arrays of same length
               && allTrue(zip(things).map(allDeepEqual)); // elements recursively equal

    //else if( this.every(isObject) )
    //  return {all have exactly same keys, and for 
    //          each key k, allDeepEqual([o1[k],o2[k],...])}
    //  e.g. ... && allTrue(objectZip(objects).map(allDeepEqual)) 

    //else if( ... )
    //  extend some more

        return allCompareEqual(things);


allDeepEqual([ [], [], [] ])
allDeepEqual([ [1], [1], [1] ])
allDeepEqual([ [1,2], [1,2] ])
allDeepEqual([ [[1,2],[3]], [[1,2],[3]] ])

allDeepEqual([ [1,2,3], [1,2,3,4] ])
allDeepEqual([ [[1,2],[3]], [[1,2],[],3] ])
allDeepEqual([ [[1,2],[3]], [[1],[2,3]] ])
allDeepEqual([ [[1,2],3], [1,[2,3]] ])

To use this like a regular function, do:

function allDeepEqual2() {
    return allDeepEqual([];


allDeepEqual2([[1,2],3], [[1,2],3])

Options 3

edit: It’s 2016 and my previous overcomplicated answer was bugging me. This recursive, imperative “recursive programming 101” implementation keeps the code really simple, and furthermore fails at the earliest possible point (giving us efficiency). It also doesn’t generate superfluous ephemeral datastructures (not that there’s anything wrong with functional programming in general, but just keeping it clean here).

If we wanted to apply this to a non-empty arrays of arrays, we could do seriesOfArrays.reduce(arraysEqual).

This is its own function, as opposed to using Object.defineProperties to attach to Array.prototype, since that would fail with a key error if we passed in an undefined value (that is however a fine design decision if you want to do so).

This only answers OPs original question.

function arraysEqual(a,b) {
        Array-aware equality checker:
        Returns whether arguments a and b are == to each other;
        however if they are equal-lengthed arrays, returns whether their 
        elements are pairwise == to each other recursively under this
    if (a instanceof Array && b instanceof Array) {
        if (a.length!=b.length)  // assert same length
            return false;
        for(var i=0; i<a.length; i++)  // assert each element equal
            if (!arraysEqual(a[i],b[i]))
                return false;
        return true;
    } else {
        return a==b;  // if not both arrays, should be the same


arraysEqual([[1,2],3], [[1,2],3])
arraysEqual([1,2,3], [1,2,3,4])
arraysEqual([[1,2],[3]], [[1,2],[],3])
arraysEqual([[1,2],[3]], [[1],[2,3]])
arraysEqual([[1,2],3], undefined)
arraysEqual(undefined, undefined)
arraysEqual(1, 2)
arraysEqual(null, null)
arraysEqual(1, 1)
arraysEqual([], 1)
arraysEqual([], undefined)
arraysEqual([], [])

If you wanted to apply this to JSON-like data structures with js Objects, you could do so. Fortunately we’re guaranteed that all objects keys are unique, so iterate over the objects OwnProperties and sort them by key, then assert that both the sorted key-array is equal and the value-array are equal, and just recurse. We can extend this to include Maps as well (where the keys are also unique). (However if we extend this to Sets, we run into the tree isomorphism problem – fortunately it’s not as hard as general graph isomorphism; there is in fact an O(#vertices) algorithm to solve it, but it can get very complicated to do it efficiently. The pathological case is if you have a set made up of lots of seemingly-indistinguishable objects, but upon further inspection some of those objects may differ as you delve deeper into them. You can also work around this by using hashing to reject almost all cases.)

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Option 4:
(continuation of 2016 edit)

This should work with most objects:

function deepEquals(a,b) {
    if (a instanceof Array && b instanceof Array)
        return arraysEqual(a,b);
    if (Object.getPrototypeOf(a)===Object.prototype && Object.getPrototypeOf(b)===Object.prototype)
        return objectsEqual(a,b);
    if (a instanceof Map && b instanceof Map)
        return mapsEqual(a,b);        
    if (a instanceof Set && b instanceof Set)
        throw "Error: set equality by hashing not implemented."
    if ((a instanceof ArrayBuffer || ArrayBuffer.isView(a)) && (b instanceof ArrayBuffer || ArrayBuffer.isView(b)))
        return typedArraysEqual(a,b);
    return a==b;  // see note[1] -- IMPORTANT

function arraysEqual(a,b) {
    if (a.length!=b.length)
        return false;
    for(var i=0; i<a.length; i++)
        if (!deepEquals(a[i],b[i]))
            return false;
    return true;
function objectsEqual(a,b) {
    var aKeys = Object.getOwnPropertyNames(a);
    var bKeys = Object.getOwnPropertyNames(b);
    if (aKeys.length!=bKeys.length)
        return false;
    for(var i=0; i<aKeys.length; i++)
        if (aKeys[i]!=bKeys[i]) // keys must be strings
            return false;
    return deepEquals(>a[k]),>b[k]));
function mapsEqual(a,b) {
    if (a.size!=b.size)
        return false;
    var aPairs = Array.from(a);
    var bPairs = Array.from(b);
    aPairs.sort((x,y) => x[0]<y[0]);
    bPairs.sort((x,y) => x[0]<y[0]);
    for(var i=0; i<a.length; i++)
        if (!deepEquals(aPairs[i][0],bPairs[i][0]) || !deepEquals(aPairs[i][1],bPairs[i][1]))
            return false;
    return true;
function typedArraysEqual(a,b) {
    a = new Uint8Array(a);
    b = new Uint8Array(b);
    if (a.length != b.length)
        return false;
    for(var i=0; i<a.length; i++)
        if (a[i]!=b[i])
            return false;
    return true;

Demo (not extensively tested):

var nineTen = new Float32Array(2);
nineTen[0]=9; nineTen[1]=10;
    [[1,[2,3]], 4, {a:5,b:6}, new Map([['c',7],['d',8]]), nineTen],
    [[1,[2,3]], 4, {b:6,a:5}, new Map([['d',8],['c',7]]), nineTen]

(sidenote: Maps are es6 dictionaries. I can’t tell if they have O(1) or O(log(N)) lookup performance, but in any case they are ‘ordered’ in the sense that they keep track of the order in which key-value pairs were inserted into them. However, the semantic of whether two Maps should be equal if elements were inserted in a different order into them is ambiguous. I give a sample implementation below of a deepEquals that considers two maps equal even if elements were inserted into them in a different order.)

(note [1]: IMPORTANT: NOTION OF EQUALITY: You may want to override the noted line with a custom notion of equality, which you’ll also have to change in the other functions anywhere it appears. For example, do you or don’t you want NaN==NaN? By default this is not the case. There are even more weird things like 0==’0′. Do you consider two objects to be the same if and only if they are the same object in memory? See . You should document the notion of equality you use. )

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You should be able to extend the above to WeakMaps, WeakSets. Not sure if it makes sense to extend to DataViews. Should also be able to extend to RegExps probably, etc.

As you extend it, you realize you do lots of unnecessary comparisons. This is where the type function that I defined way earlier (solution #2) can come in handy; then you can dispatch instantly. Whether that is worth the overhead of (possibly? not sure how it works under the hood) string representing the type is up to you. You can just then rewrite the dispatcher, i.e. the function deepEquals, to be something like:

var dispatchTypeEquals = {
    number: function(a,b) {...a==b...},
    array: function(a,b) {...deepEquals(x,y)...},
function deepEquals(a,b) {
    var typeA = extractType(a);
    var typeB = extractType(a);
    return typeA==typeB && dispatchTypeEquals[typeA](a,b);

Solution 3:

jQuery does not have a method for comparing arrays. However the Underscore library (or the comparable Lodash library) does have such a method: isEqual, and it can handle a variety of other cases (like object literals) as well. To stick to the provided example:

var a=[1,2,3];
var b=[3,2,1];
var c=new Array(1,2,3);

alert(_.isEqual(a, b) + "|" + _.isEqual(b, c));

By the way: Underscore has lots of other methods that jQuery is missing as well, so it’s a great complement to jQuery.

EDIT: As has been pointed out in the comments, the above now only works if both arrays have their elements in the same order, ie.:

_.isEqual([1,2,3], [1,2,3]); // true
_.isEqual([1,2,3], [3,2,1]); // false

Fortunately Javascript has a built in method for for solving this exact problem, sort:

_.isEqual([1,2,3].sort(), [3,2,1].sort()); // true

Solution 4:

For primitive values like numbers and strings this is an easy solution:

a = [1,2,3]

b = [3,2,1]

a.sort().toString() == b.sort().toString() 

The call to sort() will ensure that the order of the elements does not matter. The toString() call will create a string with the values comma separated so both strings can be tested for equality.

Solution 5:

With JavaScript version 1.6 it’s as easy as this:

Array.prototype.equals = function( array ) {
  return this.length == array.length && 
         this.every( function(this_i,i) { return this_i == array[i] } )  

For example, [].equals([]) gives true, while [1,2,3].equals( [1,3,2] ) yields false.

Solution 6:

Even if this would seem super simple, sometimes it’s really useful. If all you need is to see if two arrays have the same items and they are in the same order, try this:

[1, 2, 3].toString() == [1, 2, 3].toString()
[1, 2, 3,].toString() == [1, 2, 3].toString()
[1,2,3].toString() == [1, 2, 3].toString()

However, this doesn’t work for mode advanced cases such as:

[[1,2],[3]].toString() == [[1],[2,3]].toString()

It depends what you need.