or, since you don't need to give a name to the second element of the list:
f :: [a] -> [a] f (x:_:xs) = x : f xs f x = x On 7 June 2010 20:11, Ozgur Akgun <ozgurak...@gmail.com> wrote: > 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 <tphya...@gmail.com> wrote: > >> maybe this? >> >> map snd . filter (odd . fst) . zip [1,2..] $ [1,2,3,4,5] >> >> 2010/6/6 R J <rj248...@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]? >> > >> > ________________________________ >> > The New Busy is not the old busy. Search, chat and e-mail from your >> inbox. >> > Get started. >> > _______________________________________________ >> > Haskell-Cafe mailing list >> > Haskell-Cafe@haskell.org >> > http://www.haskell.org/mailman/listinfo/haskell-cafe >> > >> > >> _______________________________________________ >> Haskell-Cafe mailing list >> Haskell-Cafe@haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> > > > > -- > Ozgur Akgun > -- Ozgur Akgun
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