Create Range with inclusive end value when stepping

Is there any way to create a range which includes the end value when using a step which doesn't align?

For instance the following yields:

scala> Range.inclusive(0, 35, 10)
res3: scala.collection.immutable.Range.Inclusive = Range(0, 10, 20, 30)


But I would also like the end value (35) included like so:

scala> Range.inclusive(0, 35, 10)
res3: scala.collection.immutable.Range.Inclusive = Range(0, 10, 20, 30, 35)


3 Answers
3

No, not with the current definition/ implementation. It would be strange behaviour to have the step the same for all intermediate elements but different from the last.

Yes that makes sense - however I was considering a use case where a collection (table) could be effectively partitioned into chunks using column indexes. thanks.
– Fred Smith
May 25 '16 at 19:30

Depending on whether you really need the indices, you could use .grouped (I regard explicit indices in a loop as often a code smell)
– The Archetypal Paul
May 25 '16 at 20:54

.grouped

As mentioned, not a standard semantics. A workaround,

for (i <- 0 to 35 by 10) yield if (35 % 10 != 0 && 35 - i < 10) 35 else i


where you must replace the boundary and step values as needed.

Thanks for this workaround could come in handy although I'm going to revisit my approach and come up with a cleaner solution.
– Fred Smith
May 25 '16 at 19:32

The above solution does not work because it omits the value "30". Here is a unfold-style solution that produces a list rather than a sequence.

def unfoldRange(i: Int, j: Int, s: Int): List[Int] = {
if (i >= j) List(j)
else i :: unfoldRange(i+s,j,s)
}


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