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Last active November 15, 2022 17:13
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Remove duplicates from js array (ES5/ES6)
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)));
@rravithejareddy
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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]

@xyzdata
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xyzdata commented Mar 30, 2018

@whitehorse0
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whitehorse0 commented Apr 11, 2018

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'];

@pikislabis
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@karosi12
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karosi12 commented Apr 25, 2018

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

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

@iamvanja
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@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 

@VSmain
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VSmain commented Jul 13, 2018

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}]

@macmladen
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@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 ;)

image

@joeyparrish
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joeyparrish commented Sep 6, 2018

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

Set dedups NaN correctly, though.

@nabilfreeman
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Set is so cool! Never knew it existed!

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

import _ from "lodash";

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

Documentation on uniq.

@impfromliga
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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

@brunoandradebr
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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)

@Kr3m
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Kr3m commented Feb 8, 2019

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.

@little-brother
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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);

@philihp
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philihp commented Feb 26, 2019

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);

@indatawetrust
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indatawetrust commented Mar 15, 2019

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)];
}

@patrickmichalina
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patrickmichalina commented Mar 16, 2019

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], [])

@pankajkrr
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Find brief article here : Click here to view

@fosteev
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fosteev commented Mar 10, 2020

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

@ioness
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ioness commented Apr 27, 2020

Excelent, thanks!

@gevera
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gevera commented Nov 15, 2022

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

@guillaumegarcia13
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Hi @gevera
Beware of problems when using JSON.stringify and JSON.parse with:

Being bitten more than once... 🤕

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