[Haskell-cafe] Removing alternate items from a list

[Bill Atkins]

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 <ozgurakgun at 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 <watkins at alum.rpi.edu> wrote:
>
> f :: [a] -> [a]
> f = filter snd $ zip a (cycle [True, False])
>
> On Monday, June 7, 2010, Ozgur Akgun <ozgurakgun at 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 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
>>
>
>
> --
> Ozgur Akgun
>