f :: [a] -> [a] f = filter snd $ zip a (cycle [True, False])
On Monday, June 7, 2010, Ozgur Akgun <ozgurak...@gmail.com> wrote: > 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 xsf 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 > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
