javascript - two - lodash compare arrays

How to compare arrays in JavaScript? (20)

The Practical Way

I think it's wrong to say a particular implementation is "The Right Way™" if it's only "right" ("correct") in contrast to a "wrong" solution. Tomáš's solution is a clear improvement over string-based array comparison, but that doesn't mean it's objectively "right". What is right anyway? Is it the fastest? Is it the most flexible? Is it the easiest to comprehend? Is it the quickest to debug? Does it use the least operations? Does it have any side effects? No one solution can have the best of all the things.

Tomáš's could say his solution is fast but I would also say it is needlessly complicated. It tries to be an all-in-one solution that works for all arrays, nested or not. In fact, it even accepts more than just arrays as an input and still attempts to give a "valid" answer.

Generics offer reusability

My answer will approach the problem differently. I'll start with a generic arrayCompare procedure that is only concerned with stepping through the arrays. From there, we'll build our other basic comparison functions like arrayEqual and arrayDeepEqual, etc

// arrayCompare :: (a -> a -> Bool) -> [a] -> [a] -> Bool
const arrayCompare = f => ([x,...xs]) => ([y,...ys]) =>
  x === undefined && y === undefined
    ? true
    : Boolean (f (x) (y)) && arrayCompare (f) (xs) (ys)

In my opinion, the best kind of code doesn't even need comments, and this is no exception. There's so little happening here that you can understand the behaviour of this procedure with almost no effort at all. Sure, some of the ES6 syntax might seem foreign to you now, but that's only because ES6 is relatively new.

As the type suggests, arrayCompare takes comparison function, f, and two input arrays, xs and ys. For the most part, all we do is call f (x) (y) for each element in the input arrays. We return an early false if the user-defined f returns false – thanks to &&'s short-circuit evaluation. So yes, this means the comparator can stop iteration early and prevent looping through the rest of the input array when unnecessary.

Strict comparison

Next, using our arrayCompare function, we can easily create other functions we might need. We'll start with the elementary arrayEqual

// equal :: a -> a -> Bool
const equal = x => y =>
  x === y // notice: triple equal

// arrayEqual :: [a] -> [a] -> Bool
const arrayEqual =
  arrayCompare (equal)

const xs = [1,2,3]
const ys = [1,2,3]
console.log (arrayEqual (xs) (ys))      //=> true
// (1 === 1) && (2 === 2) && (3 === 3)  //=> true

const zs = ['1','2','3']
console.log (arrayEqual (xs) (zs))      //=> false
// (1 === '1')                          //=> false

Simple as that. arrayEqual can be defined with arrayCompare and a comparator function that compares a to b using === (for strict equality).

Notice that we also define equal as it's own function. This highlights the role of arrayCompare as a higher-order function to utilize our first order comparator in the context of another data type (Array).

Loose comparison

We could just as easily defined arrayLooseEqual using a == instead. Now when comparing 1 (Number) to '1' (String), the result will be true

// looseEqual :: a -> a -> Bool
const looseEqual = x => y =>
  x == y // notice: double equal

// arrayLooseEqual :: [a] -> [a] -> Bool
const arrayLooseEqual =
  arrayCompare (looseEqual)

const xs = [1,2,3]
const ys = ['1','2','3']
console.log (arrayLooseEqual (xs) (ys))    //=> true
// (1 == '1') && (2 == '2') && (3 == '3')  //=> true

Deep comparison (recursive)

You've probably noticed that this is only shallow comparison tho. Surely Tomáš's solution is "The Right Way™" because it does implicit deep comparison, right ?

Well our arrayCompare procedure is versatile enough to use in a way that makes a deep equality test a breeze …

// isArray :: a -> Bool
const isArray =

// arrayDeepCompare :: (a -> a -> Bool) -> [a] -> [a] -> Bool
const arrayDeepCompare = f =>
  arrayCompare (a => b =>
    isArray (a) && isArray (b)
      ? arrayDeepCompare (f) (a) (b)
      : f (a) (b))

const xs = [1,[2,[3]]]
const ys = [1,[2,['3']]]
console.log (arrayDeepCompare (equal) (xs) (ys)) //=> false
// (1 === 1) && (2 === 2) && (3 === '3')         //=> false

console.log (arrayDeepCompare (looseEqual) (xs) (ys)) //=> true
// (1 == 1) && (2 == 2) && (3 == '3')                 //=> true

Simple as that. We build a deep comparator using another higher-order function. This time we're wrapping arrayCompare using a custom comparator that will check if a and b are arrays. If so, reapply arrayDeepCompare otherwise compare a and b to the user-specified comparator (f). This allows us to keep the deep comparison behavior separate from how we actually compare the individual elements. Ie, like the example above shows, we can deep compare using equal, looseEqual, or any other comparator we make.

Because arrayDeepCompare is curried, we can partially apply it like we did in the previous examples too

// arrayDeepEqual :: [a] -> [a] -> Bool
const arrayDeepEqual =
  arrayDeepCompare (equal)

// arrayDeepLooseEqual :: [a] -> [a] -> Bool
const arrayDeepLooseEqual =
  arrayDeepCompare (looseEqual)

To me, this already a clear improvement over Tomáš's solution because I can explicitly choose a shallow or deep comparison for my arrays, as needed.

Object comparison (example)

Now what if you have an array of objects or something ? Maybe you want to consider those arrays as "equal" if each object has the same id value …

// idEqual :: {id: Number} -> {id: Number} -> Bool
const idEqual = x => y => !== undefined && ===

// arrayIdEqual :: [a] -> [a] -> Bool
const arrayIdEqual =
  arrayCompare (idEqual)

const xs = [{id:1}, {id:2}]
const ys = [{id:1}, {id:2}]
console.log (arrayIdEqual (xs) (ys)) //=> true
// (1 === 1) && (2 === 2)            //=> true

const zs = [{id:1}, {id:6}]
console.log (arrayIdEqual (xs) (zs)) //=> false
// (1 === 1) && (2 === 6)            //=> false

Simple as that. Here I've used vanilla JS objects, but this type of comparator could work for any object type; even your custom objects. Tomáš's solution would need to be completely reworked to support this kind of equality test

Deep array with objects? Not a problem. We built highly versatile, generic functions, so they'll work in a wide variety of use cases.

const xs = [{id:1}, [{id:2}]]
const ys = [{id:1}, [{id:2}]]
console.log (arrayCompare (idEqual) (xs) (ys))     //=> false
console.log (arrayDeepCompare (idEqual) (xs) (ys)) //=> true

Arbitrary comparison (example)

Or what if you wanted to do some other kind of kind of completely arbitrary comparison ? Maybe I want to know if each x is greater than each y

// gt :: Number -> Number -> Bool
const gt = x => y =>
  x > y

// arrayGt :: [a] -> [a] -> Bool
const arrayGt = arrayCompare (gt)

const xs = [5,10,20]
const ys = [2,4,8]
console.log (arrayGt (xs) (ys))     //=> true
// (5 > 2) && (10 > 4) && (20 > 8)  //=> true

const zs = [6,12,24]
console.log (arrayGt (xs) (zs))     //=> false
// (5 > 6)                          //=> false

Less is More

You can see we're actually doing more with less code. There's nothing complicated about arrayCompare itself and each of the custom comparators we've made have a very simple implementation.

With ease, we can define exactly how we wish for two arrays to be compared — shallow, deep, strict, loose, some object property, or some arbitrary computation, or any combination of these — all using one procedure, arrayCompare. Maybe even dream up a RegExp comparator ! I know how kids love those regexps …

Is it the fastest? Nope. But it probably doesn't need to be either. If speed is the only metric used to measure the quality of our code, a lot of really great code would get thrown away — That's why I'm calling this approach The Practical Way. Or maybe to be more fair, A Practical Way. This description is suitable for this answer because I'm not saying this answer is only practical in comparison to some other answer; it is objectively true. We've attained a high degree of practicality with very little code that's very easy to reason about. No other code can say we haven't earned this description.

Does that make it the "right" solution for you ? That's up for you to decide. And no one else can do that for you; only you know what your needs are. In almost all cases, I value straightforward, practical, and versatile code over clever and fast kind. What you value might differ, so pick what works for you.


My old answer was more focused on decomposing arrayEqual into tiny procedures. It's an interesting exercise, but not really the best (most practical) way to approach this problem. If you're interested, you can see this revision history.

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?

Building off Tomáš Zato's answer, I agree that just iterating through the arrays is the fastest. Additionally (like others have already stated), the function should be called equals/equal, not compare. In light of this, I modified the function to handle comparing arrays for similarity - i.e. they have the same elements, but out of order - for personal use, and thought I'd throw it on here for everyone to see.

Array.prototype.equals = function (array, strict) {
    if (!array)
        return false;

    if (arguments.length == 1)
        strict = true;

    if (this.length != array.length)
        return false;

    for (var i = 0; i < this.length; i++) {
        if (this[i] instanceof Array && array[i] instanceof Array) {
            if (!this[i].equals(array[i], strict))
                return false;
        else if (strict && this[i] != array[i]) {
            return false;
        else if (!strict) {
            return this.sort().equals(array.sort(), true);
    return true;

This function takes an additional parameter of strict that defaults to true. This strict parameter defines if the arrays need to be wholly equal in both contents and the order of those contents, or simply just contain the same contents.


var arr1 = [1, 2, 3, 4];
var arr2 = [2, 1, 4, 3];  // Loosely equal to 1
var arr3 = [2, 2, 3, 4];  // Not equal to 1
var arr4 = [1, 2, 3, 4];  // Strictly equal to 1

arr1.equals(arr2);         // false
arr1.equals(arr2, false);  // true
arr1.equals(arr3);         // false
arr1.equals(arr3, false);  // false
arr1.equals(arr4);         // true
arr1.equals(arr4, false);  // true

I've also written up a quick jsfiddle with the function and this example:

Comparing 2 arrays:

var arr1 = [1,2,3];
var arr2 = [1,2,3];

function compare(arr1,arr2)
  if((arr1 == arr2) && (arr1.length == arr2.length))
    return true;
    return false;

calling function

var isBool = compare(arr1.sort().join(),arr2.sort().join());

Even though this has a lot of answers, one that I believe to be of help:

const newArray = [ Set( [...arr1, ...arr2] ) ]

It is not stated in the question how the structure of the array is going to look like, so If you know for sure that you won't have nested arrays nor objects in you array (it happened to me, that's why I came to this answer) the above code will work.

What happens is that we use spread operator ( ... ) to concat both arrays, then we use Set to eliminate any duplicates. Once you have that you can compare their sizes, if all three arrays have the same size you are good to go.

This answer also ignores the order of elements, as I said, the exact situation happened to me, so maybe someone in the same situation might end up here (as I did).


Answering Dmitry Grinko's question: "Why did you use spread operator ( ... ) here - Set ? It doesn't work"

Consider this code:

const arr1 = [ 'a', 'b' ]
const arr2 = [ 'a', 'b', 'c' ]
const newArray = [ new Set( [...arr1, ...arr2] ) ]

You'll get

[ Set { 'a', 'b', 'c' } ]

In order to work with that value you'd need to use some Set properties (see On the other hand, when you use this code:

const arr1 = [ 'a', 'b' ]
const arr2 = [ 'a', 'b', 'c' ]
const newArray = [ Set( [...arr1, ...arr2] ) ]

You'll get

[ 'a', 'b', 'c' ]

That's the difference, the former would give me a Set, it would work too as I could get the size of that Set, but the latter gives me the array I need, what's more direct to the resolution.

Here is a Typescript version:

export function arraysEqual<T>(a: Array<T>, b: Array<T>): boolean {
    if (a === b) return true
    if (a == null || b == null) return false
    if (a.length != b.length) return false

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

export function arraysDeepEqual<T>(a: Array<T>, b: Array<T>): boolean {
    return JSON.stringify(a) === JSON.stringify(b)

Some test cases for mocha:

it('arraysEqual', function () {
    let a = [1,2]
    let b = [1,2]
    let c = [2,3]
    let d = [2, 3]
    let e = ['car','apple','banana']
    let f = ['car','apple','banana']
    let g = ['car','apple','banan8']

    expect(arraysEqual(a, b)).to.equal(true)
    expect(arraysEqual(c, d)).to.equal(true)
    expect(arraysEqual(a, d)).to.equal(false)
    expect(arraysEqual(e, f)).to.equal(true)
    expect(arraysEqual(f, g)).to.equal(false)

it('arraysDeepEqual', function () {
    let a = [1,2]
    let b = [1,2]
    let c = [2,3]
    let d = [2, 3]
    let e = ['car','apple','banana']
    let f = ['car','apple','banana']
    let g = ['car','apple','banan8']
    let h = [[1,2],'apple','banan8']
    let i = [[1,2],'apple','banan8']
    let j = [[1,3],'apple','banan8']

    expect(arraysDeepEqual(a, b)).to.equal(true)
    expect(arraysDeepEqual(c, d)).to.equal(true)
    expect(arraysDeepEqual(a, d)).to.equal(false)
    expect(arraysDeepEqual(e, f)).to.equal(true)
    expect(arraysDeepEqual(f, g)).to.equal(false)
    expect(arraysDeepEqual(h, i)).to.equal(true)
    expect(arraysDeepEqual(h, j)).to.equal(false)

Here's a CoffeeScript version, for those who prefer that:

Array.prototype.equals = (array) ->
  return false if not array # if the other array is a falsy value, return
  return false if @length isnt array.length # compare lengths - can save a lot of time

  for item, index in @
    if item instanceof Array and array[index] instanceof Array # Check if we have nested arrays
      if not item.equals(array[index]) # recurse into the nested arrays
        return false
    else if this[index] != array[index]
      return false # Warning - two different object instances will never be equal: {x:20} != {x:20}

All credits goes to @tomas-zato.

I like to use the Underscore library for array/object heavy coding projects ... in Underscore and Lodash whether you're comparing arrays or objects it just looks like this:

_.isEqual(array1, array2)   // returns a boolean
_.isEqual(object1, object2) // returns a boolean

If the array is plain and the order is matter so this two lines may help

var a = ['a','b', 'c']; var b = ['a','e', 'c'];  

if(a.length !== b.length) return false;
return !a.reduce(
  function(prev,next,idx, arr){ return prev || next != b[idx] },false

Reduce walks through one of array and returns 'false' if at least one element of 'a' is nor equial to element of 'b' Just wrap this into function

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


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

In my case compared arrays contain only numbers and strings. This function will show you if arrays contain same elements.

function are_arrs_match(arr1, arr2){
    return arr1.sort().toString() === arr2.sort().toString()

Let's test it!

arr1 = [1, 2, 3, 'nik']
arr2 = ['nik', 3, 1, 2]
arr3 = [1, 2, 5]

console.log (are_arrs_match(arr1, arr2)) //true
console.log (are_arrs_match(arr1, arr3)) //false

It's unclear what you mean by "identical". For example, are the arrays a and b below identical (note the nested arrays)?

var a = ["foo", ["bar"]], b = ["foo", ["bar"]];

Here's an optimized array comparison function that compares corresponding elements of each array in turn using strict equality and does not do recursive comparison of array elements that are themselves arrays, meaning that for the above example, arraysIdentical(a, b) would return false. It works in the general case, which JSON- and join()-based solutions will not:

function arraysIdentical(a, b) {
    var i = a.length;
    if (i != b.length) return false;
    while (i--) {
        if (a[i] !== b[i]) return false;
    return true;

My solution compares Objects, not Arrays. This would work in the same way as Tomáš's as Arrays are Objects, but without the Warning:

Object.prototype.compare_to = function(comparable){

    // Is the value being compared an object
    if(comparable instanceof Object){

        // Count the amount of properties in @comparable
        var count_of_comparable = 0;
        for(p in comparable) count_of_comparable++;

        // Loop through all the properties in @this
        for(property in this){

            // Decrements once for every property in @this

            // Prevents an infinite loop
            if(property != "compare_to"){

                // Is the property in @comparable
                if(property in comparable){

                    // Is the property also an Object
                    if(this[property] instanceof Object){

                        // Compare the properties if yes

                            // Return false if the Object properties don't match
                            return false;
                    // Are the values unequal
                    } else if(this[property] !== comparable[property]){

                        // Return false if they are unequal
                        return false;
                } else {

                    // Return false if the property is not in the object being compared
                    return false;
    } else {

        // Return false if the value is anything other than an object
        return false;

    // Return true if their are as many properties in the comparable object as @this
    return count_of_comparable == 0;

Hope this helps you or anyone else searching for an answer.

The reason is that identity or strict operator (===), it compares with no type conversion, that means if both values doesn’t have the same value and the same type, they won’t be considered equal.

take a look this link, it takes you out of doubt easy way to understand how identity operator works

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.

This function compares two arrays of arbitrary shape and dimesionality:

function equals(a1, a2) {

    if (!Array.isArray(a1) || !Array.isArray(a2)) {
        throw new Error("Arguments to function equals(a1, a2) must be arrays.");

    if (a1.length !== a2.length) {
        return false;

    for (var i=0; i<a1.length; i++) {
        if (Array.isArray(a1[i]) && Array.isArray(a2[i])) {
            if (equals(a1[i], a2[i])) {
            } else {
                return false;
        } else {
            if (a1[i] !== a2[i]) {
                return false;

    return true;

To compare arrays, loop through them and compare every value:

Comparing arrays:

// Warn if overriding existing method
    console.warn("Overriding existing Array.prototype.equals. Possible causes: New API defines the method, there's a framework conflict or you've got double inclusions in your code.");
// attach the .equals method to Array's prototype to call it on any array
Array.prototype.equals = function (array) {
    // if the other array is a falsy value, return
    if (!array)
        return false;

    // compare lengths - can save a lot of time 
    if (this.length != array.length)
        return false;

    for (var i = 0, l=this.length; i < l; i++) {
        // Check if we have nested arrays
        if (this[i] instanceof Array && array[i] instanceof Array) {
            // recurse into the nested arrays
            if (!this[i].equals(array[i]))
                return false;       
        else if (this[i] != array[i]) { 
            // Warning - two different object instances will never be equal: {x:20} != {x:20}
            return false;   
    return true;
// Hide method from for-in loops
Object.defineProperty(Array.prototype, "equals", {enumerable: false});


[1, 2, [3, 4]].equals([1, 2, [3, 2]]) === false;
[1, "2,3"].equals([1, 2, 3]) === false;
[1, 2, [3, 4]].equals([1, 2, [3, 4]]) === true;
[1, 2, 1, 2].equals([1, 2, 1, 2]) === true;

You may say "But it is much faster to compare strings - no loops..." well, then you should note there ARE loops. First recursive loop that converts Array to string and second, that compares two strings. So this method is faster than use of string.

I believe that larger amounts of data should be always stored in arrays, not in objects. However if you use objects, they can be partially compared too.
Here's how:

Comparing objects:

I've stated above, that two object instances will never be equal, even if they contain same data at the moment:

({a:1, foo:"bar", numberOfTheBeast: 666}) == ({a:1, foo:"bar", numberOfTheBeast: 666})  //false

This has a reason, since there may be, for example private variables within objects.

However, if you just use object structure to contain data, comparing is still possible:

Object.prototype.equals = function(object2) {
    //For the first loop, we only check for types
    for (propName in this) {
        //Check for inherited methods and properties - like .equals itself
        //Return false if the return value is different
        if (this.hasOwnProperty(propName) != object2.hasOwnProperty(propName)) {
            return false;
        //Check instance type
        else if (typeof this[propName] != typeof object2[propName]) {
            //Different types => not equal
            return false;
    //Now a deeper check using other objects property names
    for(propName in object2) {
        //We must check instances anyway, there may be a property that only exists in object2
            //I wonder, if remembering the checked values from the first loop would be faster or not 
        if (this.hasOwnProperty(propName) != object2.hasOwnProperty(propName)) {
            return false;
        else if (typeof this[propName] != typeof object2[propName]) {
            return false;
        //If the property is inherited, do not check any more (it must be equa if both objects inherit it)

        //Now the detail check and recursion

        //This returns the script back to the array comparing
        /**REQUIRES Array.equals**/
        if (this[propName] instanceof Array && object2[propName] instanceof Array) {
                   // recurse into the nested arrays
           if (!this[propName].equals(object2[propName]))
                        return false;
        else if (this[propName] instanceof Object && object2[propName] instanceof Object) {
                   // recurse into another objects
                   //console.log("Recursing to compare ", this[propName],"with",object2[propName], " both named \""+propName+"\"");
           if (!this[propName].equals(object2[propName]))
                        return false;
        //Normal value comparison for strings and numbers
        else if(this[propName] != object2[propName]) {
           return false;
    //If everything passed, let's say YES
    return true;

However, remember that this one is to serve in comparing JSON like data, not class instances and other stuff. If you want to compare mor complicated objects, look at this answer and it's superlong function.
To make this work with Array.equals you must edit the original function a little bit:

    // Check if we have nested arrays
    if (this[i] instanceof Array && array[i] instanceof Array) {
        // recurse into the nested arrays
        if (!this[i].equals(array[i]))
            return false;
    else if (this[i] instanceof Object && array[i] instanceof Object) {
        // recurse into another objects
        //console.log("Recursing to compare ", this[propName],"with",object2[propName], " both named \""+propName+"\"");
        if (!this[i].equals(array[i]))
            return false;
    else if (this[i] != array[i]) {

I made a little test tool for both of the functions.

Bonus: Nested arrays with indexOf and contains

Samy Bencherif has prepared useful functions for the case you're searching for a specific object in nested arrays, which are available here:

While this only works for scalar arrays (see note below), it is short:

array1.length === array2.length && array1.every(function(value, index) { return value === array2[index]})

Rr, in ECMAScript 6 / CoffeeScript / TypeScript with Arrow Functions:

array1.length === array2.length && array1.every((value, index) => value === array2[index])

(Note: 'scalar' here means values that can be compared directly using === . So: numbers, strings, objects by reference, functions by reference. See the MDN reference for more info about the comparison operators).


From what I read from the comments, sorting the array and comparing may give accurate result:

array1.length === array2.length && array1.sort().every(function(value, index) { return value === array2.sort()[index]});


array1 = [2,3,1,4];
array2 = [1,2,3,4];

Then the above code would give true

for single dimension array you can simply use:

arr1.sort().toString() == arr2.sort().toString()

this will also take care of array with mismatched index.

JSON.stringify(collectionNames).includes(JSON.stringify(sourceNames)) ?  array.push(collection[i]) : null

This is how i did it.

function compareArrays(arrayA, arrayB) {
    if (arrayA.length != arrayB.length) return true;
    for (i = 0; i < arrayA.length; i++)
        if (arrayB.indexOf(arrayA[i]) == -1) {
            return true;
    for (i = 0; i < arrayB.length; i++) {
        if (arrayA.indexOf(arrayB[i]) == -1) {
            return true;
    return false;