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?
This question is related to
javascript
arrays
json
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
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
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
}
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.
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)
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);
}
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!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
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" ]]],
));;
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.
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.
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;
}
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
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;
}
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;
}
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)
})
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.
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
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"]]]
));;
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.
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
Récursive cmp function working with number/string/array/object
<script>_x000D_
var cmp = function(element, target){_x000D_
_x000D_
if(typeof element !== typeof target)_x000D_
{_x000D_
return false;_x000D_
}_x000D_
else if(typeof element === "object" && (!target || !element))_x000D_
{_x000D_
return target === element;_x000D_
}_x000D_
else if(typeof element === "object")_x000D_
{_x000D_
var keys_element = Object.keys(element);_x000D_
var keys_target = Object.keys(target);_x000D_
_x000D_
if(keys_element.length !== keys_target.length)_x000D_
{_x000D_
return false;_x000D_
}_x000D_
else_x000D_
{_x000D_
for(var i = 0; i < keys_element.length; i++)_x000D_
{_x000D_
if(keys_element[i] !== keys_target[i])_x000D_
return false;_x000D_
if(!cmp(element[keys_element[i]], target[keys_target[i]]))_x000D_
return false;_x000D_
}_x000D_
return true;_x000D_
}_x000D_
}_x000D_
else_x000D_
{_x000D_
return element === target;_x000D_
_x000D_
}_x000D_
};_x000D_
_x000D_
console.log(cmp({_x000D_
key1: 3,_x000D_
key2: "string",_x000D_
key3: [4, "45", {key4: [5, "6", false, null, {v:1}]}]_x000D_
}, {_x000D_
key1: 3,_x000D_
key2: "string",_x000D_
key3: [4, "45", {key4: [5, "6", false, null, {v:1}]}]_x000D_
})); // true_x000D_
_x000D_
console.log(cmp({_x000D_
key1: 3,_x000D_
key2: "string",_x000D_
key3: [4, "45", {key4: [5, "6", false, null, {v:1}]}]_x000D_
}, {_x000D_
key1: 3,_x000D_
key2: "string",_x000D_
key3: [4, "45", {key4: [5, "6", undefined, null, {v:1}]}]_x000D_
})); // false_x000D_
</script>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;
}
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')
}
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;
}
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
How about using 'includes' with 'forEach'?
var array1 = [1,2,3];
var array2 = [1,2,4];
var array_diff = 0;
// loop through each element in first array
array1.forEach(function(elem){
// check if second array contains it, return false if not
if(!array2.includes(elem)){
array_diff = 1;
return false;
}else{
array_diff = 0;
}
});
// loop through each element in second array
array2.forEach(function(elem){
// check if first array contains it, return false if not
if(!array1.includes(elem)){
array_diff = 1;
return false;
}else{
array_diff = 0;
}
});
// variable array_diff determines the result
if(array_diff == 1){
alert('Arrays are different');
}else{
alert('Arrays are same');
}
Note: This should work on both sorted and unsorted arrays
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;
}
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/
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) {}
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
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.
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;
}
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!
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
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"].
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>_x000D_
let corrette = [1];_x000D_
let risposte = [];_x000D_
_x000D_
function areEqual(a, b){_x000D_
let x = 0;_x000D_
for (n in a){_x000D_
if (a[n] == b[n]){x = 1;}_x000D_
else {x = 0};_x000D_
}_x000D_
if (x){console.log("The 2 arrays are equal")}_x000D_
return x;_x000D_
}_x000D_
_x000D_
</script>_x000D_
_x000D_
Apporto di attrezzatura per cucina da parte del proprietario_x000D_
<button onclick="risposte[0]=1">Capitale proprio</button>_x000D_
<button onclick="risposte[0]=0">Capitale di debito</button>_x000D_
<button onclick="risposte[0]=0">Debito commerciale</button>_x000D_
_x000D_
<br><hr>_x000D_
<button onclick="if(areEqual(corrette,risposte)){controlla.innerHTML='Esatto'}else{controlla.innerHTML='No'}">Controlla le risposte</button>_x000D_
<div id="controlla"></div>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.
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.
=)
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.
.lengthcomparison has been removed for a common benchmark (adda1.length === a2.lengthto 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.
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());
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});
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.
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.
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).
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
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.
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
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
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.
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.
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;
}
});
}
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.
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;
}
You can simply use isEqual from lodash library. It is very efficient and clean.
import {isEqual} from "lodash";
const isTwoArraysEqual = isEqual(array1, array2);
I use this code with no issues so far:
if(a.join() == b.join())
...
It works even if there are commas in an item.
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
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;
}
}
hope it will help someone.
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
}
An alternative way using filter and arrow functions
arrOne.length === arrTwo.length && arrOne.filter((currVal, idx) => currVal !== arrTwo[idx]).length === 0
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
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;
}
Hope this helps you or anyone else searching for an answer.
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"));
JSON.stringify(collectionNames).includes(JSON.stringify(sourceNames)) ? array.push(collection[i]) : null
This is how i did it.
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
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()
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;
}
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;
};
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
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;
}
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]))
})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
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];_x000D_
var a2 = [1,2,3];_x000D_
const arraysAreEqual = a1.sort().toString()==a2.sort().toString();_x000D_
// true if both arrays have same elements else false_x000D_
console.log(arraysAreEqual);Source: Stackoverflow.com