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Cover image for 1 line of code: How to remove all duplicates from an Array
Martin Krause
Martin Krause

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1 line of code: How to remove all duplicates from an Array

#javascript #webdev #productivity #codequality 
const removeDuplicates = arr => [...new Set(arr)];
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Assumes that the given argument is an Array and removes duplicate entries, keep in mind that it works only for entries with primitive values (string, number, bigint, boolean, undefined, symbol, and null). Preserves the order of the entries and returns a copy of the array.

The repository & npm package

You can find the all the utility functions from this series at github.com/martinkr/onelinecode
The library is also published to npm as @onelinecode for your convenience.

The code and the npm package will be updated every time I publish a new article.

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Top comments (10)

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michaelcurrin profile image
Michael Currin •

Worth explaining this separately when used with array instead of sets

const x = ["abc", "def"]
const y = [...x]
// ["abc", "def"]
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And modifying x and y won't affect the other.

Then bringing in set

const x = ["abc", "def", "abc"]
const y = new Set(x)
// Set(["abc", "def"])
const z = [...y]
// ["abc", "def"]
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And the the original post does that in one line in a function

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michaelcurrin profile image
Michael Currin •

Also worth adding that you might want to stop once you have a set and not convert it back to an array.

Why?

Lookup time from a set is constant to check if a value is in a set while in an array of 1000 elements worst case you have to go through 1000 elements until you find the one at the end.

Secondly when you convert from array to set, the order will be lost. Sets are inherently unordered. So when you convert from set to array you will not get the order of the original array. So you might as well keep the set.

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qm3ster profile image
Mihail Malo •

It's actually specced to preserve insertion order: developer.mozilla.org/en-US/docs/W...

(...spread operator calls y[Symbol.iterator], which is specced to be the same as what is returned by .values() on Set)

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michaelcurrin profile image
Michael Currin •

Oh interesting, in Python it is unordered.

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martinkr profile image
Martin Krause •

Thank you for the explanation.

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humansprout profile image
Erik Waters • • Edited

I think they mean it won't behave as expected if the array contains similar objects that have different references

const originalObject = {}
const boundToOriginalObject = originalObject
const newObjectThatLooksLikeOriginal = {}
const arr = [originalObject, boundToOriginalObject, newObjectThatLooksLikeOriginal]
const dedupe = [ ...new Set(arr)]
dedupe // [{}, {}]
// deduping removes the duplicate reference but not similar objects
const mutation = {a: 1}
Object.assign(dedupe[1], mutation)
originalObject // {}
newObjectThatLooksLikeOriginal // { a: 1 }
// this proves which references remained in the deduped array prior to mutation
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It might be helpful to clarify that dedupe will remove duplicate bindings but not similar objects with a different binding

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captainyossarian profile image
yossarian •

I think it worth saying that it works only on primitives

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martinkr profile image
Martin Krause •

Thank you for your contribution. I'll add it to the description.

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michaelcurrin profile image
Michael Currin •

I just learned of an alternative for set to array

const myArr = Array.from(mySet1)
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And if you want to iterate over a set without converting to an array.

for (const item of mySet1) { 
  console.log(item)
}
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developer.mozilla.org/en-US/docs/W...

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qm3ster profile image
Mihail Malo •

Array.from goes through iterator protocol for Set and Map, so it's exactly equivalent to the [... sugar.

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