if we add 'a' to the definition of this function, (to make it work), the type of it turns out to be: [a] -> [(a, Bool)]
you might have forgotten the "map fst $" part. Best, On 8 June 2010 14:51, Bill Atkins <watk...@alum.rpi.edu> wrote: > 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 > > > -- Ozgur Akgun
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