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

[Galaxy Being]

So basically I can see that the type definitions would seem to deliver the
same thing. I test it

> (concat . (map (map (*5)))) [[1],[2],[3]]
[5,10,15]
> (map (*5) . concat) [[1],[2],[3]]
[5,10,15]

and can also conclude they give the same answer. So is this an example of
referential transparency, i.e., the ability to substitute code and be
assured both forms/expressions deliver the same answer?


On Wed, Apr 7, 2021 at 12:07 PM Akhra Gannon <tanuki at gmail.com> wrote:

> 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
>>
>>
>>
>>
>> _______________________________________________
>> Beginners mailing list
>> Beginners at haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
> _______________________________________________
> Beginners mailing list
> Beginners at haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.haskell.org/pipermail/beginners/attachments/20210408/e6c57fde/attachment.html>