Re: [Haskell-cafe] Removing alternate items from a list

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 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 wrote:

f :: [a] -> [a] f = filter snd $ zip a (cycle [True, False])

On Monday, June 7, 2010, Ozgur Akgun 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 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 wrote: maybe this?

map snd . filter (odd . fst) . zip [1,2..] $ [1,2,3,4,5]

2010/6/6 R J :

...

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