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

Asked 2010-01-08T10:32:04.893
34

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 ...] then getting every 3rd element would result in this list [0,0,0,0,0 ...]

list haskell functional-programming
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2 Answers

16

I don't have anything to test this with at work, but something like:

extractEvery m = map snd . filter (\(x,y) -> (mod x m) == 0) . zip [1..]

should work even on infinite lists.

(Edit: tested and corrected.)

answered 2010-01-08T10:59:11.103
-1

An uglier, and more limited version of the accepted answer

every :: Eq a => Int -> [a] -> [a]
every n xs = if rest == [] 
                then [] 
                else head rest : every n (tail rest)
    where rest = drop (n-1) xs

For "line golfing" it can be written like this:

every n xs = if rest == [] then [] else head rest : every n (tail rest) 
    where rest = drop (n-1) xs

(It's more limited because it has an unnecessary Eq a constraint.)

answered 2011-02-04T21:23:46.040

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