[Haskell-cafe] Removing alternate items from a list

[Ozgur Akgun]

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 at gmail.com> wrote:

> maybe this?
>
> map snd . filter (odd . fst) . zip [1,2..] $ [1,2,3,4,5]
>
> 2010/6/6 R J <rj248842 at 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
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