Scala generate unique pairs from list

Question

Input :

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

Desired output :

Iterator((1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4))

This code works :

for (cur1 <- 0 until list.size; cur2 <- (cur1 + 1) until list.size)
  yield (list(cur1), list(cur2))

but it not seems optimal, is there any better way of doing it ?

score 26 · Accepted Answer · edited May 23 '17 at 12:00

26

There's a .combinations method built-in:

scala> List(1,2,3,4).combinations(2).toList
res0: List[List[Int]] = List(List(1, 2), List(1, 3), List(1, 4), List(2, 3), List(2, 4), List(3, 4))

It returns an Iterator, but I added .toList just for the purpose of printing the result. If you want your results in tuple form, you can do:

scala> List(1,2,3,4).combinations(2).map{ case Seq(x, y) => (x, y) }.toList
res1: List[(Int, Int)] = List((1,2), (1,3), (1,4), (2,3), (2,4), (3,4))

You mentioned uniqueness as well, so you could apply .distinct to your input list uniqueness isn't a precondition of your function, because .combination will not deduplicate for you.

edited May 23 '17 at 12:00

Community

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1

answered Oct 29 '14 at 10:41

acjay

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Exactly what I looking for thx. By uniqueness I meant not having (1, 2) and (2, 1) in the result – Co_42 Oct 29 '14 at 10:45
3

@Co_42: alternative solution: `list.tails.collect{ case h :: tail => tail map {(h, _)} }.flatten.toList`. – senia Oct 29 '14 at 10:48
@senia: you should make that an answer – Juh_ May 30 '16 at 09:13

Gianmario Spacagna · Answer 2 · 2016-07-16T06:13:15.187

.combinations is the proper way for generating unique arbitrary groups of any size, another alternative solution that does not check the uniqueness in first place is using foldLeft that way:

val list = (1 to 10).toList

val header :: tail = list
tail.foldLeft((header, tail, List.empty[(Int, Int)])) {
  case ((header, tail, res), elem) =>
    (elem, tail.drop(1), res ++ tail.map(x => (header, x)))
}._3

Will produce:

res0: List[(Int, Int)] = List((1,2), (1,3), (1,4), (1,5), (1,6), (1,7), (1,8), (1,9), (1,10), (2,3), (2,4), (2,5), (2,6), (2,7), (2,8), (2,9), (2,10), (3,4), (3,5), (3,6), (3,7), (3,8), (3,9), (3,10), (4,5), (4,6), (4,7), (4,8), (4,9), (4,10), (5,6), (5,7), (5,8), (5,9), (5,10), (6,7), (6,8), (6,9), (6,10), (7,8), (7,9), (7,10), (8,9), (8,10), (9,10))

If you expect there to be duplicates then you can turn the output list into a set and bring it back into a list, but you will lose the ordering then. Thus not the recommended way if you want to have uniqueness, but should be preferred if you want to generate all of the pairs included equal elements.

E.g. I used it in the field of machine learning for generating all of the products between each pair of variables in the feature space and if two or more variables have the same value I still want to produce a new variable corresponding to their product even though those newly generated "interaction variables" will have duplicates.

Scala generate unique pairs from list

2 Answers2