[Haskell-cafe] Removing alternate items from a list

[Sebastiaan Visser]

Or, when lists had a decent eliminator defined in the Prelude (just like maybe for Maybe and either for Either):

 list :: b -> (a -> [a] -> b) -> [a] -> b
 list d _ []     = d 
 list _ f (x:xs) = f x xs

 fromList = list []

we could write the alternate function like this:

 alt :: [a] -> [a] 
 alt = list [] $ \a -> (a:)
     . list [] (const alt)


--
Sebastiaan

  
On Jun 6, 2010, at 4:46 PM, R J wrote:
> 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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