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?
Herer's my solution:
/**
* Tests two data structures for equality
* @param {object} x
* @param {object} y
* @returns {boolean}
*/
var equal = function(x, y) {
if (typeof x !== typeof y) return false;
if (x instanceof Array && y instanceof Array && x.length !== y.length) return false;
if (typeof x === 'object') {
for (var p in x) if (x.hasOwnProperty(p)) {
if (typeof x[p] === 'function' && typeof y[p] === 'function') continue;
if (x[p] instanceof Array && y[p] instanceof Array && x[p].length !== y[p].length) return false;
if (typeof x[p] !== typeof y[p]) return false;
if (typeof x[p] === 'object' && typeof y[p] === 'object') { if (!equal(x[p], y[p])) return false; } else
if (x[p] !== y[p]) return false;
}
} else return x === y;
return true;
};
Works with any nested data structure, and obviously ignores objects' methods. Don't even think of extending Object.prototype with this method, when I tried this once, jQuery broke ;)
For most arrays it's still faster than most of serialization solutions. It's probably the fastest compare method for arrays of object records.
equal({}, {a:1}) and equal({}, null) and this errors out: equal({a:2}, null)JSON.stringify(collectionNames).includes(JSON.stringify(sourceNames)) ? array.push(collection[i]) : null
This is how i did it.
An alternative way using filter and arrow functions
arrOne.length === arrTwo.length && arrOne.filter((currVal, idx) => currVal !== arrTwo[idx]).length === 0
I know that JSON.stringfy is slow when dealing with large datasets but what if you used template literals?
Example:
const a = [1, 2, 3];
const b = [1, 2, 'test'];
const a_string = Array.isArray(a) && `${a}`;
const b_string = Array.isArray(b) && `${b}`;
const result = (a === b);
console.log(result);
Taking into consideration you're using ES6 of course.
=)
a = ["1,1", "2"] and b = [1,1,2] (or b = ["1", "1,2"])... I think it will be hard to get something robust with this approach. ^^
JSON.stringify won't perform a shallow comparison either.It's a tricky implicit array equality checking but can handle the job right after coherence arrays to string.
var a1 = [1, 2, 3];
var a2 = [1, 2, 3];
var isEqual = a1 <= a2 && a1 >= a2; // true
Extending Tomáš Zato idea. Tomas's Array.prototype.compare should be infact called Array.prototype.compareIdentical.
It passes on:
[1, 2, [3, 4]].compareIdentical ([1, 2, [3, 2]]) === false;
[1, "2,3"].compareIdentical ([1, 2, 3]) === false;
[1, 2, [3, 4]].compareIdentical ([1, 2, [3, 4]]) === true;
[1, 2, 1, 2].compareIdentical ([1, 2, 1, 2]) === true;
But fails on:
[[1, 2, [3, 2]],1, 2, [3, 2]].compareIdentical([1, 2, [3, 2],[1, 2, [3, 2]]])
Here is better (in my opinion) version:
Array.prototype.compare = 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;
this.sort();
array.sort();
for (var i = 0; i < this.length; i++) {
// Check if we have nested arrays
if (this[i] instanceof Array && array[i] instanceof Array) {
// recurse into the nested arrays
if (!this[i].compare(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;
}
var a1 = [1,2,3,6];
var a2 = [1,2,3,5];
function check(a, b) {
return (a.length != b.length) ? false :
a.every(function(row, index) {
return a[index] == b[index];
});
}
check(a1, a2);
////// OR ///////
var a1 = [1,2,3,6];
var a2 = [1,2,3,6];
function check(a, b) {
return (a.length != b.length) ? false :
!(a.some(function(row, index) {
return a[index] != b[index];
}));
}
check(a1, a2)
Recursive & works on NESTED arrays:
function ArrEQ(a1,a2){
return(
//:Are both elements arrays?
Array.isArray(a1)&&Array.isArray(a2)
?
//:Yes: Test each entry for equality:
a1.every((v,i)=>(ArrEQ(v,a2[i])))
:
//:No: Simple Comparison:
(a1===a2)
);;
};;
console.log( "Works With Nested Arrays:" );
console.log( ArrEQ(
[1,2,3,[4,5,[6,"SAME/IDENTICAL"]]],
[1,2,3,[4,5,[6,"SAME/IDENTICAL"]]]
));;
console.log( ArrEQ(
[1,2,3,[4,5,[6,"DIFFERENT:APPLES" ]]],
[1,2,3,[4,5,[6,"DIFFERENT:ORANGES"]]]
));;
Works with MULTIPLE arguments with NESTED arrays:
//:Return true if all of the arrays equal.
//:Works with nested arrays.
function AllArrEQ(...arrays){
for(var i = 0; i < (arrays.length-1); i++ ){
var a1 = arrays[i+0];
var a2 = arrays[i+1];
var res =(
//:Are both elements arrays?
Array.isArray(a1)&&Array.isArray(a2)
?
//:Yes: Compare Each Sub-Array:
//:v==a1[i]
a1.every((v,i)=>(AllArrEQ(v,a2[i])))
:
//:No: Simple Comparison:
(a1===a2)
);;
if(!res){return false;}
};;
return( true );
};;
console.log( AllArrEQ(
[1,2,3,[4,5,[6,"ALL_EQUAL" ]]],
[1,2,3,[4,5,[6,"ALL_EQUAL" ]]],
[1,2,3,[4,5,[6,"ALL_EQUAL" ]]],
[1,2,3,[4,5,[6,"ALL_EQUAL" ]]],
));;
I believe in plain JS and with ECMAScript 2015, which is sweet and simple to understand.
var is_arrays_compare_similar = function (array1, array2) {
let flag = true;
if (array1.length == array2.length) {
// check first array1 object is available in array2 index
array1.every( array_obj => {
if (flag) {
if (!array2.includes(array_obj)) {
flag = false;
}
}
});
// then vice versa check array2 object is available in array1 index
array2.every( array_obj => {
if (flag) {
if (!array1.includes(array_obj)) {
flag = false;
}
}
});
return flag;
} else {
return false;
}
}
Surprisingly, nobody brought up a solution with find
const a = [1, 2, 3]
const b = [1, 2, 3, 4]
a.find((v,i) => v !== b[i])
The benefit here is that instead of comparing all values it looks for the first occurrence and ends the loop as early as possible. Or in other words, instead of asking "are two arrays equal?" it asks "is one array different from another?".
3rd by performance benchmark https://jsben.ch/TgFrA
Keep in mind, the order matters, a.find(...) !== b.find(...) and could be checked by a.length === b.length
if (a.length === b.length && a.find((v,i) => v !== b[i]) === undefined) {
// equal
}
a.find((v,i) => v !== b[i]) is exactly equivalent to a.some((v,i) => v !== b[i]) with the difference that .some() returns a boolean directly. .some() and .every() are also equivalent to each other through negation: a.every((v,i) => v -== b[i]). When I ran the benchmark the .every() code was in second place in front of the .find() variant.Actually, in the Lodash documentation, they give two pretty good examples for comparing and return fresh arrays for both differences and similarities (respectively in the examples below):
import { differenceWith, intersectionWith, isEqual } from 'lodash'
differenceWith(
[{ a: 1 }, { b: 1 }],
[{ a: 1 }, { b: 1 }, { c: 1 }],
isEqual
) // []... 💀the bigger array needs to go first!
differenceWith(
[{ a: 1 }, { b: 1 }, { c: 1 }],
[{ a: 1 }, { b: 1 }],
isEqual,
) // [{ c: 1 }] 🎉
intersectionWith(
[{ a: 1 }, { b: 1 }],
[{ a: 1 }, { b: 1 }, { c: 1 }],
isEqual,
) // [{ a: 1 }, { b: 1 }] 🎉this one doesn't care about which is bigger
If you won't always know which array will be bigger, you can write a helper function for it like so:
const biggerFirst = (arr1, arr2) => {
return arr1.length > arr2.length ? [arr1, arr2] : [arr2, arr1]
}
const [big, small] = biggerFirst(
[{ a: 1 }, { b: 1 }],
[{ a: 1 }, { b: 1 }, { c: 1 }],
)
differenceWith(big, small, isEqual) // 🎉even though we have no idea which is bigger when they are fed to biggerFirst()
From what I can tell, these match deeply as well so that's pretty nice.
I know relying on libraries for everything shouldn't be applauded, but this is the most concise/clean solution I've found to a really common problem. Hope it helps someone!
All the other solutions look complicated. This might not be the most efficient with or handle all edge cases but it works great for me.
Array.prototype.includesArray = function(arr) {
return this.map(i => JSON.stringify(i)).includes(JSON.stringify(arr))
}
Usage
[[1,1]].includesArray([1,1])
// true
[[1,1]].includesArray([1,1,2])
// false
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
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])) {
continue;
} else {
return false;
}
} else {
if (a1[i] !== a2[i]) {
return false;
}
}
}
return true;
}
With an option to compare the order or not:
function arraysEqual(a1, a2, compareOrder) {
if (a1.length !== a2.length) {
return false;
}
return a1.every(function(value, index) {
if (compareOrder) {
return value === a2[index];
} else {
return a2.indexOf(value) > -1;
}
});
}
Récursive cmp function working with number/string/array/object
<script>
var cmp = function(element, target){
if(typeof element !== typeof target)
{
return false;
}
else if(typeof element === "object" && (!target || !element))
{
return target === element;
}
else if(typeof element === "object")
{
var keys_element = Object.keys(element);
var keys_target = Object.keys(target);
if(keys_element.length !== keys_target.length)
{
return false;
}
else
{
for(var i = 0; i < keys_element.length; i++)
{
if(keys_element[i] !== keys_target[i])
return false;
if(!cmp(element[keys_element[i]], target[keys_target[i]]))
return false;
}
return true;
}
}
else
{
return element === target;
}
};
console.log(cmp({
key1: 3,
key2: "string",
key3: [4, "45", {key4: [5, "6", false, null, {v:1}]}]
}, {
key1: 3,
key2: "string",
key3: [4, "45", {key4: [5, "6", false, null, {v:1}]}]
})); // true
console.log(cmp({
key1: 3,
key2: "string",
key3: [4, "45", {key4: [5, "6", false, null, {v:1}]}]
}, {
key1: 3,
key2: "string",
key3: [4, "45", {key4: [5, "6", undefined, null, {v:1}]}]
})); // false
</script>This method is one that only works on scalar arrays, like the second voted answer on this question.
var arrs = [
[[1, 2, 3], [1, 2, 3]], // true
[[1, 2, 3, 4], [1, 2, 3]], // false
[[1, 2, 3], [1, 2, 3, 4]], // false
]
const arraysEqual = (one, two) => (one.filter((i, n) => two[n] === i).length === one.length) && (two.filter((i, n) => one[n] === i).length === two.length)
arrs.forEach(arr => {
console.log(arraysEqual(arr[0], arr[1]))
})Without ES6 syntax:
var arrs = [
[[1, 2, 3], [1, 2, 3]], // true
[[1, 2, 3, 4], [1, 2, 3]], // false
[[1, 2, 3], [1, 2, 3, 4]], // false
]
function arraysEqual(one, two) {
return (one.filter((i, n) => two[n] === i).length === one.length) && (two.filter((i, n) => one[n] === i).length === two.length)
}
arrs.forEach(arr => {
console.log(arraysEqual(arr[0], arr[1]))
})I answered this question at https://stackoverflow.com/a/10316616/711085 (which has since been marked a duplicate of this answer). There you will find a Deep Equals implementation that handles numerous cases, such as Map and Set and arbitrary nesting of arrays and objects. The discussion therein of non-transitivity of == and documenting == vs === is particularly important.
For OP's particular problem, if the arrays consist only of numbers and strings and booleans, and no NaNs, then the most efficient method for sufficiently large arrays is a precompiled function:
function areSimpleArraysEqual(a,b) {
// requires inputs be arrays of only Number, String, Boolean, and no NaN.
// will propagate error if either array is undefined.
if (a.length!=b.length)
return false;
for(let i=0; i<a.length; i++)
if (a[i]!==b[i]) // using === equality
return false;
return true;
}
One may achieve average-case O(1) and worst-case O(N) in some rare instances if one's business logic keeps appending to the ends of the arrays, by also checking if (a.length>0 && a[a.length-1]!==b[b.length-1]) return false; .
If you like simple savagery:
const a1 = ['a', 'b', 'e', 'd', 'c'];
const a2 = ['a', 'b', 'c', 'd', 'e'];
a1.length === a2.length && [...a1, ...a2].filter((item) => a2.indexOf(item) === -1).length === 0
Note that you can perform "indexOf" on any of the 2 arrays.
indexOf(item) === -1 fell out of favor when ! includes(item) was introduced in 2016.If the array is plain and the order is matter so this two lines may help
//Assume
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
Array.prototype.reduce will walk through every element in a even if the first compared elements do not match. Also using !a and != in the loop is a double negative which makes this answer more complicated (and hard to read) than it needs to be
tried deep-equal and it worked
var eq = require('deep-equal');
eq({a: 1, b: 2, c: [3, 4]}, {c: [3, 4], a: 1, b: 2});
let equals = (LHS, RHS) => {
if (!(LHS instanceof Array)) return "false > L.H.S is't an array";
if (!(RHS instanceof Array)) return "false > R.H.S is't an array";
if (LHS.length != RHS.length) return false;
let to_string = x => JSON.stringify(x.sort((a, b) => a - b));
return to_string(LHS) == to_string(RHS);
};
let l = console.log
l(equals([5,3,2],[3,2,5])) // true
l(equals([3,2,5,3],[3,2,5])) // false
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
count_of_comparable--;
// 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
if(!(this[property].compare_to(comparable[property]))){
// 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;
}
vary short and useful .. you know length are equal and arrays are simple
const imageData = [1,2,3,4] ;
const imageData2 = [1,2,3,4] ;
// true = not equal
return imageData.find((x, i) => imageData2[i] != x)
// or true = equal
return imageData.every((x, i) => imageData2[i] == x)
and when you do not know that
return JSON.stringify(imageData) == JSON.stringify(imageData2)
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}
true
All credits goes to @tomas-zato.
I would do like this:
[2,3,4,5] == [2,3,4,5].toString()
When you use the "==" operator, JavaScript checks if the values(left and right) is the same type, if it's different JavaScript try to convert both side in the same type.
Array == String
Array has toString method so JavaScript uses it to convert them to the same type, work the same way writing like this:
[2,3,4,5].toString() == [2,3,4,5].toString()
[1,2,3,4].toString() === ["1,2,3",4].toString() // => true
([] == []) == false.