How to compare arrays in JavaScript?
JavascriptArraysJsonJavascript Problem Overview
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?
Javascript Solutions
Solution 1 - Javascript
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;
// 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/
Solution 2 - Javascript
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).
UPDATE
From what I read from 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 give true
Solution 3 - Javascript
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
Solution 4 - Javascript
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.
Solution 5 - Javascript
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.
>
>Unrelated note: it is interesting to see people getting all trigger-happy John Waynes on the down vote button on perfectly legitimate answers to this question.
Solution 6 - Javascript
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.
Solution 7 - Javascript
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;
};
Solution 8 - Javascript
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/
Solution 9 - Javascript
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
Solution 10 - Javascript
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
Solution 11 - Javascript
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.
Solution 12 - Javascript
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"].
Solution 13 - Javascript
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);
Solution 14 - Javascript
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.
Solution 15 - Javascript
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)
})
Solution 16 - Javascript
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.
Solution 17 - Javascript
You can simply use isEqual from lodash library. It is very efficient and clean.
import {isEqual} from "lodash";
const isTwoArraysEqual = isEqual(array1, array2);
Solution 18 - Javascript
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.
Solution 19 - Javascript
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
Solution 20 - Javascript
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"
Solution 21 - Javascript
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)
Solution 22 - Javascript
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
Solution 23 - Javascript
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
lengthcheck 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
Solution 24 - Javascript
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;
}
Solution 25 - Javascript
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;
}
Solution 26 - Javascript
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
Solution 27 - Javascript
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.
Solution 28 - Javascript
JSON.stringify(collectionNames).includes(JSON.stringify(sourceNames)) ? array.push(collection[i]) : null
This is how i did it.
Solution 29 - Javascript
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
Solution 30 - Javascript
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
Solution 31 - Javascript
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.
=)
Solution 32 - Javascript
An alternative way using filter and arrow functions
arrOne.length === arrTwo.length && arrOne.filter((currVal, idx) => currVal !== arrTwo[idx]).length === 0
Solution 33 - Javascript
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
Solution 34 - Javascript
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
Solution 35 - Javascript
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;
}
Solution 36 - Javascript
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)
Solution 37 - Javascript
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;
else
return false;
}
calling function
var isBool = compare(arr1.sort().join(),arr2.sort().join());
Solution 38 - Javascript
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"]]]
));;
Solution 39 - Javascript
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" ]]],
));;
Solution 40 - Javascript
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!
Solution 41 - Javascript
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;
}
}
Solution 42 - Javascript
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
}
Solution 43 - Javascript
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).
Solution 44 - Javascript
We can use every() and includes() method to compare two arrays.
function check(a1,a2){
let result = a1.every((x)=>{
return a2.includes(x);
});
return result;
}
Solution 45 - Javascript
With the functional Programming approach we can achieve with the following code.
const a1 = [1, 2, 22, 4];
const a2 = [4, 2, 22, 1];
const checkValue = (firstValues: number[]) => {
return (value: number) => {
const isEqual = firstValues.includes(value);
return isEqual
}
}
const notUniq = a1.map(checkValue(a2)).some((x) => !x) // false
Solution 46 - Javascript
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
Solution 47 - Javascript
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
Solution 48 - Javascript
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;
}
Solution 49 - Javascript
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});
Solution 50 - Javascript
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;
}
});
}
Solution 51 - Javascript
Only works for one level Arrays, String or Numbers type
function isArrayEqual(ar1, ar2) {
return !ar1.some(item => ar2.indexOf(item) === -1) && ar1.length === ar2.length;
}
Solution 52 - Javascript
Here is a very short way to do it
function arrEquals(arr1, arr2){
return arr1.length == arr2.length &&
arr1.filter(elt=>arr1.filter(e=>e===elt).length == arr2.filter(e=>e===elt).length).length == arr1.length
}
Solution 53 - Javascript
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>
Solution 54 - Javascript
This simple solution works for me
function areEqual(a, b){
let x = 0;
for (n in a){
if (a[n] == b[n]){x = 1;}
else {x = 0};
}
return x;
}
a = [1,2,3];
b = [1,2,3];
> console.log(areEqual(a,b))
>true
I used it in this simple real code application
<script>
let corrette = [1];
let risposte = [];
function areEqual(a, b){
let x = 0;
for (n in a){
if (a[n] == b[n]){x = 1;}
else {x = 0};
}
if (x){console.log("The 2 arrays are equal")}
return x;
}
</script>
Apporto di attrezzatura per cucina da parte del proprietario
<button onclick="risposte[0]=1">Capitale proprio</button>
<button onclick="risposte[0]=0">Capitale di debito</button>
<button onclick="risposte[0]=0">Debito commerciale</button>
<br><hr>
<button onclick="if(areEqual(corrette,risposte)){controlla.innerHTML='Esatto'}else{controlla.innerHTML='No'}">Controlla le risposte</button>
<div id="controlla"></div>
Solution 55 - Javascript
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]))
})
Solution 56 - Javascript
I think the simplest way is to turn each array into a string like you tried, and compare the strings.
To convert the arrays into string, just put this string of methods onto the arrays. These are the arrays:
var arr1 = [1, 2, "foo", 3, "bar", 3.14]
var arr2 = [1, 2, "foo", 3, "bar", 3.14]
Now, you have to convert them into the strings. The list of methods are:
arr1.toString().replace(/,/gi, "")
arr2.toString().replace(/,/gi, "")
The methods do:
**.toString()** -
Turns array into a string, concatenating the elements of the array.
Ex.
["tree", "black hole"] -> "tree,black hole"
Sadly, it includes the commas. That is why we have to do:
***.replace(a, b)***
It finds and replaces the first argument (a) with the second argument (b) in the string you are doing it on.
Ex.
"0000010000010000000".replace("1", "2")
will return: "0000020000010000000"
It only replaces the first instance of parameter 1, so we can do regex instead.
Ex.
"0000010000010000000".replace(/1/gi, "2")
will return: "0000020000020000000"
You wrap what you want to replace with /. Say what you want to replace is 1. You make it: /1/. But then you have to add the gi at the end so that it selects every instance. So, you have to put /1/gi with a comma at the end, and then you can put what you want to replace it with.
Now, your two arrays are:
arr1: "12foo3bar3.14"
arr2: "12foo3bar3.14"
Now you say this:
if(arr1 === arr2) {
// Now the code you put inside of this if statement will only run if arr1 and arr2 have the same contents.
} else {
// This code will run if arr1 and arr2 have any differences.
}
If you want to check if arr1 CONTAINS arr2 instead of having the same exact contents, you do this.
if(arr1.indexOf(arr2) !== -1) {
//This code will happen if arr2 is inside of arr1. If there is one extra array
//item in arr1, it doesn't matter. But, if arr2 has an extra array item, nothing in
//this if will run. If you want arr2 to contain arr1, just make arr1 in the
//condition of this if arr2, and make arr2 arr1.
}
Basically, if you want the arrays to be the EXACT SAME, do this:
if(arr1.toString().replace(/,/gi, "") === arr2.toString().replace(/,/gi, "")) {
//arrays are the same
} else {
//arrays are different
}
And if you want to know if an array contains another, just do this:
arrayThatWillHoldAnotherArray = arrayThatWillHoldAnotherArray.toString().replace(/,/gi)
arrayThatWillBeInsideAnotherArray = arrayThatWillBeInsideAnotherArray.toString().replace(/,/gi)
if(arrayThatWillHoldAnotherArray.indexOf(arrayThatWillBeInsideAnotherArray) !== -1) {
//arrayThatWillHoldAnotherArray has arrayThatWillBeInsideAnotherArray inside of it
} else {
//it doesn't
}
console.log("Read the code to understand this.")
var arr1 = [1,2,"foo",3,"bar",3.14]
var arr2 = [1,2,"foo",3,"bar",3.14]
function checkIfArraysAreTheSame(a,b) {
if(a.toString().replace(/,/gi,"") === b.toString().replace(/,/gi,"")) {
console.log("A and B are the same!")
return true;
}
console.log("A and B are NOT the same!")
return false
}
checkIfArraysAreTheSame(arr1,arr2)
//expected output: A and B are the same!
//Now, let's add another item to arr2.
arr2.push("Lorem")
checkIfArraysAreTheSame(arr1,arr2)
//expected output: A and B are NOT the same!
function checkIfArrayIsNestedInsideAnother(a,b) {
//If this returns true, b is nested inside a.
if (a.toString().replace(/,/gi,"").indexOf(b.toString().replace(/,/gi,"")) > -1) {
console.log("B is nested inside of A!")
} else if(b.toString().replace(/,/gi,"").indexOf(a.toString().replace(/,/gi,"")) > -1) {
console.log("A is nested inside of B!")
}
}
checkIfArrayIsNestedInsideAnother(arr1, arr2)
//expected output: A is nested inside of B! because:
//arr1 (a): [1,2,"foo",3,"bar",3.14]
//arr2 (b): [1,2,"foo",3,"bar",3.14, "Lorem"]
//We added Lorem at line 15.
//Now, let's check if arr2 is nested inside arr1, which it is not.
checkIfArrayIsNestedInsideAnother(arr2, arr1)
//expected output: B is nested inside of A!
Solution 57 - Javascript
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; .
Solution 58 - Javascript
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;
}
Solution 59 - Javascript
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.
Solution 60 - Javascript
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
Solution 61 - Javascript
I would do like this:
[2,3,4,5] == [2,3,4,5].toString()
When you use the "==" operator, javascript check 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 use 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()
Solution 62 - Javascript
I have used : to join array and create a string to compare. for scenarios complex than this example you can use some other separator.
var a1 = [1,2,3];
var a2 = [1,2,3];
if (a1.length !== a2.length) {
console.log('a1 and a2 are not equal')
}else if(a1.join(':') === a2.join(':')){
console.log('a1 and a2 are equal')
}else{
console.log('a1 and a2 are not equal')
}
Solution 63 - Javascript
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
Solution 64 - Javascript
I needed something similar, comparing two arrays containing identifiers but in random order. In my case: "does this array contain at least one identifier from the other list?" The code is quite simple, using the reduce-function.
function hasFullOverlap(listA, listB){
return listA.reduce((allIdsAreFound, _id) => {
// We return true until an ID has not been found in the other list
return listB.includes(_id) && allIdsAreFound;
}, true);
}
if(hasFullOverlap(listA, listB) && hasFullOverlap(listB, listA)){
// Both lists contain all the values
}
Solution 65 - Javascript
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;
}
Solution 66 - Javascript
Additionally, I have converted Thomas' solution to order free comparison as I needed.
Array.prototype.equalsFreeOrder = function (array) {
var isThisElemExist;
if (!array)
return false;
if (this.length != array.length)
return false;
for (var i = 0; i < this.length; i++) {
isThisElemExist = false;
for (var k = 0; k < this.length; k++) {
if (this[i] instanceof Array && array[k] instanceof Array) {
if (this[i].equalsFreeOrder(array[k]))
isThisElemExist = true;
}
else if (this[i] == array[k]) {
isThisElemExist = true;
}
}
if (!isThisElemExist)
return false;
}
return true;
}
Solution 67 - Javascript
You can disqualify "sameness" if the number of elements do not match or if one of the elements is not in the other's array. Here is simple function that worked for me.
function isSame(arr1,arr2) {
var same=true;
for(var i=0;i < arr1.length;i++) {
if(!~jQuery.inArray(arr1[i],arr2) || arr1.length!=arr2.length){
same=false;
}
}
return same;
}
Solution 68 - Javascript
While the top answer to this question is correct and good, the code provided could use some improvement.
Below is my own code for comparing arrays and objects. The code is short and simple:
Array.prototype.equals = function(otherArray) {
if (!otherArray || this.length != otherArray.length) return false;
return this.reduce(function(equal, item, index) {
var otherItem = otherArray[index];
var itemType = typeof item, otherItemType = typeof otherItem;
if (itemType !== otherItemType) return false;
return equal && (itemType === "object" ? item.equals(otherItem) : item === otherItem);
}, true);
};
if(!Object.prototype.keys) {
Object.prototype.keys = function() {
var a = [];
for (var key in this) {
if (this.hasOwnProperty(key)) a.push(key);
}
return a;
}
Object.defineProperty(Object.prototype, "keys", {enumerable: false});
}
Object.prototype.equals = function(otherObject) {
if (!otherObject) return false;
var object = this, objectKeys = object.keys();
if (!objectKeys.equals(otherObject.keys())) return false;
return objectKeys.reduce(function(equal, key) {
var value = object[key], otherValue = otherObject[key];
var valueType = typeof value, otherValueType = typeof otherValue;
if (valueType !== otherValueType) return false;
// this will call Array.prototype.equals for arrays and Object.prototype.equals for objects
return equal && (valueType === "object" ? value.equals(otherValue) : value === otherValue);
}, true);
}
Object.defineProperty(Object.prototype, "equals", {enumerable: false});
This code supports arrays nested in objects and objects nested in arrays.
You can see a full suite of tests and test the code yourself at this repl: https://repl.it/Esfz/3
Solution 69 - Javascript
I quite like this approach in that it is substantially more succinct than others. It essentially contrasts all items to an accumulator which maintains a same value which is replaced with NaN if it reaches one that is distinct. As NaN cannot be equal to any value, including NaN itself, the value would be converted into a boolean (!!) and be false. Otherwise, the value should be true. To prevent an array of zeros to return false, the expression is converted to its absolute value and added to 1, thus !!(Math.abs(0) + 1) would be true. The absolute value was added for the case -1, which, when added to 1 would be equal to 0 and so, false.
function areArrayItemsEqual(arr) {
return !!(Math.abs(arr.reduce((a, b) => a === b ? b : NaN)) + 1);
}
Solution 70 - Javascript
If you want to compare two arrays and check if any object is same in both arrays it will works. Example :
Array1 = [a,b,c,d] Array2 = [d,e,f,g]
Here, 'd' is common in both array so this function will return true value.
cehckArray(array1, array2) {
for (let i = 0; i < array1.length; i++) {
for (let j = 0; j < array2.length; j++) {
if (array1[i] === array2[j]) {
return true;
}
}
}
// Return if no common element exist
return false;
}
Solution 71 - Javascript
I came up with another way to do it. Use join('') to change them to string, and then compare 2 strings:
var a1_str = a1.join(''),
a2_str = a2.join('');
if (a2_str === a1_str) {}
Solution 72 - Javascript
function palindrome(text)
{
var Res1 = new Array();
var Res2 = new Array();
for (i = 0; i < text.length; i++)
{
Res1[i] = text.substr(i, 1);
}
j=0;
for (k = (text.length-1); k>=0; k--)
{
Res2[j] = text.substr(k, 1);
j=j+1;
}
if(JSON.stringify(Res1)==JSON.stringify(Res2)){
return true;
}else{
return false;
}
}
document.write(palindrome("katak"));
Solution 73 - Javascript
If you are writing a test code, then
import chai from 'chai';
const arr1 = [2, 1];
const arr2 = [2, 1];
chai.expect(arr1).to.eql(arr2); // Will pass. `eql` is data compare instead of object compare.
Solution 74 - Javascript
var er = [{id:"23",name:"23222"}, {id:"222",name:"23222222"}];
var er2 = [{id:"23",name:"23222"}, {id:"222",name:"23222222"}];
var result = (JSON.stringify(er) == JSON.stringify(er2)); // true
It works json objects well if the order of the property of each entry is not changed.
var er = [{name:"23222",id:"23"}, {id:"222",name:"23222222"}];
var er2 = [{id:"23",name:"23222"}, {id:"222",name:"23222222"}];
var result = (JSON.stringify(er) == JSON.stringify(er2)); // false
But there is only one property or value in each entry of the array, this will work fine.
Solution 75 - Javascript
I use this code with no issues so far:
if(a.join() == b.join())
...
It works even if there are commas in an item.