Define functions with foldl foldr - haskell

I understand the definitions of foldl, foldr, but I have problems with functions defined by them.
For example map with foldr:
map f [] = []
map f l = foldr (\x xs -> f x : xs) [] l
I don't understand the (\x xs -> f x : xs). It is the map function, which foldr takes? But shouldn't it be (\x xs -> f x : f xs), because map f (x:xs) = f x : map f xs?
Example with foldl:
concat (x:xs) = x ++ concat xs
concat' xs = foldl (++) [] xs
concat'' xs = foldl (\ys y -> ys ++ y) [] xs
Of course I understand (++), but what's the logic behind (\ys y -> ys ++ y)? Is it ys = [] and y = xs?
So the function takes [] as ys and y is the first element of xs and concates the [] with the y?
Concrete example:
concat'' [1,2,3] = foldl (\ys y -> ys ++ y) [] [1,2,3]
=> foldl (\ys y -> ys ++ y) ((\ys y -> ys ++ y) [] [1]) [2,3]
=> foldl (\ys y -> ys ++ y) [1] [2,3]
=> foldl (\ys y -> ys ++ y) ((\ys y -> ys ++ y) [1] [2]) [3]
=> foldl (\ys y -> ys ++ y) [1,2] [3]
=> foldl (\ys y -> ys ++ y) ((\ys y -> ys ++ y) [1,2] [3]) []
=> foldl (\ys y -> ys ++ y) [1,2,3] []
=> [1,2,3]
Another thing: concat only takes 1 list xs, so if I want to concat 2 lists?
concat (x:xs) ys = x ++ concat xs ys
concat [1,2,3] [4,5,6] with foldl?
Reverse:
reverse (x:xs) = reverse xs ++ [x]
reverse' l = foldl (\xs x -> [x] : xs) [] l
reverse'' l = foldr (\x xs -> xs ++ [x]) [] l
The foldr is intuitive clear (with the questions from above), but what's behind the reverse order in foldl (\xs x -> [x] : xs)? This foldl (\x xs -> xs ++ [x]) [] l would be wrong, wouldn't it?
Thanks a lot!

The code
foldr (\x xs -> ...) end list
could be read, roughly, as follows
scan the whole list
if it's empty, just return end end
otherwise:
let x be the element at hand
let xs be the rest of the list, after having been processed
apply the ... operation
The emphasized part is crucial. xs is not the rest of the list, but the result of the "recursive call" on it.
Indeed, xs is a bad name for that. In thee general case, it's not even a list! E.g. one would never write (silly example)
foldr (\x xs -> x + xs) 0 [1..100] -- sum 1..100
but rather prefer something like
foldr (\x partialSum -> x + partialSum) 0 [1..100] -- sum 1..100
(Actually, one would not sum using foldr, but let's leave that aside.)
So, just read it like this:
map f l = foldr (\x mappedTail -> f x : mappedTail) [] l

Related

What does map' f xs = foldr (\y ys -> (f y):ys) [] xs. mean? [duplicate]

This question already has answers here:
Use foldr to define map in Haskell,
(3 answers)
Closed 3 months ago.
I'm new to Haskell and I want to understand what this syntax means.
This is the context of the function:
map' :: (a -> b) -> [a] -> [b]
map' f xs = foldr (\y ys -> (f y):ys) [] xs
It's defining the map function from the prelude in terms of foldr. I'm confused by what foldr (\y ys -> (f y):ys) [] xs means. Especially the (\y ys -> (f y): ys) part.
In my understanding, y is the first value in the list ys which is the input and the function f is being applied to the y value in the list ys. I understand these values are taken in the context of pattern matching. Kindly correct me if I'm wrong.
Take a look at a simplified definition of foldr (adapted from https://wiki.haskell.org/Foldr_Foldl_Foldl'):
foldr _ z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
The second argument passed to f (ys, in your case) is basically the result of folding the rest of the list, without having to explicitly make the recursive call.
We can compare an explicitly recursive definition of map':
map' f [] = []
map' f (x:xs) = f x : map' xs
to the expansion of your definition of map' using equational reasoning.
map' f (x:xs)
-- by definition of map'
== foldr (\y ys -> f y : ys) [] (x:xs)
-- by definition of foldr
== (\y ys -> f y : ys) x (foldr (\y ys -> f y : ys) [] xs)
-- application of folding function to x...
== (\ys -> f x : ys) (foldr (\y ys -> f y : ys) [] xs)
-- ... and then the recursive fold
== f x : foldr (\y ys -> f y : ys) [] xs
-- by definition of map'
== f x : map' f xs

Using foldr to define map (develop)

Having a hard time understanding fold... Is the expansion correct ? Also would appreciate any links, or analogies that would make fold more digestible.
foldMap :: (a -> b) -> [a] -> [b]
foldMap f [] = []
foldMap f xs = foldr (\x ys -> (f x) : ys) [] xs
b = (\x ys -> (f x):ys)
foldMap (*2) [1,2,3]
= b 1 (b 2 (foldr b [] 3))
= b 1 (b 2 (b 3 ( b [] [])))
= b 1 (b 2 ((*2 3) : []))
= b 1 ((*2 2) : (6 :[]))
= (* 2 1) : (4 : (6 : []))
= 2 : (4 : (6 : []))
First, let's not use the name foldMap since that's already a standard function different from map. If you want to re-implement an existing function with the same or similar semantics, convention is to give it the same name but either in a separate module, or with a prime ' appended to the name. Also, we can omit the empty-list case, since you can just pass that to the fold just as well:
map' :: (a -> b) -> [a] -> [b]
map' f xs = foldr (\x ys -> f x : ys) [] xs
Now if you want to evaluate this function by hand, first just use the definition without inserting anything more:
map' (*2) [1,2,3,4]
≡ let f = (*2)
xs = [1,2,3,4]
in foldr (\x ys -> (f x) : ys) [] xs
≡ foldr (\x ys -> (*2) x : ys) [] [1,2,3,4]
Now just prettify a bit:
≡ foldr (\x ys -> x*2 : ys) [] [1,2,3,4]
Now to evaluate this through, you also need the definition of foldr. It's actually a bit different in GHC, but effectively
foldr _ z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
So with your example
...
≡ foldr (\x ys -> x*2 : ys) [] (1:[2,3,4])
≡ (\x ys -> x*2 : ys) 1 (foldr (\x ys -> x*2 : ys) [] [2,3,4])
Now we can perform a β-reduction:
≡ 1*2 : foldr (\x ys -> x*2 : ys) [] [2,3,4]
≡ 2 : foldr (\x ys -> x*2 : ys) [] [2,3,4]
...and repeat for the recursion.
foldr defines a family of equations,
foldr g n [] = n
foldr g n [x] = g x (foldr g n []) = g x n
foldr g n [x,y] = g x (foldr g n [y]) = g x (g y n)
foldr g n [x,y,z] = g x (foldr g n [y,z]) = g x (g y (g z n))
----- r ---------
and so on. g is a reducer function,
g x r = ....
accepting as x an element of the input list, and as r the result of recursively processing the rest of the input list (as can be seen in the equations).
map, on the other hand, defines a family of equations
map f [] = []
map f [x] = [f x] = (:) (f x) [] = ((:) . f) x []
map f [x,y] = [f x, f y] = ((:) . f) x (((:) . f) y [])
map f [x,y,z] = [f x, f y, f z] = ((:) . f) x (((:) . f) y (((:) . f) z []))
= (:) (f x) ( (:) (f y) ( (:) (f z) []))
The two families simply exactly match with
g = ((:) . f) = (\x -> (:) (f x)) = (\x r -> f x : r)
and n = [], and thus
foldr ((:) . f) [] xs == map f xs
We can prove this rigorously by mathematical induction on the input list's length, following the defining laws of foldr,
foldr g n [] = []
foldr g n (x:xs) = g x (foldr g n xs)
which are the basis for the equations at the top of this post.
Modern Haskell has Fodable type class with its basic fold following the laws of
fold(<>,n) [] = n
fold(<>,n) (xs ++ ys) = fold(<>,n) xs <> fold(<>,n) ys
and the map is naturally defined in its terms as
map f xs = foldMap (\x -> [f x]) xs
turning [x, y, z, ...] into [f x] ++ [f y] ++ [f z] ++ ..., since for lists (<>) == (++). This follows from the equivalence
f x : ys == [f x] ++ ys
This also lets us define filter along the same lines easily, as
filter p xs = foldMap (\x -> [x | p x]) xs
To your specific question, the expansion is correct, except that (*2 x) should be written as ((*2) x), which is the same as (x * 2). (* 2 x) is not a valid Haskell (though valid Lisp :) ).
Functions like (*2) are known as "operator sections" -- the missing argument goes into the empty slot: (* 2) 3 = (3 * 2) = (3 *) 2 = (*) 3 2.
You also asked for some links: see e.g. this, this and this.

Haskell takeWhile + 1

How can I write a takeWhile that would keep the first element that doesn't match the condition?
Example (obviously my example is trickier than this) :
Instead of takeWhile (\× - > x! = 3) [1..10] to return [1,2] I need [1,2,3].
I thought of (takeWhile myFunc myList) ++ [find myFunc myList] but it means I need to go through my list 2 times...
Any idea?
You can use span or break.
λ> span (/=3) [1..10]
([1,2],[3,4,5,6,7,8,9,10])
So you can do something like this:
takeWhileInc :: (a -> Bool) -> [a] -> [a]
takeWhileInc p xs = case zs of [] -> error "not found"
(z:_) -> ys ++ [z]
where
(ys, zs) = span p xs
(Or whatever you want to happen when zs is empty because no 3
was found.)
You can roll your own.
takeWhileOneMore :: (a -> Bool) -> [a] -> [a]
takeWhileOneMore p = foldr (\x ys -> if p x then x:ys else [x]) []
Compare it with
takeWhile :: (a -> Bool) -> [a] -> [a]
takeWhile p = foldr (\x ys -> if p x then x:ys else []) []
Explicit recursion would also be fine for this.
takeWhileOneMore :: (a -> Bool) -> [a] -> [a]
takeWhileOneMore p [] = []
takeWhileOneMore p (x:xs) =
if p x
then x : takeWhileOneMore p xs
else [x]
I like to use the base function more than many people do, such as re-using takeWhile in an intelligent way to get the desired result. For example, you can create a new list of predicates with the first element being True and takeWhile this list is true:
takeWhileP1 p xs = map snd (takeWhile fst (zip (True:map p xs) xs)
This generalizes nicely as well (not necessarily efficient in this form):
takeWhilePlusN n p xs = map snd (takeWhile fst (zip (replicate n True ++ map p xs) xs))
Or perhaps easier to read:
takeWhilePlusN n p xs =
let preds = replicate n True ++ map p xs
annotated = zip preds xs
in map snd (takeWhile fst annotated)
And the result:
*Main> takeWhilePlusN 3 (<5) [1..10]
[1,2,3,4,5,6,7]
*Main> takeWhilePlusN 1 (<5) [1..10]
[1,2,3,4,5]
*Main> takeWhileP1 (<5) [1..10]
[1,2,3,4,5]
*Main> takeWhile (<5) [1..10]
[1,2,3,4]
When the condition fails for a element, instead of terminating with empty list, we can return the element.
takeWhileInclusive :: (a->Bool) -> [a] -> [a]
takeWhileInclusive _ [] = []
takeWhileInclusive predicate (x:xs) = if predicate x
then do (x: takeWhileInclusive predicate xs)
else [x]

Haskell - output a type which takes 2 parametes

I'm trying to make an array of my own type letterCount in form of [('letter',occurance),(),...]. How can I make an output of a type which takes 2 parameters. Here's my code:
type LetterCount = (Char,Int)
letterOccur :: Char->[Char] -> Int
letterOccur c [] = 0
letterOccur c (x:xs) = if (c == x) then ((letterOccur c xs) + 1)
else letterOccur c xs
letterStats :: [Char] -> [LetterCount]
letterStats :: [] = []
letterStats (x:xs) = [x,(letterOccur x (x:xs))] ++ letterStats xs
I'm guessing you're trying to do something like this:
letterStats :: [Char] -> [LetterCount]
letterStats :: [] = []
letterStats (x:xs) = (x, (+1) $ letterOccur x xs) : letterStats xs
All you want to do is add +1 as you are not counting x when checking xs for all x occurances. Also what you want is to return a list of the letterCount type which is a Tuple not a list, therefore I changed
[x, letterOccur x xs] ++ letterStats xs
to
(x, (+1) $ letterOccur x xs) : letterStats xs
Although you could also do this:
[(x, (+1) $ letterOccur x xs)] ++ letterStats xs
But is unnecessary.

Enumerating all pairs of possibly infinite lists [duplicate]

I have a function for finite lists
> kart :: [a] -> [b] -> [(a,b)]
> kart xs ys = [(x,y) | x <- xs, y <- ys]
but how to implement it for infinite lists? I have heard something about Cantor and set theory.
I also found a function like
> genFromPair (e1, e2) = [x*e1 + y*e2 | x <- [0..], y <- [0..]]
But I'm not sure if it helps, because Hugs only gives out pairs without ever stopping.
Thanks for help.
Your first definition, kart xs ys = [(x,y) | x <- xs, y <- ys], is equivalent to
kart xs ys = xs >>= (\x ->
ys >>= (\y -> [(x,y)]))
where
(x:xs) >>= g = g x ++ (xs >>= g)
(x:xs) ++ ys = x : (xs ++ ys)
are sequential operations. Redefine them as alternating operations,
(x:xs) >>/ g = g x +/ (xs >>/ g)
(x:xs) +/ ys = x : (ys +/ xs)
[] +/ ys = ys
and your definition should be good to go for infinite lists as well:
kart_i xs ys = xs >>/ (\x ->
ys >>/ (\y -> [(x,y)]))
testing,
Prelude> take 20 $ kart_i [1..] [101..]
[(1,101),(2,101),(1,102),(3,101),(1,103),(2,102),(1,104),(4,101),(1,105),(2,103)
,(1,106),(3,102),(1,107),(2,104),(1,108),(5,101),(1,109),(2,105),(1,110),(3,103)]
courtesy of "The Reasoned Schemer". (see also conda, condi, conde, condu).
another way, more explicit, is to create separate sub-streams and combine them:
kart_i2 xs ys = foldr g [] [map (x,) ys | x <- xs]
where
g a b = head a : head b : g (tail a) (tail b)
this actually produces exactly the same results. But now we have more control over how we combine the sub-streams. We can be more diagonal:
kart_i3 xs ys = g [] [map (x,) ys | x <- xs]
where -- works both for finite
g [] [] = [] -- and infinite lists
g a b = concatMap (take 1) a
++ g (filter (not . null) (take 1 b ++ map (drop 1) a))
(drop 1 b)
so that now we get
Prelude> take 20 $ kart_i3 [1..] [101..]
[(1,101),(2,101),(1,102),(3,101),(2,102),(1,103),(4,101),(3,102),(2,103),(1,104)
,(5,101),(4,102),(3,103),(2,104),(1,105),(6,101),(5,102),(4,103),(3,104),(2,105)]
With some searching on SO I've also found an answer by Norman Ramsey with seemingly yet another way to generate the sequence, splitting these sub-streams into four areas - top-left tip, top row, left column, and recursively the rest. His merge there is the same as our +/ here.
Your second definition,
genFromPair (e1, e2) = [x*e1 + y*e2 | x <- [0..], y <- [0..]]
is equivalent to just
genFromPair (e1, e2) = [0*e1 + y*e2 | y <- [0..]]
Because the list [0..] is infinite there's no chance for any other value of x to come into play. This is the problem that the above definitions all try to avoid.
Prelude> let kart = (\xs ys -> [(x,y) | ls <- map (\x -> map (\y -> (x,y)) ys) xs, (x,y) <- ls])
Prelude> :t kart
kart :: [t] -> [t1] -> [(t, t1)]
Prelude> take 10 $ kart [0..] [1..]
[(0,1),(0,2),(0,3),(0,4),(0,5),(0,6),(0,7),(0,8),(0,9),(0,10)]
Prelude> take 10 $ kart [0..] [5..10]
[(0,5),(0,6),(0,7),(0,8),(0,9),(0,10),(1,5),(1,6),(1,7),(1,8)]
you can think of the sequel as
0: (0, 0)
/ \
1: (1,0) (0,1)
/ \ / \
2: (2,0) (1, 1) (0,2)
...
Each level can be expressed by level n: [(n,0), (n-1, 1), (n-2, 2), ..., (0, n)]
Doing this to n <- [0..]
We have
cartesianProducts = [(n-m, m) | n<-[0..], m<-[0..n]]

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