Is there a nice way in Scala to convert a List of numbers into a List of ranges?

Question

I haven't found so far a way to do the opposite operation of List.range:

Existent range function in Scala collections:

List.range(1, 4) = [1, 2, 3]

Looking for writing kind of toRanges function that could give me the following results:

List(1,3,4,5,7,8,9).toRanges() = [(1,1), (3,5), (7,9)]

I'm trying to see how's the functional way of achieving it

If you look into How to transform a list of numbers into a list of ranges of consecutive numbers The answer given is too large and imperative.

Thanks

For resolving my current problem the list is ordered and it cannot contain duplicates, — Abdoul Cisse, Feb 28 '16 at 15:18

score 5 · Accepted Answer · answered Feb 27 '16 at 19:11

5

toRanges funciton may look like:

def toRanges(list:List[Int]) = list.foldLeft(List[(Int,Int)]()) {
  case (Nil, i) => (i,i)::Nil
  case ((a,b)::t, i) => if (b+1==i) (a, i)::t else (i,i)::(a,b)::t
}.reverse

answered Feb 27 '16 at 19:11

Nyavro

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Great answer! Though I would recommend `foldRight` instead of `foldLeft` and `reverse`. – Paul Draper Feb 27 '16 at 19:27
@PaulDraper Possible stack overflow for long lists with `foldRight` – Victor Moroz Feb 27 '16 at 20:20
2

@VictorMoroz, not true. The standard library implementation of `foldRight` just reverses the list and calls `foldLeft` – The Archetypal Paul Feb 27 '16 at 22:14
@TheArchetypalPaul You are right, somehow I was using 2.9.2 and it did blow stack. `foldRight` you mentioned appeared in 2.10 – Victor Moroz Feb 27 '16 at 23:55

Is there a nice way in Scala to convert a List of numbers into a List of ranges?

1 Answers1