Remove duplicates from js array (ES5/ES6)

uniq.js

var uniqueArray = function(arrArg) {
  return arrArg.filter(function(elem, pos,arr) {
    return arr.indexOf(elem) == pos;
  });
};

var uniqEs6 = (arrArg) => {
  return arrArg.filter((elem, pos, arr) => {
    return arr.indexOf(elem) == pos;
  });
}

var test = ['mike','james','james','alex'];
var testBis = ['alex', 'yuri', 'jabari'];
console.log(uniqueArray(test.concat(testBis)));

If the array elements are objects, indexOf won't help. Here is a soln for that (in es6)

const uniqList = fullList.filter((s1, pos, arr) => arr.findIndex((s2)=>s2._id === s1._id) === pos});

This is what I came up with as I needed to filter results from an elastic search

let array1 = [{"id":1,"name":"apple","id":2,"name":"orange"}];
let array2 = [{"id":1,"name":"apple","id":4,"name":"purple"}];

let unique = array1.concat(array2).filter((obj, key, array) => array.map((obj2) => obj.id !== obj2.id));

will return a unique array filtered on any key...

const uniqueArray = (arr, objKey) => arr.filter((obj, key, array) => array.map((obj2) => obj[`${objKey}`] !== obj2[`${objKey}`] ));
this.state = uniqueArray(array1, "id");

Hope this can save others some time!

For generating a unique array of objects:

uniqueArray = a => [...new Set(a.map(o => JSON.stringify(o)))].map(s => JSON.parse(s))

@timrsmith Just a small word of caution with the JSON.parse(JSON.stringify(...)) approach. If you have properties with Date they will be replaced with their JSON representation (toISOString()).
uniqueArray([ {a: new Date()}, {b: new Date(1978,3,29)}, {a: new Date()}, {b: new Date(1978,3,29)} ]);

(2) [{…}, {…}]
0: {a: "2018-02-20T15:50:53.516Z"}
1: {b: "1978-04-28T22:00:00.000Z"}

@timrsmith beautiful!

I too feel compelled to say: +1 for @guidobouman's [...new Set(arr)]

Check with include before add, like below
var Array = [1, 2, 3]

Add()
{
if(!Array. include(4))
{
Array.push(4);
console.log(Array);
}

}

Output: [1,2,3,4]

ES6 Set & Unique keys

let left = ["research", "stockfast01", "news", "stockfast01"];

left = [...new Set(left)];

https://stackoverflow.com/questions/9229645/remove-duplicate-values-from-js-array

https://codereview.stackexchange.com/questions/60128/removing-duplicates-from-an-array-quickly

ES6 version

var names = ['mike','james','james','alex'];

let output = names.filter((value) => {
  return !this[value] && (this[value] = true)
}, Object.create(null))

// output is ['mike','james','alex'];

@timrsmith +1

var arr = [1,2,4,13,1];
Array.from(new Set(arr))

So how does this work when merging should happen using a property of the array object?

@seanmavley You can merge array before passing it to Set.

const a = [1, 2, 3]
cont b = [2, 3, 4, 5]
const uniqueMerged = [...new Set([...a, ...b])] // 1, 2, 3, 4, 5 

do it with ladosh

const _ = require('lodash');
...
_.uniq([1,2,3,3,'a','a','x'])//1,2,3,'a','x'
_.uniqBy([{a:1,b:2},{a:1,b:2},{a:1,b:3}], v=>v.a.toString()+v.b.toString())//[{a:1,b:2},{a:1,b:3}]

@mordentware I was puzzled with your results so I made a CodePen with my own data sample where I needed to remove duplicates.

Data array has 6.288 items, 5.284 of them are unique. Results are nearly the same for both sorted and unsorted arrays.

My findings are that filter and reduce are similar while Set was much faster. Reduce with spread was much slower due to a large number of deconstruction/construction.

See the Pen Deduplicating speed test by Mladen Đurić (@macmladen) on CodePen.

(times may vary due to system, browser, CPU, memory...)

Results on MacBook Pro i7 form 2011, using Firefox (usually with few hundred open tabs ;)

Just discovered this thread and found that [0, 1, NaN, 3].indexOf(NaN) yields -1. :-( Probably because NaN != NaN.

Set dedups NaN correctly, though.

Set is so cool! Never knew it existed!

As with @VSmain, I'm tossing Lodash into the ring:

import _ from "lodash";

_.uniq([2, 1, 2]);
// => [2, 1]

Documentation on uniq.

Set is shorter but filter with indexof is faster. example

it's just because the better complexity will faster on bigger counts. And be faster greatly be on them, instead of 10 elements array...
you have to check it at least on hundreds

If you need to filter by object value and a performance way :

// creates an object only once - garbage be glad ^^
let cachedObject = {};

// array to be filtered
let arr = [
    {id : 0, prop : 'blah'},
    {id : 1, prop : 'foo'},
    {id : 0, prop : 'bar'}
]

// "filter" to object - keep original array - garbage be glad too ^^
arr.map((item)=> cachedObject[item.id] = item)

// optional, turns object to array
arr = Object.values(cachedObject)

What's wrong with

let arr1 = ["apples", "apples", "oranges", "bananas"]; arr1 = Array.from(new Set(arr1));

Surely this is a lot simpler if you're just trying to remove duplicates. Obviously this would work better as the return statement in a function. I'm just posting this as an example.

let arr = [1, 2, 3, 4, 3, 2];
arr.filter((e, i, arr) => arr.indexOf(e) == i);

---

let arr = [{x:10, y: 20}, {x:10, y: 30}, {x:10, y: 20}]; // each element has x and y props
arr.filter((e, i, arr) => arr.findIndex(e2 => Object.keys(e2).every(prop => e2[prop] == e[prop])) == i);

I find this reads a little bit better if you stash your reducer function in a named variable.

const duplicates = (e, i, arr) => arr.indexOf(e) === i

let arr = [1, 2, 3, 4, 3, 2];
arr.filter(duplicates);

Array.prototype.uniq = function(key) {
  return key
    ? this.map(e => e[key])
        .map((e, i, final) => final.indexOf(e) === i && i)
        .filter(e => this[e])
        .map(e => this[e])
    : [...new Set(this)];
}

In Typescript

export const dedupeByProperty =
  <T>(arr: ReadonlyArray<T>, objKey: keyof T) =>
    arr.reduce<ReadonlyArray<T>>((acc, curr) =>
      acc.some(a => a[objKey] === curr[objKey])
        ? acc
        : [...acc, curr], [])

Find brief article here : Click here to view

const uniq = elements.reduce((acc, value) => acc.some(i => i.id === value.id) ? acc : acc.concat(value), []); // id your uniq key

Excelent, thanks!

const deduplicateArrays = ( arr1, arr2 = [] ) => [
...new Set([
           ...arr1.map(i => JSON.stringify(i)),
           ...arr2.map(i => JSON.stringify(i))
          ])
].map(i => JSON.parse(i))

Hi @gevera
Beware of problems when using JSON.stringify and JSON.parse with:

• number: https://github.com/josdejong/lossless-json
• date: https://stackoverflow.com/questions/31096130/how-to-json-stringify-a-javascript-date-and-preserve-timezone

Being bitten more than once... 🤕