Re: [Haskell-beginners] question about list processing

With the join it's the same thing as: computeHead :: (a -> [b]) -> [a] -> [b] computeHead f xs = take 1 xs >>= f On Thu, 12 Nov 2015 at 06:58 Dennis Raddle wrote:

Oh I needed join.

computeHead :: (a -> [b]) -> [a] -> [b] computeHead f = join . take 1 . fmap f

That's because

f :: a -> [b]

not

f: a -> b

On Thu, Nov 12, 2015 at 5:22 AM, Martin Vlk wrote:

...

How about:

computeHead f = take 1 . fmap f

Martin

Dennis Raddle:

...

What would be an elegant way of writing this

computeHead :: (a -> [b]) -> [a] -> [b]

Where when [a] is null, it returns a null list, but when [a] contains one or more elements, it applies the given function to the head of a and returns that? Is there some existing typeclass operator that facilitates this?

You can write

computeHead _ [] = [] computeHead f (x:_) = f x

But that first line seems suspicious to me... it makes me think about how in the list Monad, an empty list falls through. But I can't quite make it work.

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