How to get every Nth element of an infinite list in Haskell?

Question

More specifically, how do I generate a new list of every Nth element from an existing infinite list?

E.g. if the list is [5, 3, 0, 1, 8, 0, 3, 4, 0, 93, 211, 0

Answer 1

This seems a slightly better use of unfold:

everyNth :: Int -> [a] -> [a]
everyNth n = unfoldr g
  where
    g [] = Nothing
    g xs = let (ys, zs) = splitAt n xs in Just (head ys, zs)

Or even better, use Data.List.Spit:

everyNth n = (map head) . chunksOf n

Answer 2

A lot of answers here already use Data.List.unfoldr, but I wanted to bring up an interesting way of writing the somewhat annoying unfold that may be helpful in other contexts:

{-# LANGUAGE TupleSections #-}
import Data.List (unfoldr)
import Data.Maybe (listToMaybe)

every n = unfoldr f . drop (n - 1)
    where f xs = (,drop n xs) <$> listToMaybe xs

When the list is null, listToMaybe returns a Nothing, and fmap similarly will then be Nothing. However, if a Just is produced, then the proper tuple is constructed by turning the head of the list into a tuple of the head and the rest of the values to use in the next iteration.

Answer 3

My version using drop:

every n xs = case drop (n-1) xs of
              (y:ys) -> y : every n ys
              [] -> []

Edit: this also works for finite lists. For infinite lists only, Charles Stewart's solution is slightly shorter.

Answer 4

Alternate solution to get rid of mod:

extractEvery n = map snd . filter ((== n) . fst) . zip (cycle [1..n])

Answer 5

I nuked my version of Haskell the other day, so untested, but the following seems a little simpler than the others (leverages pattern matching and drop to avoid zip and mod):

everynth :: Int -> [a] -> [a]
everynth n xs = y : everynth n ys
         where y : ys = drop (n-1) xs

Answer 6

Starting at the first element:

everyf n [] = []
everyf n as  = head as : everyf n (drop n as)

Starting at the nth element:

every n = everyf n . drop (n-1)