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

[Ut Primum]

Yes, thanks to referential transparency you can substitute
concat . map (map f)
with
map f . concat
in an expression and be sure you'll always get the same result


Il gio 8 apr 2021, 07:04 Galaxy Being <borgauf at gmail.com> ha scritto:

> 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
>>>
>>>
>>>
>>>
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