Is there a way to return the difference between two arrays in JavaScript?

For example:

```
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
```

Any advice greatly appreciated.

Is there a way to return the difference between two arrays in JavaScript?

For example:

```
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
```

Any advice greatly appreciated.

javascript
arrays
array-difference
93
0
John Adawan
2020-03-17 12:30:25 +0000 UTC

I assume you are comparing a normal array. If not, you need to change the *for* loop to a *for .. in* loop.

```
function arr_diff (a1, a2) {
var a = [], diff = [];
for (var i = 0; i < a1.length; i++) {
a[a1[i]] = true;
}
for (var i = 0; i < a2.length; i++) {
if (a[a2[i]]) {
delete a[a2[i]];
} else {
a[a2[i]] = true;
}
}
for (var k in a) {
diff.push(k);
}
return diff;
}
console.log(arr_diff(['a', 'b'], ['a', 'b', 'c', 'd']));
console.log(arr_diff("abcd", "abcde"));
console.log(arr_diff("zxc", "zxc"));
```

A better solution, if you don't care about backward compatibility, is using filter. But still, this solution works.

There is a better way using ES7:

**Intersection**

```
let intersection = arr1.filter(x => arr2.includes(x));
```

For `[1,2,3] [2,3]`

it will yield `[2,3]`

. On the other hand, for `[1,2,3] [2,3,5]`

will return the same thing.

**Difference**

```
let difference = arr1.filter(x => !arr2.includes(x));
```

For `[1,2,3] [2,3]`

it will yield `[1]`

. On the other hand, for `[1,2,3] [2,3,5]`

will return the same thing.

For a **symmetric difference**, you can do:

```
let difference = arr1
.filter(x => !arr2.includes(x))
.concat(arr2.filter(x => !arr1.includes(x)));
```

This way, you will get an array containing all the elements of arr1 that are not in arr2 and vice-versa

As @Joshaven Potter pointed out on his answer, you can add this to Array.prototype so it can be used like this:

```
Array.prototype.diff = function(arr2) { return this.filter(x => arr2.includes(x)); }
[1, 2, 3].diff([2, 3])
```

```
Array.prototype.diff = function(a) {
return this.filter(function(i) {return a.indexOf(i) < 0;});
};
////////////////////
// Examples
////////////////////
[1,2,3,4,5,6].diff( [3,4,5] );
// => [1, 2, 6]
["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
// => ["test5", "test6"]
```

```
Array.prototype.diff = function(a) {
return this.filter(function(i) {return a.indexOf(i) < 0;});
};
////////////////////
// Examples
////////////////////
var dif1 = [1,2,3,4,5,6].diff( [3,4,5] );
console.log(dif1); // => [1, 2, 6]
var dif2 = ["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
console.log(dif2); // => ["test5", "test6"]
```

**Note** indexOf and filter are not available in ie before ie9.

This is by far the easiest way to get exactly the result you are looking for, using jQuery:

```
var diff = $(old_array).not(new_array).get();
```

`diff`

now contains what was in `old_array`

that is not in `new_array`

The difference method in Underscore (or its drop-in replacement, Lo-Dash) can do this too:

```
(R)eturns the values from array that are not present in the other arrays
_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]
```

As with any Underscore function, you could also use it in a more object-oriented style:

```
_([1, 2, 3, 4, 5]).difference([5, 2, 10]);
```

There are two possible intepretations for "difference". I'll let you choose which one you want. Say you have:

```
var a1 = ['a', 'b' ];
var a2 = [ 'b', 'c'];
```

If you want to get

`['a']`

, use this function:`function difference(a1, a2) { var result = []; for (var i = 0; i < a1.length; i++) { if (a2.indexOf(a1[i]) === -1) { result.push(a1[i]); } } return result; }`

If you want to get

`['a', 'c']`

(all elements contained in*either*`a1`

or`a2`

, but not both -- the so-called*symmetric difference*), use this function:`function symmetricDifference(a1, a2) { var result = []; for (var i = 0; i < a1.length; i++) { if (a2.indexOf(a1[i]) === -1) { result.push(a1[i]); } } for (i = 0; i < a2.length; i++) { if (a1.indexOf(a2[i]) === -1) { result.push(a2[i]); } } return result; }`

If you are using lodash, you can use `_.difference(a1, a2)`

(case 1 above) or `_.xor(a1, a2)`

(case 2).

If you are using Underscore.js, you can use the `_.difference(a1, a2)`

function for case 1.

The code above works on all browsers. However, for large arrays of more than about 10,000 items, it becomes quite slow, because it has O(n²) complexity. On many modern browsers, we can take advantage of the ES6 `Set`

object to speed things up. Lodash automatically uses `Set`

when it's available. If you are not using lodash, use the following implementation, inspired by Axel Rauschmayer's blog post:

```
function difference(a1, a2) {
var a2Set = new Set(a2);
return a1.filter(function(x) { return !a2Set.has(x); });
}
function symmetricDifference(a1, a2) {
return difference(a1, a2).concat(difference(a2, a1));
}
```

The behavior for all examples may be surprising or non-obvious if you care about -0, +0, NaN or sparse arrays. (For most uses, this doesn't matter.)

To get the **symmetric difference** you need to compare the arrays in both ways (or in all the ways in case of multiple arrays)

```
// diff between just two arrays:
function arrayDiff(a, b) {
return [
...a.filter(x => !b.includes(x)),
...b.filter(x => !a.includes(x))
];
}
// diff between multiple arrays:
function arrayDiff(...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter(x => !unique.includes(x));
}));
}
```

```
// diff between just two arrays:
function arrayDiff(a, b) {
return [
...a.filter(x => b.indexOf(x) === -1),
...b.filter(x => a.indexOf(x) === -1)
];
}
// diff between multiple arrays:
function arrayDiff(...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter(x => unique.indexOf(x) === -1);
}));
}
```

```
// diff between just two arrays:
function arrayDiff(a, b) {
var arrays = Array.prototype.slice.call(arguments);
var diff = [];
arrays.forEach(function(arr, i) {
var other = i === 1 ? a : b;
arr.forEach(function(x) {
if (other.indexOf(x) === -1) {
diff.push(x);
}
});
})
return diff;
}
// diff between multiple arrays:
function arrayDiff() {
var arrays = Array.prototype.slice.call(arguments);
var diff = [];
arrays.forEach(function(arr, i) {
var others = arrays.slice(0);
others.splice(i, 1);
var otherValues = Array.prototype.concat.apply([], others);
var unique = otherValues.filter(function (x, j) {
return otherValues.indexOf(x) === j;
});
diff = diff.concat(arr.filter(x => unique.indexOf(x) === -1));
});
return diff;
}
```

**Example:**

```
// diff between two arrays:
const a = ['a', 'd', 'e'];
const b = ['a', 'b', 'c', 'd'];
arrayDiff(a, b); // (3) ["e", "b", "c"]
// diff between multiple arrays
const a = ['b', 'c', 'd', 'e', 'g'];
const b = ['a', 'b'];
const c = ['a', 'e', 'f'];
arrayDiff(a, b, c); // (4) ["c", "d", "g", "f"]
```

```
function arrayDiffByKey(key, ...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter( x =>
!unique.some(y => x[key] === y[key])
);
}));
}
```

**Example:**

```
const a = [{k:1}, {k:2}, {k:3}];
const b = [{k:1}, {k:4}, {k:5}, {k:6}];
const c = [{k:3}, {k:5}, {k:7}];
arrayDiffByKey('k', a, b, c); // (4) [{k:2}, {k:4}, {k:6}, {k:7}]
```

You could use a Set in this case. It is optimized for this kind of operation (union, intersection, difference).

Make sure it applies to your case, once it allows no duplicates.

```
var a = new JS.Set([1,2,3,4,5,6,7,8,9]);
var b = new JS.Set([2,4,6,8]);
a.difference(b)
// -> Set{1,3,5,7,9}
```

```
function diff(a1, a2) {
return a1.concat(a2).filter(function(val, index, arr){
return arr.indexOf(val) === arr.lastIndexOf(val);
});
}
```

Merge both the arrays, unique values will appear only once so indexOf() will be the same as lastIndexOf().

to subtract one array from another, simply use the snippet below:

```
var a1 = ['1','2','3','4','6'];
var a2 = ['3','4','5'];
var items = new Array();
items = jQuery.grep(a1,function (item) {
return jQuery.inArray(item, a2) < 0;
});
```

It will returns ['1,'2','6'] that are items of first array which don't exist in the second.

Therefore, according to your problem sample, following code is the exact solution:

```
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var _array = new Array();
_array = jQuery.grep(array2, function (item) {
return jQuery.inArray(item, array1) < 0;
});
```

A cleaner approach in ES6 is the following solution.

```
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
```

```
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
```

```
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
```

```
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]
```

Computing the `difference`

between two arrays is one of the `Set`

operations. The term already indicates that the native `Set`

type should be used, in order to increase the lookup speed. Anyway, there are three permutations when you compute the difference between two sets:

```
[+left difference] [-intersection] [-right difference]
[-left difference] [-intersection] [+right difference]
[+left difference] [-intersection] [+right difference]
```

Here is a functional solution that reflects these permutations.

`difference`

:```
// small, reusable auxiliary functions
const apply = f => x => f(x);
const flip = f => y => x => f(x) (y);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
// left difference
const differencel = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? false
: true
) (xs);
};
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// run the computation
console.log( differencel(xs) (ys) );
```

`difference`

:`differencer`

is trivial. It is just `differencel`

with flipped arguments. You can write a function for convenience: `const differencer = flip(differencel)`

. That's all!

`difference`

:Now that we have the left and right one, implementing the symmetric `difference`

gets trivial as well:

```
// small, reusable auxiliary functions
const apply = f => x => f(x);
const flip = f => y => x => f(x) (y);
const concat = y => xs => xs.concat(y);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
// left difference
const differencel = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? false
: true
) (xs);
};
// symmetric difference
const difference = ys => xs =>
concat(differencel(xs) (ys)) (flip(differencel) (xs) (ys));
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// run the computation
console.log( difference(xs) (ys) );
```

I guess this example is a good starting point to obtain an impression what functional programming means:

**Programming with building blocks that can be plugged together in many different ways.**

A solution using `indexOf()`

will be ok for small arrays but as they grow in length the performance of the algorithm approaches `O(n^2)`

. Here's a solution that will perform better for very large arrays by using objects as associative arrays to store the array entries as keys; it also eliminates duplicate entries automatically but only works with string values (or values which can be safely stored as strings):

```
function arrayDiff(a1, a2) {
var o1={}, o2={}, diff=[], i, len, k;
for (i=0, len=a1.length; i<len; i++) { o1[a1[i]] = true; }
for (i=0, len=a2.length; i<len; i++) { o2[a2[i]] = true; }
for (k in o1) { if (!(k in o2)) { diff.push(k); } }
for (k in o2) { if (!(k in o1)) { diff.push(k); } }
return diff;
}
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
arrayDiff(a1, a2); // => ['c', 'd']
arrayDiff(a2, a1); // => ['c', 'd']
```

With the arrival of ES6 with sets and splat operator (at the time of being works only in Firefox, check compatibility table), you can write the following one liner:

```
var a = ['a', 'b', 'c', 'd'];
var b = ['a', 'b'];
var b1 = new Set(b);
var difference = [...new Set([...a].filter(x => !b1.has(x)))];
```

which will result in `[ "c", "d" ]`

.

The above answer by Joshaven Potter is great. But it returns elements in array B that are not in array C, but not the other way around. For example, if `var a=[1,2,3,4,5,6].diff( [3,4,5,7]);`

then it will output: ==> `[1,2,6]`

, but **not** `[1,2,6,7]`

, which is the actual difference between the two. You can still use Potter's code above but simply redo the comparison once backwards too:

```
Array.prototype.diff = function(a) {
return this.filter(function(i) {return !(a.indexOf(i) > -1);});
};
////////////////////
// Examples
////////////////////
var a=[1,2,3,4,5,6].diff( [3,4,5,7]);
var b=[3,4,5,7].diff([1,2,3,4,5,6]);
var c=a.concat(b);
console.log(c);
```

This should output: `[ 1, 2, 6, 7 ]`

Another way to solve the problem

```
function diffArray(arr1, arr2) {
return arr1.concat(arr2).filter(function (val) {
if (!(arr1.includes(val) && arr2.includes(val)))
return val;
});
}
diffArray([1, 2, 3, 7], [3, 2, 1, 4, 5]); // return [7, 4, 5]
```

Also, you can use arrow function syntax:

```
const diffArray = (arr1, arr2) => arr1.concat(arr2)
.filter(val => !(arr1.includes(val) && arr2.includes(val)));
diffArray([1, 2, 3, 7], [3, 2, 1, 4, 5]); // return [7, 4, 5]
```

```
Array.prototype.difference = function(e) {
return this.filter(function(i) {return e.indexOf(i) < 0;});
};
eg:-
[1,2,3,4,5,6,7].difference( [3,4,5] );
=> [1, 2, 6 , 7]
```

How about this:

```
Array.prototype.contains = function(needle){
for (var i=0; i<this.length; i++)
if (this[i] == needle) return true;
return false;
}
Array.prototype.diff = function(compare) {
return this.filter(function(elem) {return !compare.contains(elem);})
}
var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));
```

So this way you can do `array1.diff(array2)`

to get their difference (Horrible time complexity for the algorithm though - O(array1.length x array2.length) I believe)

Very Simple Solution with the filter function of JavaScript:

```
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
function diffArray(arr1, arr2) {
var newArr = [];
var myArr = arr1.concat(arr2);
newArr = myArr.filter(function(item){
return arr2.indexOf(item) < 0 || arr1.indexOf(item) < 0;
});
alert(newArr);
}
diffArray(a1, a2);
```

Using http://phrogz.net/JS/ArraySetMath.js you can:

```
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var array3 = array2.subtract( array1 );
// ["test5", "test6"]
var array4 = array1.exclusion( array2 );
// ["test5", "test6"]
```

```
function diffArray(arr1, arr2) {
var newArr = arr1.concat(arr2);
return newArr.filter(function(i){
return newArr.indexOf(i) == newArr.lastIndexOf(i);
});
}
```

this is works for me

- Pure JavaScript solution (no libraries)
- Compatible with older browsers (doesn't use
`filter`

) - O(n^2)
- Optional
`fn`

callback parameter that lets you specify how to compare array items

```
function diff(a, b, fn){
var max = Math.max(a.length, b.length);
d = [];
fn = typeof fn === 'function' ? fn : false
for(var i=0; i < max; i++){
var ac = i < a.length ? a[i] : undefined
bc = i < b.length ? b[i] : undefined;
for(var k=0; k < max; k++){
ac = ac === undefined || (k < b.length && (fn ? fn(ac, b[k]) : ac == b[k])) ? undefined : ac;
bc = bc === undefined || (k < a.length && (fn ? fn(bc, a[k]) : bc == a[k])) ? undefined : bc;
if(ac == undefined && bc == undefined) break;
}
ac !== undefined && d.push(ac);
bc !== undefined && d.push(bc);
}
return d;
}
alert(
"Test 1: " +
diff(
[1, 2, 3, 4],
[1, 4, 5, 6, 7]
).join(', ') +
"\nTest 2: " +
diff(
[{id:'a',toString:function(){return this.id}},{id:'b',toString:function(){return this.id}},{id:'c',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}],
[{id:'a',toString:function(){return this.id}},{id:'e',toString:function(){return this.id}},{id:'f',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}],
function(a, b){ return a.id == b.id; }
).join(', ')
);
```

This is working: basically merge the two arrays, look for the duplicates and push what is not duplicated into a new array which is the difference.

```
function diff(arr1, arr2) {
var newArr = [];
var arr = arr1.concat(arr2);
for (var i in arr){
var f = arr[i];
var t = 0;
for (j=0; j<arr.length; j++){
if(arr[j] === f){
t++;
}
}
if (t === 1){
newArr.push(f);
}
}
return newArr;
}
```

**Symmetric** and **linear complexity**. Requires ES6.

```
function arrDiff(arr1, arr2) {
var arrays = [arr1, arr2].sort((a, b) => a.length - b.length);
var smallSet = new Set(arrays[0]);
return arrays[1].filter(x => !smallSet.has(x));
}
```

Just thinking... for the sake of a challenge ;-) would this work... (for basic arrays of strings, numbers, etc.) no nested arrays

```
function diffArrays(arr1, arr2, returnUnion){
var ret = [];
var test = {};
var bigArray, smallArray, key;
if(arr1.length >= arr2.length){
bigArray = arr1;
smallArray = arr2;
} else {
bigArray = arr2;
smallArray = arr1;
}
for(var i=0;i<bigArray.length;i++){
key = bigArray[i];
test[key] = true;
}
if(!returnUnion){
//diffing
for(var i=0;i<smallArray.length;i++){
key = smallArray[i];
if(!test[key]){
test[key] = null;
}
}
} else {
//union
for(var i=0;i<smallArray.length;i++){
key = smallArray[i];
if(!test[key]){
test[key] = true;
}
}
}
for(var i in test){
ret.push(i);
}
return ret;
}
array1 = "test1", "test2","test3", "test4", "test7"
array2 = "test1", "test2","test3","test4", "test5", "test6"
diffArray = diffArrays(array1, array2);
//returns ["test5","test6","test7"]
diffArray = diffArrays(array1, array2, true);
//returns ["test1", "test2","test3","test4", "test5", "test6","test7"]
```

Note the sorting will likely not be as noted above... but if desired, call .sort() on the array to sort it.

I wanted a similar function which took in an old array and a new array and gave me an array of added items and an array of removed items, and I wanted it to be efficient (so no .contains!).

You can play with my proposed solution here: http://jsbin.com/osewu3/12.

Can anyone see any problems/improvements to that algorithm? Thanks!

Code listing:

```
function diff(o, n) {
// deal with empty lists
if (o == undefined) o = [];
if (n == undefined) n = [];
// sort both arrays (or this won't work)
o.sort(); n.sort();
// don't compare if either list is empty
if (o.length == 0 || n.length == 0) return {added: n, removed: o};
// declare temporary variables
var op = 0; var np = 0;
var a = []; var r = [];
// compare arrays and add to add or remove lists
while (op < o.length && np < n.length) {
if (o[op] < n[np]) {
// push to diff?
r.push(o[op]);
op++;
}
else if (o[op] > n[np]) {
// push to diff?
a.push(n[np]);
np++;
}
else {
op++;np++;
}
}
// add remaining items
if( np < n.length )
a = a.concat(n.slice(np, n.length));
if( op < o.length )
r = r.concat(o.slice(op, o.length));
return {added: a, removed: r};
}
```

I was looking for a simple answer that didn't involve using different libraries, and I came up with my own that I don't think has been mentioned here. I don't know how efficient it is or anything but it works;

```
function find_diff(arr1, arr2) {
diff = [];
joined = arr1.concat(arr2);
for( i = 0; i <= joined.length; i++ ) {
current = joined[i];
if( joined.indexOf(current) == joined.lastIndexOf(current) ) {
diff.push(current);
}
}
return diff;
}
```

For my code I need duplicates taken out as well, but I guess that isn't always preferred.

I guess the main downside is it's potentially comparing many options that have already been rejected.

littlebit fix for the best answer

```
function arr_diff(a1, a2)
{
var a=[], diff=[];
for(var i=0;i<a1.length;i++)
a[a1[i]]=a1[i];
for(var i=0;i<a2.length;i++)
if(a[a2[i]]) delete a[a2[i]];
else a[a2[i]]=a2[i];
for(var k in a)
diff.push(a[k]);
return diff;
}
```

this will take current type of element in consideration. b/c when we make a[a1[i]] it converts a value to string from its oroginal value, so we lost actual value.

You can use underscore.js : http://underscorejs.org/#intersection

You have needed methods for array :

```
_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]
_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2]
```

//es6 approach

```
function diff(a, b) {
var u = a.slice(); //dup the array
b.map(e => {
if (u.indexOf(e) > -1) delete u[u.indexOf(e)]
else u.push(e) //add non existing item to temp array
})
return u.filter((x) => {return (x != null)}) //flatten result
}
```

Use extra memory to do this. That way you can solve it with less time complexity, O(n) instead of o(n*n).

```
function getDiff(arr1,arr2){
let k = {};
let diff = []
arr1.map(i=>{
if (!k.hasOwnProperty(i)) {
k[i] = 1
}
}
)
arr2.map(j=>{
if (!k.hasOwnProperty(j)) {
k[j] = 1;
} else {
k[j] = 2;
}
}
)
for (var i in k) {
if (k[i] === 1)
diff.push(+i)
}
return diff
}
getDiff([4, 3, 52, 3, 5, 67, 9, 3],[4, 5, 6, 75, 3, 334, 5, 5, 6])
```