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 <watkins@alum.rpi.edu> wrote:
f :: [a] -> [a]
f = filter snd $ zip a (cycle [True, False])

On Monday, June 7, 2010, Ozgur Akgun <ozgurakgun@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 <ozgurakgun@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 <tphyahoo@gmail.com> wrote:
> maybe this?
>
> map snd . filter (odd . fst) . zip [1,2..] $ [1,2,3,4,5]
>
> 2010/6/6 R J <rj248842@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]?
>>
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> Ozgur Akgun
>
>
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> Ozgur Akgun
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--
Ozgur Akgun