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
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