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

And a very functional one without apparent conditionals:

everyNth :: Int -> [a] -> [a]
everyNth 0 = take 1
everyNth n = foldr ($) [] . zipWith ($) operations where
  operations = cycle $ (:) : replicate (n-1) (const id)

(Note, this one takes every element whose index is a multiple of n. Which is slightly different from the original question, but easy to convert.)