How to remove duplicates from an unordered Immutable.List()?

Question

How would one remove duplicates from an unordered Immutable.List()? (without using toJS() or toArray())

e.g.

Immutable.List.of("green", "blue","green","black", "blue")

Varon · Accepted Answer · 2017-04-26T00:59:22.210

39

You can convert it to a Set. A Set is a List with unique values.

Immutable.List.of("green", "blue","green","black", "blue").toSet()

If you need it as list again just convert it back then:

Immutable.List.of("green", "blue","green","black", "blue").toSet().toList()

Update:

It exists a shorter possibility to get unique values:

Immutable.List.of("green", "blue","green","black", "blue").distinct

edited Apr 26 '17 at 00:59

answered Mar 07 '16 at 20:31

Varon

3,736
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6

Using Immutable 3.7.3 the `.distinct` solution does not seem to work (it's undefined). `.toSet().toList()` does the job! – manosim Sep 18 '17 at 09:44
2

for Immutable 3.8.2 as well - ```.distinct``` returns ```undefined``` – gl03 Sep 06 '18 at 20:30
immutable@4.0.0-rc.12 `.distinct` returns `undefined` – bJacoG May 07 '19 at 09:03
Is there a way to use `Seq.Set` to lazily remove duplicates? – CMCDragonkai Aug 23 '19 at 08:22
Note that `toOrderedSet().toList()` is likely what one wants (so the orderer in the list is preserved but repeating entries are removed). – Peter Lundberg May 21 '21 at 11:01

score 30 · Answer 2 · answered Jan 04 '17 at 15:24

30

If you have a more complex type you can also use groupBy to provide your own selector to compare on. The following will remove duplicates on the property .name:

var distinctColors = duplicateColors.groupBy(x => x.name).map(x => x.first()).toList();

answered Jan 04 '17 at 15:24

Christoffer Ramqvist

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How to remove duplicates from an unordered Immutable.List()?

2 Answers2