# 99 questions/Solutions/16

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< 99 questions | Solutions(Difference between revisions)

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dropEvery :: Int -> [a] -> [a] |
dropEvery :: Int -> [a] -> [a] |
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dropEvery n xs = snd $ foldl (\acc e -> if fst acc > 1 then (fst acc - 1, snd acc ++ [e]) else (n, snd acc)) (n, []) xs |
dropEvery n xs = snd $ foldl (\acc e -> if fst acc > 1 then (fst acc - 1, snd acc ++ [e]) else (n, snd acc)) (n, []) xs |
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+ | </haskell> |
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+ | |||

+ | Another very similar approach to the previous: |
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+ | <haskell> |
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+ | dropEvery :: [a] -> Int -> [a] |
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+ | dropEvery xs n = fst $ foldr (\x (xs, i) -> (if mod i n == 0 then xs else x:xs, i - 1)) ([], length xs) xs |
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+ | </haskell> |
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+ | |||

+ | Another foldl solution: |
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+ | <haskell> |
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+ | dropEvery :: [a] -> Int -> [a] |
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+ | dropEvery lst n = snd $ foldl helper (1, []) lst |
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+ | where helper (i,acc) x = if n == i |
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+ | then (1,acc) |
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+ | else (i+1,acc++[x]) |
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</haskell> |
</haskell> |

## Revision as of 14:10, 9 September 2012

(**) 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))

A similar iterative solution but using a closure:

dropEvery :: [a] -> Int -> [a] dropEvery xs n = helper xs n where helper [] _ = [] helper (x:xs) 1 = helper xs n helper (x:xs) k = x : helper xs (k-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 = flip $ \n -> map snd . filter ((n/=) . fst) . zip (cycle [1..n])

Using zip and list comprehensions

dropEvery :: [a] -> Int -> [a] dropEvery xs n = [ i | (i,c) <- ( zip xs [1,2..]), (mod c n) /= 0]

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

First thing that came to mind:

dropEvery xs n = map fst $ filter (\(x,i) -> i `mod` n /= 0) $ zip xs [1..]

The filter function can be simplified as seen above:

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

And yet another approach using folds:

dropEvery :: Int -> [a] -> [a] dropEvery n xs = snd $ foldl (\acc e -> if fst acc > 1 then (fst acc - 1, snd acc ++ [e]) else (n, snd acc)) (n, []) xs

Another very similar approach to the previous:

dropEvery :: [a] -> Int -> [a] dropEvery xs n = fst $ foldr (\x (xs, i) -> (if mod i n == 0 then xs else x:xs, i - 1)) ([], length xs) xs

Another foldl solution:

dropEvery :: [a] -> Int -> [a] dropEvery lst n = snd $ foldl helper (1, []) lst where helper (i,acc) x = if n == i then (1,acc) else (i+1,acc++[x])