[Haskell-beginners] List comprehensions with multiple generators

[Ken Overton]

I see now, thanks to both of you; I think I was thrown off by the term
'generators' -- the 'refinement' provided by the second generator is more
like filter rather than map.

I guess if I actually wanted the case all the pairs were evaluated and
resulted in [Bool] where the first occurrence of an item were True, rather
than multiple generators I would have two nested list comprehensions?

Thanks everyone,

On Sun, Apr 26, 2020 at 9:39 AM Francesco Ariis <fa-ml at ariis.it> wrote:

> Hello Ken,
>
> On Sun, Apr 26, 2020 at 08:50:20AM -0400, Ken Overton wrote:
> > I recently came across this function which made me realize I don't
> > understand list comprehensions well. I hope someone can help me
> understand
> > them better by understanding this example better. The function takes a
> list
> > of Eq and returns the list of unique elements from it:
> >
> >     unique :: Eq a => [a] -> [a]
> >     unique xs = [x | (x,y) <- zip xs [0..], x `notElem` (take y xs)]
> >
> >
> > [...]
> >
> > So the first generator should produce [(Eq,Int)] as input to the second
> > generator? And the second generator should produce [Bool]?
>
> 1. (x,y) <- zip xs [0..] -- generates a list of pairs.
> 2. x `notElem` (take y xs) -- acts like a guard to 1., so only the `x`s
>    which are not in the first `y` elements of `xs` (in other words, the
>    previous elements of `xs`) will be returned.
>
> The examples on the wiki [1] show more way of using list
> comprehensions.
>
> [1] https://wiki.haskell.org/List_comprehension#Examples
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-- 
Ken Overton
(917) 863-3937
ken.overton at gmail.com
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