Re: [Haskell-cafe] Removing alternate items from a list

And for a few more lines of codes, you get a more flexible solution: data Consume = Take | Skip consumeBy :: [Consume] -> [a] -> [a] consumeBy [] _ = [] consumeBy _ [] = [] consumeBy (tOrS:takesAndSkips) (x:xs) = case tOrS of Take -> x : consumeBy takesAndSkips xs Skip -> consumeBy takesAndSkips xs *Main> consumeBy (cycle [Take,Take,Skip]) [1,2,3,4,5,6] [1,2,4,5] *Main> consumeBy (cycle [Take,Take,Take,Skip]) [1,2,3,4,5,6] [1,2,3,5,6] -deech On 6/8/10, Christopher Done wrote:

Can't forget fix in a game of code golf!

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(fix $ \f (x:_: xs) -> x : f xs) [1..] => [1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,4...

2010/6/8 Yitzchak Gale :

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R J wrote:

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What's the cleanest definition for a function f :: [a] -> [a] that takes a list and returns the same list, with alternate items removed?  e.g., f [0, 1, 2, 3, 4, 5] = [1,3,5]?

f = map head . takeWhile (not . null) . iterate (drop 2) . drop 1

Regards, Yitz _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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