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How in Scala to find unique items in List?

Volodymyr Bezuglyy
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8 Answers8

223

In 2.8, it's:

List(1,2,3,2,1).distinct  // => List(1, 2, 3)
Ben Kovitz
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jamesqiu
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    Imho, 1 and 2 aren't unique items in the list. Only the 3 is. You create a list of unique items from the list, which is a different thing. – user unknown May 11 '12 at 20:02
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    If that's what you want (it's usually not), go with: List(1,2,3,2,1).groupBy(x=>x).filter(_._2.lengthCompare(1) == 0).keySet – moveaway00 Jul 03 '13 at 15:56
24

The most efficient order-preserving way of doing this would be to use a Set as an ancillary data structure:

def unique[A](ls: List[A]) = {
  def loop(set: Set[A], ls: List[A]): List[A] = ls match {
    case hd :: tail if set contains hd => loop(set, tail)
    case hd :: tail => hd :: loop(set + hd, tail)
    case Nil => Nil
  }

  loop(Set(), ls)
}

We can wrap this in some nicer syntax using an implicit conversion:

implicit def listToSyntax[A](ls: List[A]) = new {
  def unique = unique(ls)
}

List(1, 1, 2, 3, 4, 5, 4).unique    // => List(1, 2, 3, 4, 5)
Daniel Spiewak
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14

Roll your own uniq filter with order retention:

scala> val l = List(1,2,3,3,4,6,5,6)
l: List[Int] = List(1, 2, 3, 3, 4, 6, 5, 6)

scala> l.foldLeft(Nil: List[Int]) {(acc, next) => if (acc contains next) acc else next :: acc }.reverse
res0: List[Int] = List(1, 2, 3, 4, 6, 5)
Synesso
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11

If you refer to the Rosetta Code: Create a Sequence of unique elements

val list = List(1,2,3,4,2,3,4,99)
val l2 = list.removeDuplicates
// l2: scala.List[scala.Int] = List(1,2,3,4,99)

Since List is immutable, you wont modify the initial List by calling removeDuplicates

Warning: as mentioned by this tweet(!), this does not preserve the order:

scala> val list = List(2,1,2,4,2,9,3)
list: List[Int] = List(2, 1, 2, 4, 2, 9, 3)

scala> val l2 = list.removeDuplicates
l2: List[Int] = List(1, 4, 2, 9, 3)

For a Seq, that method should be available in Scala2.8, according to ticket 929.
In the meantime, you will need to define an ad-hoc static method as the one seen here

VonC
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7

Imho, all the interpretations of the question are false:

How in Scala to find unique items in List?

Given this list:

val ili = List (1, 2, 3, 4, 4, 3, 1, 1, 4, 1)

the only unique item in the list is 2. The other items aren't unique. 

ili.toSet.filter (i => ili.indexOf (i) == ili.lastIndexOf (i))

will find it.

user unknown
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    most java/scala devs would translate a question of "unique items in a given list" as "all distinct values with a given list" not "all values showing up singularly within a given list"... which is how every answer before yours interpreted it, as well as the person asking the question (who approved such an answer). what does coming in 2+ years later and being extra sensitive on the parsing of an already answered question add? – mujimu Jul 02 '12 at 20:04
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    @mujimu: I don't remember why I stumbled over the question 2 years later. Maybe sombebody closed a similar question as exact duplicate and linked here. Often I try to answer such questions for myself as an exercise before looking for the solutions of others; whether they have a similar or a better solution or whether it makes sense to publish mine. So I found that the others answered a different question. After knowing how others understood the question differently I still think that the terminology is wrong. The meaning of a sentence shouldn't be judged by majority. – user unknown Jul 03 '12 at 10:13
5
list.filter { x => list.count(_ == x) == 1 }
slouc
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2

A simple ad-hoc method is just to add the List to a Set, and use from there:

  val l = List(1,2,3,3,3,4,5,5,6,7,8,8,8,9,9)
  val s = Set() ++ x
  println(s)

Produces:

> Set(5, 1, 6, 9, 2, 7, 3, 8, 4)

This works for a Seq (or any Iterable), but is not necessary in 2.8, where the removeDuplicates method will probably be more readable. Also, not sure about the runtime performance vs a more thought-out conversion.

Also, note the lost ordering.

Mitch Blevins
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0

list.toSet will do it since Set by definition only contains unique elements

Murilo
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