[Haskell-beginners] Is map (map f) just map f?

[Akhra Gannon]

Check the types!

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

Therefore:

map f :: [a] -> [b]

map . map :: (a -> b) -> [[a]] -> [[b]]

map (map f) :: [[a]] -> [[b]]

And,

concat :: [[a]] -> [a]

Put it all together and you should see how that rewrite works!


On Wed, Apr 7, 2021, 9:47 AM Galaxy Being <borgauf at gmail.com> wrote:

> I'm in Bird's *Thinking Functionally with Haskell* and the topic is
> natural transformations. He says
>
> filter p . map f = map f . filter (p . f)
>
> and he has a proof, but one step of the proof he goes from
>
> filter p . map f = concat . map (map f) . map (test (p . f))
>
> to
>
> filter p . map f = map f . concat . map (test (p . f))
>
> which means
>
> concat . map (map f) => map f . concat
>
> which means
>
> map (map f) = map f
>
> ... or I'm getting this step wrong somehow. To begin with, I'm having a
> hard time comprehending map(map f), Any ideas on how this is possible?
>
> LB
>
>
>
>
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