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How to Compare the Same Contents but Different Ordered Arrays in JavaScript

by Thanga GanapathyApril 23rd, 2024

Too Long; Didn't Read

One common challenge developers face is comparing arrays—specifically, determining if two arrays contain the same contents, regardless of the order of those contents. This task might seem straightforward at first glance, but it quickly delves into complexities. In this blog post, we’ll discuss various methods, from the simple and direct to the more sophisticated and efficient, ensuring that by the end of this article.

featured image - How to Compare the Same Contents but Different Ordered Arrays in JavaScript

Welcome,

In this blog post, we’ll discuss various methods, from the simple and direct to the more sophisticated and efficient.

Basics

This is for Beginners, in JavaScript, comparing two arrays with equality operators simply does not work.

[] == [] // false
[] === [] // false

Comparing Two Arrays

Let us begin with regular array comparisons.

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

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

  return true;
}

// Example usage:
const array1 = [1, 2, 3];
const array2 = [1, 2, 3];
const array3 = [1, 2, 4];

arraysEqual(array1, array2); // true
arraysEqual(array1, array3); // false

Level 1

Our arraysEqual function works for regular array comparisons but not for different orders.

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

arraysEqual(array1, array2); // false

Let's change it to work for different ordered arrays.

We can do this by simply sort the given arrays before comparison.

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

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

  return true;
}

Now, our function works well with different-ordered arrays.

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

arraysEqual(array1, array2); // true

Level 2

Now, we test the arraysEqual function by giving mixed types of values.

arraysEqual(['1', 1], [1, '1']); // false

As you can see, our function failed even with the two arrays containing the same values.

Let's fix this.

function compareMixedTypes(a, b) {
  if (typeof a === typeof b) {
    return a === b ? 0 : a < b ? -1 : 1;
  }
  return typeof a < typeof b ? -1 : 1;
}

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

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

  return true;
}

arraysEqual(['1', 1], [1, '1']); // true

Level 3

Let's test the arraysEqual function with an array of objects.

const a = [{a: 1}, {b: 2}];
const b = [{b: 2}, {a: 1}];

arraysEqual(a, b); // false

As you can see, our function failed because the sorting does work for primitives but simply not for objects.

Let's remove sorting and try a different method.

function inArray(array, el) {
  for (var i = array.length; i--; ) {
    if ( array[i] === el ) return true;
  }
  return false;
}

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

  for (var i = a.length; i--; ) {
    if (!inArray(b, a[i])) {
      return false;
    }
  }
  return true;
}

Let's test again.

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

arraysEqual(array1, array2); // true
arraysEqual(['1', 1], [1, '1']); // true
arraysEqual([1, 1, 2, 2], [1, 2, 1, 2]); // true

const a = [{a: 1}, {b: 2}];
const b = [{b: 2}, {a: 1}];

arraysEqual(a, b); // false

As you can see the modified function works for primitives but not for objects, let's fix it in the next level.

Level 4

The problem with the inArray function is that we are comparing two values using === operator; we need to find a way to compare objects also with deep nested.

We can use the popular lodash library's isEqual function here. Let's change it.

function inArray(array, el) {
  for (var i = array.length; i--; ) {
    if (_.isEqual(array[i], el)) return true;
  }
  return false;
}

Now, let's try it.

let a = [{a: 1}, {b: 2}];
let b = [{b: 2}, {a: 1}];

arraysEqual(a, b); // true

a = [{a: {b: {c: ['a', 'b', 'c']}}}, [1, 2, 3]];
b = [[1, 2, 3], {a: {b: {c: ['a', 'b', 'c']}}}];
arraysEqual(a, b); // true

Level 5

Now what? If our code is working fine for all cases, then why do we need another level here?

There are some problems in the _.isEqual function.

https://github.com/lodash/lodash/issues/5401

https://github.com/lodash/lodash/issues/3640

https://github.com/lodash/lodash/issues/3428

Let's verify it now.

const a = [{m: new Map([['x', 'y']])}]
const b = [{m: new Map([['y', 'x']])}]

arraysEqual(a, b); // true

console.log(a, b);
/* 
[ { m: Map(1) { 'x' => 'y' } } ] 
[ { m: Map(1) { 'y' => 'x' } } ]
*/

So, what is next?

Let me introduce our organization's Standard Library function isEqlArr here.

import { isEqlArr, shuffle } from '@opentf/std';

let a = [1, 2, 3];
let b = [2, 3, 1];
isEqlArr(a, b); // true

a = ['1', 1];
b = [1, '1'];
isEqlArr(a, b); // true

a = [1, 1, 2, 2];
b = [1, 2, 1, 2];
isEqlArr(a, b); // true

a = 'Apple'.split('')
b = shuffle('Apple')
isEqlArr(a, b); // true

a = [{a: 1}, {b: 2}];
b = [{b: 2}, {a: 1}];
isEqlArr(a, b); // true

a = [{a: {b: {c: ['a', 'b', 'c']}}}, [1, 2, 3]];
b = [[1, 2, 3], {a: {b: {c: ['a', 'b', 'c']}}}];
isEqlArr(a, b); // true

a = [{m: new Map([['x', 'y']])}]
b = [{m: new Map([['y', 'x']])}]
isEqlArr(a, b); // false

Example: Real-World Use Case

In this example, we are going to find out how many sets of the same products were in the invoices of a company.

Note: If it doesn't make sense, just ignore this example.

import { isEqlArr } from '@opentf/std';

const invoices = [
  {
    id: 'c2607dc9-bd74-446d-ac83-de201d158b87',
    products: [
      {
        id: '4e5fe7a0-0fd2-44f3-bdcd-6a3b2eea4b50',
        name: 'Product A'
      },
       {
        id: '863cb499-1943-473b-85f2-285ed9128c74',
        name: 'Product B'
      }
    ]
  },
  {
    id: '6e523056-1601-41ba-b89e-6ae41164a843',
    products: [
      {
        id: '4e5fe7a0-0fd2-44f3-bdcd-6a3b2eea4b50',
        name: 'Product A'
      },
       {
        id: '401cc691-92f1-4e9e-bd78-995cb9663c5e',
        name: 'Product C'
      },
      
    ]
  },
  {
    id: '3d375953-4c60-433c-b8c2-c45628d8f2d6',
    products: [
      {
        id: '863cb499-1943-473b-85f2-285ed9128c74',
        name: 'Product B'
      },
      {
        id: '4e5fe7a0-0fd2-44f3-bdcd-6a3b2eea4b50',
        name: 'Product A'
      }
    ]
  }
]

let count = 0

for (let i = 0; i < invoices.length; i += 1) {
  for (let j = 0; j < invoices.length; j += 1) {
    if (i !== j) {
      if (isEqlArr(invoices[i].products, invoices[j].products)) {
        count++
      }
    }
  }
}

console.log(count); // 2

Note: You can try out these examples on our online Node.js REPL