Yes, this was an old draft I accidentally sent out. My post higher up the thread is correct. :)
On Tuesday, June 8, 2010, Ozgur Akgun <ozgurak...@gmail.com> wrote: > 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 > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
