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Claire Parker-Jones
Claire Parker-Jones

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How to create an array of unique values in JavaScript using Sets

TL;DR

let uniqueArray = [...new Set([5,5,2,2,2,4,2])];
// [5,2,4]
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Pardon?

Sorry for the verbose title - sometimes things can be explained better with a code example.

Imagine you have a JavaScript array that contains many elements, some of which are duplicated:

let dupeArray = [1,1,4,5,4,4,2,1,5];
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Your goal is to remove the duplicates and leave only one entry per value:

let uniqueArray = [1,4,5,2];
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You might write a for-loop, or use a map or filter to get this result. There are a million different ways to tackle a problem in computer programming (which is what makes it so fun!). ES6 introduces a bunch of new features, including Sets, which we can also use to solve this problem in one line of code.

Note: the element type could be strings, or a mixture of numbers and strings, but I'm using integers here because they are quicker to type!

What's a Set?

A Set is a new data structure introduced in ES6. Quoting directly from MDN:

Set objects are collections of values. You can iterate through the elements of a set in insertion order. A value in the Set may only occur once; it is unique in the Set's collection.

The rule for unique values is one we can use to our advantage here.

Let's create a Set, add some values to it and query the size. (You can follow along in the developer tools console of a modern browser):

let mySet = new Set().add(1).add(3).add(2).add(1);
// Set(3) {1, 3, 2}
mySet.size
// 3
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Notice how the final 1 wasn't added and the size of the Set remained at 3 instead of 4. In an array, it would have been added and the length would be 4.

There are two ways to add values to a Set. Firstly by using the add method as above. Secondly by passing an array to the constructor (new Set()):

let arraySet1 = new Set([3,2,1]);
// Set(3) {3, 2, 1}
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Duplicated values are also removed if they are included in the initial array:

let arraySet2 = new Set([8,8,9,2]);
// Set(3) {8,9,2}
arraySet2.size;
// 3
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You can probably see where this is going. All that’s left to do is to convert this Set into an array and we’ve achieved our original goal. There are two ways to do this: both using ES6 methods.

Array.from()

The Array.from method creates a new array from an array-like structure:

let dupeArray = [3,2,3,3,5,2];
let uniqueArray = Array.from(new Set(dupeArray));
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Spread operator ...

Those three dots are ubiquitous in ES6. They crop up everywhere and have several uses (and they're a right pain to google). When we use them as the spread operator they can be used to create an array:

let uniqueArray = [...new Set(dupeArray)];
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Which of these two methods should you use? The spread syntax looks cool, but Array.from is more explicit in its purpose and easier to read. They both accomplish the same thing here so the choice is yours!

Conclusion

Something that would have took many lines of code and variables can now be executed in a one-liner in ES6. What a time to be alive.

let uniqueArray = [...new Set([5,5,2,4,2])];
// [5,2,4]
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Top comments (14)

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xngwng profile image
Xing Wang

Good highlight this feature.

However, it maybe worth pointing out that with Set, you can't control the equality operator.

It uses '===', which only work only off same actual object or same value for a primitive.

Which limits the usage somewhat compares to other methods such as map or filter.

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tohodo profile image
Tommy

If you find yourself needing to push unique items to an existing array that was declared with const (i.e., you cannot use new Set()), then here are two patterns you can use:

// Given this...
const arr = ['A', 'B', 'C'];
const itemsToAdd = ['C', 'D', 'E'];

// Both methods will produce arr = ['A', 'B', 'C', 'D', 'E']

// Method 1:
itemsToAdd.forEach((i) => {
  if (!arr.includes(i)) {
    arr.push(i);
  }
});

// Method 2:
arr.push(...itemsToAdd.filter((i) => !arr.includes(i)));
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Method 1 is about 30% faster in my Chrome:
jsbench.me/jzlhpp4thr/1

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itsmeseb profile image
Sebastian L. K. Sørensen

I will implement this right away in a project I am working on. Although I like the Array.from() better, since it's easier to read. I like easy-to-read code :) Thank you for sharing!

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mehraas profile image
Ashish Mehra • Edited

I used to filter and map over an array to remove duplicates. Thanks for sharing this method, Mam.

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jochemstoel profile image
Jochem Stoel

@claireparker I have been hanging around in this neighborhood for little over a year now and appreciated a lot of quality content from many different sources but this is the first time I am Googling something and it leads to finding the answer here on this website.

You have a new follower.

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seth10 profile image
Seth T

Is there any difference in performance between Array.from and using the spread operator?

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kimlongbtc profile image
KimLongUn

"They both accomplish the same thing here so the choice is yours!"

So i dont know if there is some missunderstanding here but as far as i see it they dont work the same as in my case using the spread operator returns

[Set(7)]
0: Set(7) {"anthropological", "security", "surveillance", "technology", "biology", …}

while using Array.from returns

(7) ["anthropological", "security", "surveillance", "technology", "biology", "extra-algorithmic experience", "randomness"]

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neovortex profile image
Neocortexxx • Edited

"They both accomplish the same thing here so the choice is yours!"
Well, actually she refers to the spread operator ('...') and Array.from.
In both cases you still have to use new Set.

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javascripterika profile image
JavaScriptErika

Great article and explanations! Thank you for your awesome insight and clear examples!

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ayoalfonso profile image
AyoAlfonso

What if the elements are deep objects themselves rather than just basic elements like integers and alphabets.

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nutondev profile image
nuton.dev

We will read about it in your post, right? :)

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evangelistaagil profile image
Marcelo Faundez

Great tip. Thanks

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sait profile image
Sai gowtham
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wbarath profile image
William Barath

// ES5:
Object.keys([1,1,2,2,3,4].reduce((prev,val)=>(prev[val]=1,prev),{}))
// ['1', '2', '3', '4']