Difference between revisions of "99 questions/Solutions/16"
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(a slight variation using guards) |
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</haskell> |
</haskell> |
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− | + | An alternative iterative solution: |
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+ | |||
<haskell> |
<haskell> |
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dropEvery :: [a] -> Int -> [a] |
dropEvery :: [a] -> Int -> [a] |
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</haskell> |
</haskell> |
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− | + | Yet another iterative solution which divides lists using Prelude: |
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+ | |||
<haskell> |
<haskell> |
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dropEvery :: [a] -> Int -> [a] |
dropEvery :: [a] -> Int -> [a] |
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Line 29: | Line 31: | ||
</haskell> |
</haskell> |
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− | + | A similar approach using guards: |
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+ | |||
+ | <haskell> |
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+ | dropEvery :: [a] -> Int -> [a] |
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+ | dropEvery xs n |
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+ | | length xs < n = xs |
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+ | | otherwise = take (n-1) xs ++ dropEvery (drop n xs) n |
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+ | </haskell> |
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+ | |||
+ | Using zip: |
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+ | |||
<haskell> |
<haskell> |
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dropEvery n = map snd . filter ((n/=) . fst) . zip (cycle [1..n]) |
dropEvery n = map snd . filter ((n/=) . fst) . zip (cycle [1..n]) |
Revision as of 20:01, 15 July 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])