i think explicit recursion is quite clean?


f :: [a] -> [a]
f (x:y:zs) = x : f zs
f x = x



On 7 June 2010 19:42, Thomas Hartman <tphyahoo@gmail.com> wrote:
maybe this?

map snd . filter (odd . fst) . zip [1,2..] $ [1,2,3,4,5]

2010/6/6 R J <rj248842@hotmail.com>:
> 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]?
>
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Ozgur Akgun