[Haskell-cafe] Removing alternate items from a list

[Ozgur Akgun]

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 xs
f x = x



On 7 June 2010 20:11, Ozgur Akgun <ozgurakgun at 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 at gmail.com> wrote:
>
>> maybe this?
>>
>> map snd . filter (odd . fst) . zip [1,2..] $ [1,2,3,4,5]
>>
>> 2010/6/6 R J <rj248842 at 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
>



-- 
Ozgur Akgun
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