How to find an array in an object of arrays [duplicate] - arrays

Question

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?

Answer 1

To compare arrays, loop through them and compare every value:
Comparing arrays:
// Warn if overriding existing method
if(Array.prototype.equals)
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;
// if the argument is the same array, we can be sure the contents are same as well
if(array === this)
return true;
// 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});
Usage:
[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
//https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/hasOwnProperty
//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)
if(!this.hasOwnProperty(propName))
continue;
//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 more complicated objects, look at this answer and it's super long 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;
}
/**REQUIRES OBJECT COMPARE**/
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: https://jsfiddle.net/SamyBencherif/8352y6yw/

Answer 2

While this only works for scalar arrays (see note below), it is short code:
array1.length === array2.length && array1.every(function(value, index) { return value === array2[index]})
Same as above but 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).
UPDATE
From what I read in the comments, sorting the array and comparing may give accurate result:
const array2Sorted = array2.slice().sort();
array1.length === array2.length && array1.slice().sort().every(function(value, index) {
return value === array2Sorted[index];
});
Eg:
array1 = [2,3,1,4];
array2 = [1,2,3,4];
Then the above code would return true

Answer 3

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
Underscore isEqual docs
Lodash isEqual docs

Answer 4

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.

Answer 5

In the spirit of the original question:
I'd like to compare two arrays... ideally, efficiently. Nothing
fancy, just true if they are identical, and false if not.
I have been running performance tests on some of the more simple suggestions proposed here with the following results (fast to slow):
while (67%) by Tim Down
var i = a1.length;
while (i--) {
if (a1[i] !== a2[i]) return false;
}
return true
every (69%) by user2782196
a1.every((v,i)=> v === a2[i]);
reduce (74%) by DEIs
a1.reduce((a, b) => a && a2.includes(b), true);
join & toString (78%) by Gaizka Allende & vivek
a1.join('') === a2.join('');
a1.toString() === a2.toString();
half toString (90%) by Victor Palomo
a1 == a2.toString();
stringify (100%) by radtek
JSON.stringify(a1) === JSON.stringify(a2);
Note the examples below assumes the arrays are sorted, single-dimensional arrays. .length comparison has been removed for a common benchmark (add a1.length === a2.length to any of the suggestions and you will get a ~10% performance boost). Choose whatever solutions that works best for you knowing the speed and limitation of each.

Answer 6

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 =
Array.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 =>
x.id !== undefined && x.id === y.id
// 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.
Edit
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.

Answer 7

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;
};

Answer 8

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.
Example:
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:
http://jsfiddle.net/Roundaround/DLkxX/

Answer 9

On the same lines as JSON.encode is to use join().
function checkArrays( arrA, arrB ){
//check if lengths are different
if(arrA.length !== arrB.length) return false;
//slice so we do not effect the original
//sort makes sure they are in order
//join makes it a string so we can do a string compare
var cA = arrA.slice().sort().join(",");
var cB = arrB.slice().sort().join(",");
return cA===cB;
}
var a = [1,2,3,4,5];
var b = [5,4,3,2,1];
var c = [1,2,3,4];
var d = [1,2,3,4,6];
var e = ["1","2","3","4","5"]; //will return true
console.log( checkArrays(a,b) ); //true
console.log( checkArrays(a,c) ); //false
console.log( checkArrays(a,d) ); //false
console.log( checkArrays(a,e) ); //true
Only problem is if you care about types which the last comparison tests.
If you care about types, you will have to loop.
function checkArrays( arrA, arrB ){
//check if lengths are different
if(arrA.length !== arrB.length) return false;
//slice so we do not effect the orginal
//sort makes sure they are in order
var cA = arrA.slice().sort();
var cB = arrB.slice().sort();
for(var i=0;i<cA.length;i++){
if(cA[i]!==cB[i]) return false;
}
return true;
}
var a = [1,2,3,4,5];
var b = [5,4,3,2,1];
var c = [1,2,3,4];
var d = [1,2,3,4,6];
var e = ["1","2","3","4","5"];
console.log( checkArrays(a,b) ); //true
console.log( checkArrays(a,c) ); //false
console.log( checkArrays(a,d) ); //false
console.log( checkArrays(a,e) ); //false
If the order should remain the same, than it is just a loop, no sort is needed.
function checkArrays( arrA, arrB ){
//check if lengths are different
if(arrA.length !== arrB.length) return false;
for(var i=0;i<arrA.length;i++){
if(arrA[i]!==arrB[i]) return false;
}
return true;
}
var a = [1,2,3,4,5];
var b = [5,4,3,2,1];
var c = [1,2,3,4];
var d = [1,2,3,4,6];
var e = ["1","2","3","4","5"];
console.log( checkArrays(a,a) ); //true
console.log( checkArrays(a,b) ); //false
console.log( checkArrays(a,c) ); //false
console.log( checkArrays(a,d) ); //false
console.log( checkArrays(a,e) ); //false

Answer 10

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

Answer 11

Even though this has a lot of answers, one that I believe to be of help:
const newArray = [ ...new 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).
Edit1.
Answering Dmitry Grinko's question: "Why did you use spread operator ( ... ) here - ...new Set ? It doesn't work"
Consider this code:
const arr1 = [ 'a', 'b' ]
const arr2 = [ 'a', 'b', 'c' ]
const newArray = [ new Set( [...arr1, ...arr2] ) ]
console.log(newArray)
You'll get
[ Set { 'a', 'b', 'c' } ]
In order to work with that value you'd need to use some Set properties (see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set).
On the other hand, when you use this code:
const arr1 = [ 'a', 'b' ]
const arr2 = [ 'a', 'b', 'c' ]
const newArray = [ ...new Set( [...arr1, ...arr2] ) ]
console.log(newArray)
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.

Answer 12

Shortest
For an array of numbers try:
a1==''+a2
var a1 = [1,2,3];
var a2 = [1,2,3];
console.log( a1==''+a2 )
Note: this method will not work when the array also contains strings, e.g. a2 = [1, "2,3"].

Answer 13

You can simply use isEqual from lodash library. It is very efficient and clean.
import isEqual from "lodash/isEqual";
const isTwoArraysEqual = isEqual(array1, array2);

Answer 14

Your code will not handle the case appropriately when both arrays have same elements but not in same order.
Have a look at my code with your example which compares two arrays whose elements are numbers, you might modify or extend it for other element types (by utilising .join() instead of .toString()).
var a1 = [1,2,3];
var a2 = [1,2,3];
const arraysAreEqual = a1.sort().toString()==a2.sort().toString();
// true if both arrays have same elements else false
console.log(arraysAreEqual);

Answer 15

There are many complicated long answers in here, so I just want to contribute one very simple answer: use toString() to turn an array into a simple comma-separated string which you can easily compare with ===
let a = [1, 2, 3]
let b = [1, 2, 3]
let c = [4, 2, 3]
console.log(a.toString()) // this outputs "1,2,3"
console.log(a.toString() === b.toString()) // this outputs true because "1,2,3" === "1,2,3"
console.log(a.toString() === c.toString()) // this outputs false because "1,2,3" != "4,2,3"

Answer 16

Here you go,
const a = [1, 2, 3]
const b = [1, 2, 3, 4, 5]
const diff = b.filter(e => !a.includes(e))
console.log(diff)
Most of the above answers dosen't work for unordered list.
This works for unordered lists too.
const a = [3, 2, 1]
const b = [1, 2, 3, 4, 5]
const diff = b.filter(e => !a.includes(e))
console.log(diff)
If size of a is greater than b,
const a = [1, 2, 3, 4, 5]
const b = [3, 2, 1]
const diff = a.length > b.length ? a.filter(e => !b.includes(e)) : b.filter(e => !a.includes(e))
console.log(diff)

Answer 17

Code Golfing
There are plenty of answers showing how to compare arrays efficiently.
Below is the shortest way to compare two int or (string) arrays, measured in bytes of code.
const a = [1, 2, 3]
const b = [1, 2, 3]
console.log("1. ", a.join() == b.join())
console.log("2. ", a.join() == [].join())
console.log("3. ", 1 + a == 1 + b)
console.log("4. ", 1 + [] == 1 + b)
// even shorter
console.log("4. b) ", a == "" + b)
// false positives (see flaws)
console.log("5. ", 1 + ["3"] == 1 + [3]) // type differences
console.log("6. ", 1 + ["1,2"] == 1 + ["1", "2"])
Explanation
This works because when using the + operator, the types are automatically converted to allow concatenation. In this case, the 1 and the [1, 2, 3] are both converted to a string.
Internally, JavaScript uses [1, 2, 3].join() to convert the array to a string and then adds them resulting in 11,2,3. When doing this on both arrays, one can simply use === or == to compare the two strings.
Flaws
Using this technique, the comparison does not care if the elements in the arrays to be compared are of different types. [1, 2] will be equal to ["1", "2"] because of the string conversion.
EDIT: As pointed out in the comments, comparing string arrays can produce false positives, such as ["1,2"] being 'equal' to ["1", "2"]. This is of no concern if you are sure these never occur (e.g. in many code golfing challenges).
Disclaimer
While this is useful for code golfing, it should probably not be used in production code. The two flaws pointed out aren't helping that either.

Answer 18

Lot of good answers here. This is how I usually do it -
if ( arr1.length === arr2.length && arr1.every((a1) => arr2.includes(a1)) ) {
// logic
}
every() will only return true if all elements pass the given camparison
logic. If it encounters a false, in any iteration, it terminates and
returns false.
Time complexity will be O(n*m).

Answer 19

Here is a Typescript version:
//https://stackoverflow.com/a/16436975/2589276
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
}
//https://stackoverflow.com/a/16436975/2589276
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)
})

Answer 20

There is a Stage 1 proposal, introduced in 2020, to allow for the easy comparison of arrays by adding Array.prototype.equals to the language. This is how it would work, without any libraries, monkeypatching, or any other code:
[1, 2, 3].equals([1, 2, 3]) // evaluates to true
[1, 2, undefined].equals([1, 2, 3]) // evaluates to false
[1, [2, [3, 4]]].equals([1, [2, [3, 4]]]) // evaluates to true
It's only a tentative proposal so far - TC39 will now "devote time to examining the problem space, solutions and cross-cutting concerns". If it makes it to stage 2, it has a good chance of eventually being integrated into the language proper.

Answer 21

If you are using a testing framework like Mocha with the Chai assertion library, you can use deep equality to compare arrays.
expect(a1).to.deep.equal(a2)
This should return true only if the arrays have equal elements at corresponding indices.

Answer 22

If they are two arrays of numbers or strings only, this is a quick one-line one
const array1 = [1, 2, 3];
const array2 = [1, 3, 4];
console.log(array1.join(',') === array2.join(',')) //false
const array3 = [1, 2, 3];
const array4 = [1, 2, 3];
console.log(array3.join(',') === array4.join(',')) //true

Answer 23

Another approach with very few code (using Array reduce and Array includes):
arr1.length == arr2.length && arr1.reduce((a, b) => a && arr2.includes(b), true)
If you want to compare also the equality of order:
arr1.length == arr2.length && arr1.reduce((a, b, i) => a && arr2[i], true)
The length check ensures that the set of elements in one array isn't just a subset of the other one.
The reducer is used to walk through one array and search for each item in other array. If one item isn't found the reduce function returns false.
In the first example it's being tested that an element is included
The second example check for the order too

Answer 24

We could do this the functional way, using every (https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/every)
function compareArrays(array1, array2) {
if (array1.length === array2.length)
return array1.every((a, index) => a === array2[index])
else
return false
}
// test
var a1 = [1,2,3];
var a2 = [1,2,3];
var a3 = ['a', 'r', 'r', 'a', 'y', '1']
var a4 = ['a', 'r', 'r', 'a', 'y', '2']
console.log(compareArrays(a1,a2)) // true
console.log(compareArrays(a1,a3)) // false
console.log(compareArrays(a3,a4)) // false

Answer 25

This compares 2 unsorted arrays:
function areEqual(a, b) {
if ( a.length != b.length) {
return false;
}
return a.filter(function(i) {
return !b.includes(i);
}).length === 0;
}

Answer 26

A simple approach:
function equals(a, b) {
if ((a && !b) || (!a && b) || (!a && !b) || (a.length !== b.length)) {
return false;
}
var isDifferent = a.some(function (element, index) {
return element !== b[index];
});
return !isDifferent;
}

Answer 27

Already some great answers.But i would like to share anther idea which has proven to be reliable in comparing arrays. We can compare two array using JSON.stringify ( ) . It will create a string out the the array and thus compare two obtained strings from two array for equality
JSON.stringify([1,{a:1},2]) == JSON.stringify([1,{a:1},2]) //true
JSON.stringify([1,{a:1},2]) == JSON.stringify([1,{a:2},2]) //false
JSON.stringify([1,{a:1},2]) == JSON.stringify([1,{a:2},[3,4],2]) //false
JSON.stringify([1,{a:1},[3,4],2]) == JSON.stringify([1,{a:2},[3,4],2]) //false
JSON.stringify([1,{a:2},[3,4],2]) == JSON.stringify([1,{a:2},[3,4],2]) //true
JSON.stringify([1,{a:2},[3,4],2]) == JSON.stringify([1,{a:2},[3,4,[5]],2]) //false
JSON.stringify([1,{a:2},[3,4,[4]],2]) == JSON.stringify([1,{a:2},[3,4,[5]],2]) //false
JSON.stringify([1,{a:2},[3,4,[5]],2]) == JSON.stringify([1,{a:2},[3,4,[5]],2]) //true

Answer 28

In a simple way uning stringify but at same time thinking in complex arrays:
**Simple arrays**:
var a = [1,2,3,4];
var b = [4,2,1,4];
JSON.stringify(a.sort()) === JSON.stringify(b.sort()) // true
**Complex arrays**:
var a = [{id:5,name:'as'},{id:2,name:'bes'}];
var b = [{id:2,name:'bes'},{id:5,name:'as'}];
JSON.stringify(a.sort(function(a,b) {return a.id - b.id})) === JSON.stringify(b.sort(function(a,b) {return a.id - b.id})) // true
**Or we can create a sort function**
function sortX(a,b) {
return a.id -b.id; //change for the necessary rules
}
JSON.stringify(a.sort(sortX)) === JSON.stringify(b.sort(sortX)) // true

Answer 29

Here a possibility for unsorted arrays and custom comparison:
const array1 = [1,3,2,4,5];
const array2 = [1,3,2,4,5];
const isInArray1 = array1.every(item => array2.find(item2 => item===item2))
const isInArray2 = array2.every(item => array1.find(item2 => item===item2))
const isSameArray = array1.length === array2.length && isInArray1 && isInArray2
console.log(isSameArray); //true

Answer 30

Simple
type Values = number | string;
/** Not sorted array */
function compare<Values>(a1: Array<Values>, a2: Array<Values>): boolean {
if (a1.length !== a2.length) {
return false;
}
/** Unique values */
const set1 = new Set<Values>(a1);
const set2 = new Set<Values>(a2);
if (set1.size !== set2.size) {
return false;
}
return [...set1].every((value) => [...set2].includes(value));
}
compare([1, 2, 3], [1, 2, 3]); // true
compare([1, 2, 3], [1, 3, 2]); // true
compare([1, 1, 1], [1, 2, 3]); // false
compare([1, 1, 2, 3], [1, 2, 3]); // false
/** Sorted arrays, faster method */
function compareSorted<Values>(a1: Array<Values>, a2: Array<Values>): boolean {
if (a1.length !== a2.length) {
return false;
}
/** Unique values */
const set1 = new Set<Values>(a1);
const set2 = new Set<Values>(a2);
if (set1.size !== set2.size) {
return false;
}
return [...set1].every((value, index) => value === [...set2][index]);
}
compareSorted([1, 2, 3], [1, 2, 3]); // true
compareSorted([1, 2, 3], [1, 3, 2]); // false
compareSorted([1, 1, 1], [1, 2, 3]); // false
compareSorted([1, 1, 2, 3], [1, 2, 3]); // false