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Get non unique values from two arrays [duplicate]

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"]

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 //
//////////////
const dif1 = [1,2,3,4,5,6].diff( [3,4,5] );
console.log(dif1); // => [1, 2, 6]
const dif2 = ["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
console.log(dif2); // => ["test5", "test6"]
Note .indexOf() and .filter() are not available before IE9.

This answer was written in 2009, so it is a bit outdated, also it's rather educational for understanding the problem. Best solution I'd use today would be
let difference = arr1.filter(x => !arr2.includes(x));
(credits to other author here)
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"));

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]);

Plain JavaScript
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;
}
Lodash / Underscore
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.
ES6 Set, for very large arrays
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));
}
Notes
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.)

A cleaner approach in ES6 is the following solution.
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
Difference
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
Intersection
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
Disjunctive Union (Symmetric Difference)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]

To get the symmetric difference you need to compare the arrays in both ways (or in all the ways in case of multiple arrays)
ES7 (ECMAScript 2016)
// 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));
}));
}
ES6 (ECMAScript 2015)
// 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);
}));
}
ES5 (ECMAScript 5.1)
// 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"]
Difference between Arrays of Objects
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}

One Liners
const unique = (a) => [...new Set(a)];
const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{}));
const intersection = (a, b) => a.filter((v) => b.includes(v));
const diff = (a, b) => a.filter((v) => !b.includes(v));
const symDiff = (a, b) => diff(a, b).concat(diff(b, a));
const union = (a, b) => diff(a, b).concat(b);
const a = unique([1, 2, 3, 4, 5, 5]);
console.log(a);
const b = [4, 5, 6, 7, 8];
console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b));
console.log(uniqueBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
{ id: 1, name: "abc" },
],
(v) => v.id
));
const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u)));
console.log(intersectionBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u)));
console.log(diffBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
TypeScript
playground link
const unique = <T>(array: T[]) => [...new Set(array)];
const intersection = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => array2.includes(v));
const diff = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => !array2.includes(v));
const symDiff = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(diff(array2, array1));
const union = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(array2);
const intersectionBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));
const diffBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));
const uniqueBy = <T>(
array: T[],
predicate: (v: T, i: number, a: T[]) => string
) =>
Object.values(
array.reduce((acc, value, index) => {
acc[predicate(value, index, array)] = value;
return acc;
}, {} as { [key: string]: T })
);

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().

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" ].

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;
});

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]

Functional approach with ES2015
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.
Left 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) );
Right 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!
Symmetric 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']

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 ]

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);

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)

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

If you have two list of objects
const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]
let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))

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"]

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(', ')
);

To find the difference of 2 arrays without duplicates:
function difference(arr1, arr2){
let setA = new Set(arr1);
let differenceSet = new Set(arr2.filter(ele => !setA.has(ele)));
return [...differenceSet ];
}
1.difference([2,2,3,4],[2,3,3,4]) will return []
2.difference([1,2,3],[4,5,6]) will return [4,5,6]
3.difference([1,2,3,4],[1,2]) will return []
4.difference([1,2],[1,2,3,4]) will return [3,4]
Note: The above solution requires that you always send the larger array as the second parameter. To find the absolute difference, you will need to first find the larger array of the two and then work on them.
To find the absolute difference of 2 arrays without duplicates:
function absDifference(arr1, arr2){
const {larger, smaller} = arr1.length > arr2.length ?
{larger: arr1, smaller: arr2} : {larger: arr2, smaller: arr1}
let setA = new Set(smaller);
let absDifferenceSet = new Set(larger.filter(ele => !setA.has(ele)));
return [...absDifferenceSet ];
}
1.absDifference([2,2,3,4],[2,3,3,4]) will return []
2.absDifference([1,2,3],[4,5,6]) will return [4,5,6]
3.absDifference([1,2,3,4],[1,2]) will return [3,4]
4.absDifference([1,2],[1,2,3,4]) will return [3,4]
Note the example 3 from both the solutions

Here is another solution that can return the differences, just like git diff: (it has been written in typescript, if you're not using typescript version, just remove the types)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* #param { T[] } from
* #param { T[] } to
* #returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
Here is a sample of usage:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}

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};
}

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]

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;
}

//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
}

Typescript - Difference between 2 array of objects [duplicate]

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"]

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 //
//////////////
const dif1 = [1,2,3,4,5,6].diff( [3,4,5] );
console.log(dif1); // => [1, 2, 6]
const dif2 = ["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
console.log(dif2); // => ["test5", "test6"]
Note .indexOf() and .filter() are not available before IE9.

This answer was written in 2009, so it is a bit outdated, also it's rather educational for understanding the problem. Best solution I'd use today would be
let difference = arr1.filter(x => !arr2.includes(x));
(credits to other author here)
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"));

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]);

Plain JavaScript
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;
}
Lodash / Underscore
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.
ES6 Set, for very large arrays
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));
}
Notes
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.)

A cleaner approach in ES6 is the following solution.
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
Difference
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
Intersection
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
Disjunctive Union (Symmetric Difference)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]

To get the symmetric difference you need to compare the arrays in both ways (or in all the ways in case of multiple arrays)
ES7 (ECMAScript 2016)
// 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));
}));
}
ES6 (ECMAScript 2015)
// 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);
}));
}
ES5 (ECMAScript 5.1)
// 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"]
Difference between Arrays of Objects
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}

One Liners
const unique = (a) => [...new Set(a)];
const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{}));
const intersection = (a, b) => a.filter((v) => b.includes(v));
const diff = (a, b) => a.filter((v) => !b.includes(v));
const symDiff = (a, b) => diff(a, b).concat(diff(b, a));
const union = (a, b) => diff(a, b).concat(b);
const a = unique([1, 2, 3, 4, 5, 5]);
console.log(a);
const b = [4, 5, 6, 7, 8];
console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b));
console.log(uniqueBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
{ id: 1, name: "abc" },
],
(v) => v.id
));
const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u)));
console.log(intersectionBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u)));
console.log(diffBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
TypeScript
playground link
const unique = <T>(array: T[]) => [...new Set(array)];
const intersection = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => array2.includes(v));
const diff = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => !array2.includes(v));
const symDiff = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(diff(array2, array1));
const union = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(array2);
const intersectionBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));
const diffBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));
const uniqueBy = <T>(
array: T[],
predicate: (v: T, i: number, a: T[]) => string
) =>
Object.values(
array.reduce((acc, value, index) => {
acc[predicate(value, index, array)] = value;
return acc;
}, {} as { [key: string]: T })
);

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().

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" ].

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;
});

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]

Functional approach with ES2015
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.
Left 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) );
Right 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!
Symmetric 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']

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 ]

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);

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)

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

If you have two list of objects
const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]
let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))

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"]

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(', ')
);

To find the difference of 2 arrays without duplicates:
function difference(arr1, arr2){
let setA = new Set(arr1);
let differenceSet = new Set(arr2.filter(ele => !setA.has(ele)));
return [...differenceSet ];
}
1.difference([2,2,3,4],[2,3,3,4]) will return []
2.difference([1,2,3],[4,5,6]) will return [4,5,6]
3.difference([1,2,3,4],[1,2]) will return []
4.difference([1,2],[1,2,3,4]) will return [3,4]
Note: The above solution requires that you always send the larger array as the second parameter. To find the absolute difference, you will need to first find the larger array of the two and then work on them.
To find the absolute difference of 2 arrays without duplicates:
function absDifference(arr1, arr2){
const {larger, smaller} = arr1.length > arr2.length ?
{larger: arr1, smaller: arr2} : {larger: arr2, smaller: arr1}
let setA = new Set(smaller);
let absDifferenceSet = new Set(larger.filter(ele => !setA.has(ele)));
return [...absDifferenceSet ];
}
1.absDifference([2,2,3,4],[2,3,3,4]) will return []
2.absDifference([1,2,3],[4,5,6]) will return [4,5,6]
3.absDifference([1,2,3,4],[1,2]) will return [3,4]
4.absDifference([1,2],[1,2,3,4]) will return [3,4]
Note the example 3 from both the solutions

Here is another solution that can return the differences, just like git diff: (it has been written in typescript, if you're not using typescript version, just remove the types)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* #param { T[] } from
* #param { T[] } to
* #returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
Here is a sample of usage:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}

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};
}

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]

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;
}

//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
}

How to find each value of array in another array? [duplicate]

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"]

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 //
//////////////
const dif1 = [1,2,3,4,5,6].diff( [3,4,5] );
console.log(dif1); // => [1, 2, 6]
const dif2 = ["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
console.log(dif2); // => ["test5", "test6"]
Note .indexOf() and .filter() are not available before IE9.

This answer was written in 2009, so it is a bit outdated, also it's rather educational for understanding the problem. Best solution I'd use today would be
let difference = arr1.filter(x => !arr2.includes(x));
(credits to other author here)
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"));

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]);

Plain JavaScript
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;
}
Lodash / Underscore
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.
ES6 Set, for very large arrays
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));
}
Notes
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.)

A cleaner approach in ES6 is the following solution.
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
Difference
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
Intersection
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
Disjunctive Union (Symmetric Difference)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]

To get the symmetric difference you need to compare the arrays in both ways (or in all the ways in case of multiple arrays)
ES7 (ECMAScript 2016)
// 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));
}));
}
ES6 (ECMAScript 2015)
// 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);
}));
}
ES5 (ECMAScript 5.1)
// 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"]
Difference between Arrays of Objects
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}

One Liners
const unique = (a) => [...new Set(a)];
const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{}));
const intersection = (a, b) => a.filter((v) => b.includes(v));
const diff = (a, b) => a.filter((v) => !b.includes(v));
const symDiff = (a, b) => diff(a, b).concat(diff(b, a));
const union = (a, b) => diff(a, b).concat(b);
const a = unique([1, 2, 3, 4, 5, 5]);
console.log(a);
const b = [4, 5, 6, 7, 8];
console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b));
console.log(uniqueBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
{ id: 1, name: "abc" },
],
(v) => v.id
));
const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u)));
console.log(intersectionBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u)));
console.log(diffBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
TypeScript
playground link
const unique = <T>(array: T[]) => [...new Set(array)];
const intersection = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => array2.includes(v));
const diff = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => !array2.includes(v));
const symDiff = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(diff(array2, array1));
const union = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(array2);
const intersectionBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));
const diffBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));
const uniqueBy = <T>(
array: T[],
predicate: (v: T, i: number, a: T[]) => string
) =>
Object.values(
array.reduce((acc, value, index) => {
acc[predicate(value, index, array)] = value;
return acc;
}, {} as { [key: string]: T })
);

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().

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" ].

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;
});

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]

Functional approach with ES2015
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.
Left 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) );
Right 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!
Symmetric 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']

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 ]

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);

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)

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

If you have two list of objects
const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]
let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))

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"]

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(', ')
);

To find the difference of 2 arrays without duplicates:
function difference(arr1, arr2){
let setA = new Set(arr1);
let differenceSet = new Set(arr2.filter(ele => !setA.has(ele)));
return [...differenceSet ];
}
1.difference([2,2,3,4],[2,3,3,4]) will return []
2.difference([1,2,3],[4,5,6]) will return [4,5,6]
3.difference([1,2,3,4],[1,2]) will return []
4.difference([1,2],[1,2,3,4]) will return [3,4]
Note: The above solution requires that you always send the larger array as the second parameter. To find the absolute difference, you will need to first find the larger array of the two and then work on them.
To find the absolute difference of 2 arrays without duplicates:
function absDifference(arr1, arr2){
const {larger, smaller} = arr1.length > arr2.length ?
{larger: arr1, smaller: arr2} : {larger: arr2, smaller: arr1}
let setA = new Set(smaller);
let absDifferenceSet = new Set(larger.filter(ele => !setA.has(ele)));
return [...absDifferenceSet ];
}
1.absDifference([2,2,3,4],[2,3,3,4]) will return []
2.absDifference([1,2,3],[4,5,6]) will return [4,5,6]
3.absDifference([1,2,3,4],[1,2]) will return [3,4]
4.absDifference([1,2],[1,2,3,4]) will return [3,4]
Note the example 3 from both the solutions

Here is another solution that can return the differences, just like git diff: (it has been written in typescript, if you're not using typescript version, just remove the types)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* #param { T[] } from
* #param { T[] } to
* #returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
Here is a sample of usage:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}

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};
}

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]

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;
}

//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
}

How to calculate intersection of multiple arrays in JavaScript? And what does [equals: function] mean?

I am aware of this question, simplest code for array intersection but all the solutions presume the number of arrays is two, which cannot be certain in my case.
I have divs on a page with data that contains arrays. I want to find the values common to all arrays. I do not know how many divs/arrays I will have in advance. What is the best way to calculate values common to all arrays?
var array1 = ["Lorem", "ipsum", "dolor"];
var array2 = ["Lorem", "ipsum", "quick", "brown", "foo"];
var array3 = ["Jumps", "Over", "Lazy", "Lorem"];
var array4 = [1337, 420, 666, "Lorem"];
//Result should be ["Lorem"];
I found another solution elsewhere, using Underscore.js.
var arrayOfArrays = [[4234, 2323, 43], [1323, 43, 1313], [23, 34, 43]];
_.intersection.apply(_, arrayOfArrays)
//Result is [43]
I've tested this with simple dummy data at my end and it seems to work. But for some reason, some of the arrays I'm producing, which contain simple strings, also automatically include an added value, "equals: function":
["Dummy1", "Dummy2", "Dummy3", equals: function]
And whenever I use the Underscore.js intersection method, on an array of arrays, I always get [equals: function] in dev tools, and not - if "Dummy3" is common to all arrays - ["Dummy3"].
So TL;DR is there another solution to array intersection that would suit my case? And can anyone explain what [equals: function] means here? When I expand the item in the dev tools, it produces an empty array and a list of methods available on arrays (pop, push, shift etc), but these methods are all faded out, while equals: function is highlighted.

You could just use Array#reduce with Array#filter and Array#includes.
var array1 = ["Lorem", "ipsum", "dolor"],
array2 = ["Lorem", "ipsum", "quick", "brown", "foo"],
array3 = ["Jumps", "Over", "Lazy", "Lorem"],
array4 = [1337, 420, 666, "Lorem"],
data = [array1, array2, array3, array4],
result = data.reduce((a, b) => a.filter(c => b.includes(c)));
console.log(result);

I wrote a helper function for this:
function intersection() {
var result = [];
var lists;
if(arguments.length === 1) {
lists = arguments[0];
} else {
lists = arguments;
}
for(var i = 0; i < lists.length; i++) {
var currentList = lists[i];
for(var y = 0; y < currentList.length; y++) {
var currentValue = currentList[y];
if(result.indexOf(currentValue) === -1) {
var existsInAll = true;
for(var x = 0; x < lists.length; x++) {
if(lists[x].indexOf(currentValue) === -1) {
existsInAll = false;
break;
}
}
if(existsInAll) {
result.push(currentValue);
}
}
}
}
return result;
}
Use it like this:
intersection(array1, array2, array3, array4); //["Lorem"]
Or like this:
intersection([array1, array2, array3, array4]); //["Lorem"]
Full code here
UPDATE 1
A slightly smaller implementation here using filter

This can be done pretty succinctly if you fancy employing some recursion and the new ES2015 syntax:
const array1 = ["Lorem", "ipsum", "dolor"];
const array2 = ["Lorem", "ipsum", "quick", "brown", "foo"];
const array3 = ["Jumps", "Over", "Lazy", "Lorem"];
const array4 = [1337, 420, 666, "Lorem"];
const arrayOfArrays = [[4234, 2323, 43], [1323, 43, 1313], [23, 34, 43]];
// Filter xs where, for a given x, there exists some y in ys where y === x.
const intersect2 = (xs,ys) => xs.filter(x => ys.some(y => y === x));
// When there is only one array left, return it (the termination condition
// of the recursion). Otherwise first find the intersection of the first
// two arrays (intersect2), then repeat the whole process for that result
// combined with the remaining arrays (intersect). Thus the number of arrays
// passed as arguments to intersect is reduced by one each time, until
// there is only one array remaining.
const intersect = (xs,ys,...rest) => ys === undefined ? xs : intersect(intersect2(xs,ys),...rest);
console.log(intersect(array1, array2, array3, array4));
console.log(intersect(...arrayOfArrays));
// Alternatively, in old money,
var intersect2ES5 = function (xs, ys) {
return xs.filter(function (x) {
return ys.some(function (y) {
return y === x;
});
});
};
// Changed slightly from above, to take a single array of arrays,
// which matches the underscore.js approach in the Q., and is better anyhow.
var intersectES5 = function (zss) {
var xs = zss[0];
var ys = zss[1];
var rest = zss.slice(2);
if (ys === undefined) {
return xs;
}
return intersectES5([intersect2ES5(xs, ys)].concat(rest));
};
console.log(intersectES5([array1, array2, array3, array4]));
console.log(intersectES5(arrayOfArrays));

Using a combination of ideas from several contributors and the latest ES6 goodness, I arrived at
const array1 = ["Lorem", "ipsum", "dolor"];
const array2 = ["Lorem", "ipsum", "quick", "brown", "foo"];
const array3 = ["Jumps", "Over", "Lazy", "Lorem"];
const array4 = [1337, 420, 666, "Lorem"];
Array.prototype.intersect = function intersect(a, ...b) {
const c = function (a, b) {
b = new Set(b);
return a.filter((a) => b.has(a));
};
return undefined === a ? this : intersect.call(c(this, a), ...b);
};
console.log(array1.intersect(array2, array3, array4));
// ["Lorem"]

For anyone confused by this in the future,
_.intersection.apply(_, arrayOfArrays)
Is in fact the most elegant way to do this. But:
var arrayOfArrays = [[43, 34343, 23232], [43, 314159, 343], [43, 243]];
arrayOfArrays = _.intersection.apply(_, arrayOfArrays);
Will not work! Must do
var differentVariableName = _.intersection.apply(_,arrayOfArrays);

The cleanest way I've found to do this wasn't actually listed on this page, so here you are:
arrays[0].filter(elem => arrays.every(array => array.includes(elem)))
Reads like nice, clear english: every array includes the element. It assumes that you have at least 1 element in arrays, though. If you can't make this assumption, you can use optional chaining:
arrays?[0].filter(elem => arrays.every(array => array.includes(elem))) ?? []

Your code with _lodash is working fine.
As you can say in this fiddle:
this code:
var arrayOfArrays = [[4234, 2323, 43], [1323, 43, 1313], [23, 34, 43]];
var a = _.intersection.apply(_, arrayOfArrays);
console.log(a);
console.log(a.length);
Will have output:
[42]
1
Maybe you see
equals: function
because you are using kind of debugger.
Try to just print the array with console.log, you will get only 42.

Small recursive divide and conquer solution that does not rely on es6 or any library.
It accepts an array of arrays which makes the code shorter and allows you to pass arguments by using map.
function intersection(a) {
if (a.length > 2)
return intersection([intersection(a.slice(0, a.length / 2)), intersection(a.slice(a.length / 2))]);
if (a.length == 1)
return a[0];
return a[0].filter(function(item) {
return a[1].indexOf(item) !== -1;
});
}
var list1 = [ 'a', 'b', 'c' ];
var list2 = [ 'd', 'b', 'e' ];
var list3 = [ 'f', 'b', 'e' ];
console.log(intersection([list1, list2, list3]));

If you can use ES6 Maps and your arrays items are scalar values (easily usable as Map keys), then you can try this (works in my case) :
const intersect_lists = (lists) => {
const results = []
const lookup = new Map()
lists.map((list, idx) => {
list.map((element) => {
const count = lookup.get(element) || 0
if(count === idx) {
lookup.set(element, 1 + count)
} else {
lookup.delete(element)
}
})
})
// only elements present in all lists will have
// their respective counter equllling the total number of lists
Array.from(lookup.keys()).map((key) => {
if(lookup.get(key) === lists.length) {
results.push(key)
}
})
return results
}
Optionally you can pre-sort "lists" (of lists) by creasing length to avoid lots of iterations of the outer map() call, especially if lists lengths are heterogenous :
lists.sort((l1, l2) => l1.length - l2.length).map((list, idx) => { ... })

Sol with Maps
// nums1 = [1,2,2,1], nums2 = [2,2]
// n m
// O(nm) + space O(min(n, m))
// preprocess nums2 to a Map<number, count>
// O(n + m) + space(min(n, m))
// process the shorter one
let preprocessTarget = nums1
let loopTarget = nums2
if (nums1.length > nums2.length) {
preprocessTarget = nums2
loopTarget = nums1
}
// Map<element, number>
const countMap = new Map()
for (let num of preprocessTarget) {
if (countMap.has(num)) {
countMap.set(num, countMap.get(num) + 1)
} else {
countMap.set(num, 1)
}
}
const result = []
for (let num of loopTarget) {
if (countMap.has(num)) {
result.push(num)
const count = countMap.get(num)
if (count === 1) {
countMap.delete(num)
} else {
countMap.set(num, count - 1)
}
}
}
return result

const data = [array1, array2, array3, array4].filter(arr => arr.length > 0);
const result = [...new Set(data)];
let final = Array.of<any>();
for (const key of result) {
final = final.concat(key);
}
console.log(final);

Lodash pure:
_.keys(_.pickBy(_.groupBy(_.flatten(arrays)), function (e) {return e.length > 1}))
Lodash with plain js:
var elements = {}, duplicates = {};
_.each(arrays, function (array) {
_.each(array, function (element) {
if (!elements[element]) {
elements[element] = true;
} else {
duplicates[element] = true;
}
});
});
_.keys(duplicates);

I manage to accomplish this with a reduce call:
var intersected = intersect([[1, 2, 3], [2, 3, 4], [3, 4, 5]]);
console.log(intersected); // [3]
function intersect(arrays) {
if (0 === arrays.length) {
return [];
}
return arrays.reduce((intersection, array) => {
return intersection.filter(intersectedItem => array.some(item => intersectedItem === item));
}, arrays[0]);
}

Intersection of a variable number of arrays.
This is how I do it:
function getArraysIntersection(list1, list2, ...otherLists) {
const result = [];
for (let i = 0; i < list1.length; i++) {
let item1 = list1[i];
let found = false;
for (var j = 0; j < list2.length && !found; j++) {
found = item1 === list2[j];
}
if (found === true) {
result.push(item1);
}
}
if (otherLists.length) {
return getArraysIntersection(result, otherLists.shift(), ...otherLists);
}
else {
return result;
}
}
SNIPPET
function getArraysIntersection(list1, list2, ...otherLists) {
const result = [];
for (let i = 0; i < list1.length; i++) {
let item1 = list1[i];
let found = false;
for (var j = 0; j < list2.length && !found; j++) {
found = item1 === list2[j];
}
if (found === true) {
result.push(item1);
}
}
if (otherLists.length) {
return getArraysIntersection(result, otherLists.shift(), ...otherLists);
}
else {
return result;
}
}
const a = {label: "a", value: "value_A"};
const b = {label: "b", value: "value_B"};
const c = {label: "c", value: "value_C"};
const d = {label: "d", value: "value_D"};
const e = {label: "e", value: "value_E"};
const arr1 = [a,b,c];
const arr2 = [a,b,c];
const arr3 = [c];
const t0 = performance.now();
const intersection = getArraysIntersection(arr1,arr2,arr3);
const t1 = performance.now();
console.log('This took t1-t0: ' + (t1-t0).toFixed(2) + ' ms');
console.log(intersection);

const intersect = (arrayA, arrayB) => {
return arrayA.filter(elem => arrayB.includes(elem));
};
const intersectAll = (...arrays) => {
if (!Array.isArray(arrays) || arrays.length === 0) return [];
if (arrays.length === 1) return arrays[0];
return intersectAll(intersect(arrays[0], arrays[1]), ...arrays.slice(2));
};

For anyone who might need, this implements the intersection inside an array of arrays:
intersection(array) {
if (array.length === 1)
return array[0];
else {
array[1] = array[0].filter(value => array[1].includes(value));
array.shift();
return intersection(array);
}
}

function getIntersection(ar1,ar2,...arrays){
if(!ar2) return ar1
let intersection = ar1.filter(value => ar2.includes(value));
if(arrays.length ===0 ) return intersection
return getIntersection(intersection,...arrays)
}
console.log(getIntersection([1,2,3], [3,4], [5,6,3]) // [3]

Function to calculate intersection of multiple arrays in JavaScript
Write a method that creates an array of unique values that are included in all given arrays. Expected Result: ([1, 2], [2, 3]) => [2]
const arr1 = [1, 2, 1, 2, 1, 2];
const arr2 = [2, 3];
const arr3 = ["a", "b"];
const arr4 = ["b", "c"];
const arr5 = ["b", "e", "c"];
const arr6 = ["b", "b", "e"];
const arr7 = ["b", "c", "e"];
const arr8 = ["b", "e", "c"];
const intersection = (...arrays) => {
(data = [...arrays]),
(result = data.reduce((a, b) => a.filter((c) => b.includes(c))));
return [...new Set(result)];
};
console.log(intersection(arr1, arr2)); // [2]
console.log(intersection(arr3, arr4, arr5)); // ['b']
console.log(intersection(arr5, arr6, arr7, arr8)); // ['b', 'e']

Getting objects within array not present in other array [duplicate]

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"]

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 //
//////////////
const dif1 = [1,2,3,4,5,6].diff( [3,4,5] );
console.log(dif1); // => [1, 2, 6]
const dif2 = ["test1", "test2","test3","test4","test5","test6"].diff(["test1","test2","test3","test4"]);
console.log(dif2); // => ["test5", "test6"]
Note .indexOf() and .filter() are not available before IE9.

This answer was written in 2009, so it is a bit outdated, also it's rather educational for understanding the problem. Best solution I'd use today would be
let difference = arr1.filter(x => !arr2.includes(x));
(credits to other author here)
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"));

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]);

Plain JavaScript
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;
}
Lodash / Underscore
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.
ES6 Set, for very large arrays
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));
}
Notes
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.)

A cleaner approach in ES6 is the following solution.
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
Difference
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
Intersection
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
Disjunctive Union (Symmetric Difference)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]

To get the symmetric difference you need to compare the arrays in both ways (or in all the ways in case of multiple arrays)
ES7 (ECMAScript 2016)
// 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));
}));
}
ES6 (ECMAScript 2015)
// 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);
}));
}
ES5 (ECMAScript 5.1)
// 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"]
Difference between Arrays of Objects
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}

One Liners
const unique = (a) => [...new Set(a)];
const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{}));
const intersection = (a, b) => a.filter((v) => b.includes(v));
const diff = (a, b) => a.filter((v) => !b.includes(v));
const symDiff = (a, b) => diff(a, b).concat(diff(b, a));
const union = (a, b) => diff(a, b).concat(b);
const a = unique([1, 2, 3, 4, 5, 5]);
console.log(a);
const b = [4, 5, 6, 7, 8];
console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b));
console.log(uniqueBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
{ id: 1, name: "abc" },
],
(v) => v.id
));
const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u)));
console.log(intersectionBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u)));
console.log(diffBy(
[
{ id: 1, name: "abc" },
{ id: 2, name: "xyz" },
],
[
{ id: 1, name: "abc" },
{ id: 3, name: "pqr" },
],
(v, u) => v.id === u.id
));
TypeScript
playground link
const unique = <T>(array: T[]) => [...new Set(array)];
const intersection = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => array2.includes(v));
const diff = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => !array2.includes(v));
const symDiff = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(diff(array2, array1));
const union = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(array2);
const intersectionBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));
const diffBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));
const uniqueBy = <T>(
array: T[],
predicate: (v: T, i: number, a: T[]) => string
) =>
Object.values(
array.reduce((acc, value, index) => {
acc[predicate(value, index, array)] = value;
return acc;
}, {} as { [key: string]: T })
);

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().

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" ].

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;
});

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]

Functional approach with ES2015
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.
Left 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) );
Right 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!
Symmetric 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']

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 ]

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);

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)

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

If you have two list of objects
const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]
let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))

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"]

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(', ')
);

To find the difference of 2 arrays without duplicates:
function difference(arr1, arr2){
let setA = new Set(arr1);
let differenceSet = new Set(arr2.filter(ele => !setA.has(ele)));
return [...differenceSet ];
}
1.difference([2,2,3,4],[2,3,3,4]) will return []
2.difference([1,2,3],[4,5,6]) will return [4,5,6]
3.difference([1,2,3,4],[1,2]) will return []
4.difference([1,2],[1,2,3,4]) will return [3,4]
Note: The above solution requires that you always send the larger array as the second parameter. To find the absolute difference, you will need to first find the larger array of the two and then work on them.
To find the absolute difference of 2 arrays without duplicates:
function absDifference(arr1, arr2){
const {larger, smaller} = arr1.length > arr2.length ?
{larger: arr1, smaller: arr2} : {larger: arr2, smaller: arr1}
let setA = new Set(smaller);
let absDifferenceSet = new Set(larger.filter(ele => !setA.has(ele)));
return [...absDifferenceSet ];
}
1.absDifference([2,2,3,4],[2,3,3,4]) will return []
2.absDifference([1,2,3],[4,5,6]) will return [4,5,6]
3.absDifference([1,2,3,4],[1,2]) will return [3,4]
4.absDifference([1,2],[1,2,3,4]) will return [3,4]
Note the example 3 from both the solutions

Here is another solution that can return the differences, just like git diff: (it has been written in typescript, if you're not using typescript version, just remove the types)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* #param { T[] } from
* #param { T[] } to
* #returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
Here is a sample of usage:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}

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};
}

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]

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;
}

//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
}