List of every n-th item in a given list

Question

This is a simple exercise I am solving in Scala: given a list l return a new list, which contains every n-th element of l. If n > l.size return an empty list.

def skip(l: List[Int], n: Int) = 
  Range(1, l.size/n + 1).map(i => l.take(i * n).last).toList

My solution (see above) seem to work but I am looking for smth. simpler. How would you simplify it?

Answer 1

scala> def skip[A](l:List[A], n:Int) = 
         l.zipWithIndex.collect {case (e,i) if ((i+1) % n) == 0 => e} // (i+1) because zipWithIndex is 0-based
skip: [A](l: List[A], n: Int)List[A]

scala> val l = (1 to 10).toList
l: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> skip(l,3)
res2: List[Int] = List(3, 6, 9)

scala> skip(l,11)
res3: List[Int] = List()

Answer 2

Somewhat simpler:

scala> val l = (1 to 10).toList
l: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

// n == 3

scala> l.drop(2).grouped(3).map(_.head).toList
res0: List[Int] = List(3, 6, 9)

// n > l.length

scala> l.drop(11).grouped(12).map(_.head).toList
res1: List[Int] = List()

(the toList just to force the iteratot to be evaluated)

Works with infinite lists:

Stream.from(1).drop(2).grouped(3).map(_.head).take(4).toList
res2: List[Int] = List(3, 6, 9, 12)