I am having two arrays, how can i compare the two arrays at single shot.
var arr1= ["a","b","c"]; var arr2 = ["a","c","d"] if(arr1 == arr2){ console.log(true); }else{ console.log(false); } 
- Possible duplicate of How to compare arrays in JavaScript?ibodi– ibodi2019-06-10 00:17:13 +00:00Commented Jun 10, 2019 at 0:17
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8 Answers
var arr1 = ["a","b","c"]; var arr2 = ["a","c","d"]; if (arr1.length == arr2.length && arr1.every(function(u, i) { return u === arr2[i]; }) ) { console.log(true); } else { console.log(false); } Side note for edge cases:
=== is often considered slightly broken for this kind of task because NaN behaves unexpectedly:
var arr1 = ["a",NaN,"b"]; var arr2 = ["a",NaN,"b"]; if (arr1.length == arr2.length && arr1.every(function(u, i) { return u === arr2[i]; }) ) { console.log(true); } else { console.log(false); } The code above actually logs false because NaN !== NaN. In addition, === can't distinguish +0 from -0. To cover both of these cases, you could use a stronger comparison known as "egal" or "is", which can easily be implemented like so:
function is(a, b) { return a === b && (a !== 0 || 1 / a === 1 / b) // false for +0 vs -0 || a !== a && b !== b; // true for NaN vs NaN } var arr1 = ["a",NaN,"b"]; var arr2 = ["a",NaN,"b"]; if (arr1.length == arr2.length && arr1.every(function(u, i) { // Use "is" instead of "===" return is(u, arr2[i]); }) ) { console.log(true); } else { console.log(false); } 
4 Comments
Erwin Wessels
Note that this might give the wrong results in case arr1 is shorter then arr2 - try:
var arr1 = [ 'a', 'b' ], arr2 = ['a','b','c']; Include a simple length comparison before the every, and you're done: if( arr1.length === arr2.length && arr1.every( ...
Nathan Wall
Very good point, Erwin. I have updated my answer to include the length test you have suggested. Thanks!
kta
if the value exists but in different position then will this algorithm still works?
Jeremy Chone
This assumes the two arrays are sorted the same way.
[ES6]
Top answer is good & enough.
But when you just want to compare its values are same you have to sort it before. here's no need sort code.
if(arr1.length == arr2.length && arr1.every((v) => arr2.indexOf(v) >= 0)) { console.log(true); } else { console.log(false); } And.. I think using a 'some' instead of 'every' is better.
If those are not same, 'some' gives you a early exit. - very little early but early ;)
if(arr1.length == arr2.length && !arr1.some((v) => arr2.indexOf(v) < 0)) { console.log(true); } else { console.log(false); } 
Comments
I would make use of underscore for this.
var same = (_.difference(arr1, arr2).length == 0) 
1 Comment
alykoshin
This is incorrect.
_ = require('underscore'); _.difference(["a","c"], ["a","c","d"]).length; returns: 0 Same for lodash.The top answer is good, but I would also consider using Array.prototype:
Array.prototype.equals = function (arr) { return this.length == arr.length && this.every((u, i) => u === arr[i]); } console.log([1,2,3].equals([1,2,3])); // true console.log([1,2,3].equals([1,3,3])); // false // BUT! console.log(["a",NaN,"b"].equals(["a",NaN,"b"])); // false, because NaN !== NaN If you want it to work for NaNs too and distinguish +0 and -0, better use this:
Array.prototype.equals = function (arr) { function is(a, b) { // taken from the top answer return a === b && (a !== 0 || 1 / a === 1 / b) // false for +0 vs -0 || a !== a && b !== b; // true for NaN vs NaN } return this.length == arr.length && this.every((u, i) => is(u, arr[i])); } console.log(["a",NaN,"b"].equals(["a",NaN,"b"])); // true 
Comments
Here's another one, without ES5 every:
function arrEq(arr1, arr2) { for (var i = 0; i < arr1.length; i++) if (arr1[i] != arr2[i]) return false; return i == arr2.length; } 
Comments
(Although the question is much older than v9.0.0, but) since the question is about Node - starting from Node v9.0.0 you can use the built-in "util" module's isDeepStrictEqual(arr1, arr2) for comparing arrays (or objects, or anything for that matter)
const util = require('util'); let arr1 = [1,2,3]; let arr2 = [1,2,3]; let arr3 = [1,3,2]; console.log(util.isDeepStrictEqual(arr1, arr2)) // true console.log(util.isDeepStrictEqual(arr1, arr3)) // false Also works for NaN
There is also a non-strict version of this: https://nodejs.org/api/assert.html#assertdeepequalactual-expected-message
Comments
I wanted to add some modification of the code made by 'Taihwan Hah' but could not leave a comment (the system told me so)
So here is my modifs:
function ArrayEquals(arr1,arr2){ return arr1.length === arr2.length && !arr1.some((v) => arr2.indexOf(v) < 0) && !arr2.some((v) => arr1.indexOf(v) < 0); } basically, I had to check for but array because my arrays do not contains unique numbers.
Comments
I would like to improve the answer from staackuser2 a little bit:
var same = (arr1.length === arr2.length) && (_.difference(arr1, arr2).length === 0) or
var same = (_.difference(arr1, arr2).length === 0) && (_.difference(arr2, arr1).length === 0) 
