Scala filter on a list by index

Question

I wanted to write it functionally, and the best I could do was:

list.zipWithIndex.filter((tt:Tuple2[Thing,Int])=>(tt._2%3==0)).unzip._1

to g

Answer 1

Ah, how about this?

val l = List(10,9,8,7,6,5,4,3,2,1,0)
for (i <- (0 to l.size - 1 by 3).toList) yield l(i)
//res0: List[Int] = List(10, 7, 4, 1)

which can be made more general by

def seqByN[A](xs: Seq[A], n: Int): Seq[A] = for (i <- 0 to xs.size - 1 by n) yield xs(i)

scala> seqByN(List(10,9,8,7,6,5,4,3,2,1,0), 3)
res1: Seq[Int] = Vector(10,7,4,1)

scala> seqByN(List(10,9,8,7,6,5,4,3,2,1,0), 3).toList
res2: Seq[Int] = List(10,7,4,1)

scala> seqByN(List[Int](), 3)
res1: Seq[Int] = Vector()

But by functional do you mean only using the various List combinator functions? Otherwise, are Streams functional enough?

def fromByN[A](xs: List[A], n: Int): Stream[A] = if (xs.isEmpty) Stream.empty else
  xs.head #:: fromByN(xs drop n, n)

scala> fromByN(List(10,9,8,7,6,5,4,3,2,1,0), 3).toList
res17: List[Int] = List(10, 7, 4, 1)