# 99 questions/Solutions/16

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< 99 questions | Solutions(Difference between revisions)

Line 49: | Line 49: | ||

<haskell> | <haskell> | ||

− | dropEvery xs n = take (n-1) xs ++ dropEvery (drop n | + | dropEvery xs n |

+ | |n <= 0 = x | ||

+ | |x == [] = [] | ||

+ | |otherwise = take (n-1) xs ++ dropEvery (drop n xs) n | ||

</haskell> | </haskell> | ||

## Revision as of 02:49, 2 September 2010

(**) Drop every N'th element from a list.

dropEvery :: [a] -> Int -> [a] dropEvery [] _ = [] dropEvery (x:xs) n = dropEvery' (x:xs) n 1 where dropEvery' (x:xs) n i = (if (n `divides` i) then [] else [x]) ++ (dropEvery' xs n (i+1)) dropEvery' [] _ _ = [] divides x y = y `mod` x == 0

An alternative iterative solution:

dropEvery :: [a] -> Int -> [a] dropEvery list count = helper list count count where helper [] _ _ = [] helper (x:xs) count 1 = helper xs count count helper (x:xs) count n = x : (helper xs count (n - 1))

Yet another iterative solution which divides lists using Prelude:

dropEvery :: [a] -> Int -> [a] dropEvery [] _ = [] dropEvery list count = (take (count-1) list) ++ dropEvery (drop count list) count

A similar approach using guards:

dropEvery :: [a] -> Int -> [a] dropEvery xs n | length xs < n = xs | otherwise = take (n-1) xs ++ dropEvery (drop n xs) n

Using zip:

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

Using take and drop:

dropEvery xs n |n <= 0 = x |x == [] = [] |otherwise = take (n-1) xs ++ dropEvery (drop n xs) n

A more complicated approach which first divides the input list into sublists that do not contain the nth element, and then concatenates the sublists to a result list (if not apparent: the author's a novice):

dropEvery :: [a] -> Int -> [a] dropEvery [] _ = [] dropEvery xs n = concat (split n xs) where split _ [] = [] split n xs = fst splitted : split n ((safetail . snd) splitted) where splitted = splitAt (n-1) xs safetail xs | null xs = [] | otherwise = tail xs