Haskell Nested List Comprehensions - list

Question

I am studying for an exam and I am looking at an example of nested list comprehensions from the book "Learn you a Haskell", and I was hoping if anyone could explain me step by step how to analyze it and come out with its output.
let xxs = [[1,2,3],[2,3,4],[4,5]]
[ [ x | x <- xs, even x] | xs <- xxs ]]
Output: ([[2],[2,4],[4]])

Answer 1

[ [ x | x <- xs, even x] | xs <- xxs ]
[ [ x | x <- xs, even x] | xs <- [[1,2,3],[2,3,4],[4,5]] ]
[ [ x | x <- [1,2,3], even x] , [ x | x <- [2,3,4], even x] , [ x | x <- [4,5], even x] ]
[filter even [1,2,3], filter even [2,3,4], filter even [4,5]]
[[2],[2,4],[4]]
Or
[ [ x | x <- xs, even x] | xs <- xxs ]
map (\xs -> [ x | x <- xs, even x] ) xxs
map (\xs -> filter even xs) [[1,2,3],[2,3,4],[4,5]]
[filter even [1,2,3], filter even [2,3,4], filter even [4,5]]
[[2],[2,4],[4]]
Note this isn't the transformation that GHC actually does, just a way of writing it that might help you understand the output.

Answer 2

List comprehensions could have been defined by few identities:
[ f x | x <- [], ... ] === []
[ f x | x <- [y], ... ] === [ f y | {y/x}... ] -- well, actually, it's
-- case y of x -> [ f y | {y/x}... ] ; _ -> []
[ f x | x <- xs ++ ys, ...] === [ f x | x <- xs, ...] ++ [ f x | x <- ys, ...]
[ f x | True, ...] === [ f x | ... ]
[ f x | False, ...] === []
The handling of complex patterns (as opposed to simple variable patterns) is elided, only hinted at, for simplicity. {y/x}... means, y is substituted for x in .... For actual definition, see the Report.
It follows that
[ f x | xs <- xss, x <- xs] === concat [ [f x | x <- xs] | xs <- xss]
and
[ f x | x <- xs, test x ] === map f (filter test xs)
Your expression is equivalent to
[ [ x | x <- xs, even x] | xs <- xxs ] -- `]`, sic!
=
[ f xs | xs <- xxs ] where f xs = [ x | x <- xs, even x]
Which is to say, there's nothing special about a list comprehension being used as a value expression in the definition of f. It looks "nested", but actually, it isn't.
What is nested, are the generator expressions separated by commas:
[ x | xs <- xss, x <- xs ] === concat [ [x | x <- xs] | xs <- xss ]
-- ^^^ nested generator
(the equivalency like we saw above.) So then,
[ [ x | x <- xs, even x] | xs <- [[1,2,3],[2,3,4],[4,5]] ]
=
[ [ x | x <- [1,2,3], even x]] ++ [[ x | x <- [2,3,4], even x]] ++ [[ x | x <- [4,5], even x] ]
=
[ [ x | x <- [1,2,3], even x], [ x | x <- [2,3,4], even x], [ x | x <- [4,5], even x] ]
=
[ [ x | x <- [1], even x]++[ x | x <- [2], even x]++[ x | x <- [3], even x]
, [ x | x <- [2], even x]++[ x | x <- [3], even x]++[ x | x <- [4], even x]
, [ x | x <- [4], even x]++[ x | x <- [5], even x] ]
=
[ [ 1 | even 1]++[ 2 | even 2]++[ 3 | even 3]
, [ 2 | even 2]++[ 3 | even 3]++[ 4 | even 4]
, [ 4 | even 4]++[ 5 | even 5] ]
=
[ []++[ 2 ]++[], [ 2 ]++[]++[ 4 ], [ 4 ]++[] ]
=
[ [2], [2,4], [4] ]
Or, with filter if you'd prefer,
[ [ x | x <- [1,2,3], even x], [ x | x <- [2,3,4], even x], [ x | x <- [4,5], even x] ]
=
[ filter even [1,2,3], filter even [2,3,4], filter even [4,5] ]
=
[ [2], [2,4], [4] ]