How to pick elements in even index and odd index?

Question

I have a list, I want to break the list into two, one with elements in the even indexes, and the other from the odd indexes.

breakByIndexes :: [a] -> ([a], [a])

For example:

> breakByIndexes ["A", "B", "C", "D"]
(["A", "C"], ["B", "D"]

> breakByIndexes ["A", "B", "C", "D", "E"]
(["A", "C", "E"], ["B", "D"]

I got a solution like this

breakByIndexes [] = ([], [])
breakByIndexes [e] = ([e], [])
breakByIndexes (e:o:xs) =
  let (es, os) = breakByIndexes xs
   in (e : es, o : os)

But I'm curious is it possible to implement without using recursion? And is it possible to implement by composing existing functions from Data.List?

Answer 1

Yes you are right, using the partition function from Data.List.

Prelude Data.List> (s, u) = partition (even . fst) (zip [0 .. ] "ABCD")
Prelude Data.List> (_, s2) = unzip s
Prelude Data.List> (_, u2) = unzip u
Prelude Data.List> (s2, u2)
("AC","BD")

How I found this? Go to Hoogle and fill in [a] -> ([a], [a]).