I want to map a conditional function only on the first item that passes.
map (>5) [1,2,3,4,5,6,7,8,9]
would result in
[False,False,False,False,False,True,True,True,True]
I'm looking for something that would result in
[False,False,False,False,False,True,False,False,False]
So only the first occurrence of being greater than 5 results in True.
I tried scanl, various folds and tried to roll my own mapUntil kind of thing.
Seems like a simple problem but I'm drawing a blank.
break specifically separates the list in 2 parts where the first part is all False, the opposite of span.
break (>5) [1,2,3,8,2,5,1,7,9]
>>> ([1,2,3],[8,2,5,1,7,9])
Then it's just what chi did:
oneTrue f lst = map (const False) a ++ rest b
where (a,b) = break f lst
rest [] = []
rest (x:xs) = True : map (const False) xs
A basic solution:
mapUntil p = onlyOne . map p
where
onlyOne [] = []
onlyOne (x:xs)
| x = True : map (const False) xs
| otherwise = False : onlyOne xs
With library helpers:
mapUntil p = snd . mapAccumL (\x y -> (x||y, not x && y)) False . map p
Above x is a boolean standing for "have seen a true before?", as a kind-of state. y is the list element. x||y is the new state, while not x && y is the new list element.
Alternatively (using Control.Arrow.second):
mapUntil p = uncurry (++) . second go . break id . map p
where
go [] = []
go (x:xs) = x : map (const False) xs
I would use the mapAccumL tool like;
λ> Data.List.mapAccumL (\b n -> if b then (b, (not b)) else (n > 5, n > 5)) False [1,2,3,4,5,6,7,8,9]
(True,[False,False,False,False,False,True,False,False,False])
Here we carry the b as the state of our interim calculations and in every step decide according to it's previous state. Obviously you need the snd part of the final result.
Edit : After reading the new comment of #Gord under his question I decided to extend my answer to cover his true problem.
Rephrasing the case event of branch that starts with pointerPress (x,y) into...
To start with, you never use x or y from the pattern match (x,y) so lets call it c. Then...
PointerPress c -> State circleCoords circleColors circleDraggeds c
where
bools = fmap checkMouseOverlaps $ (,) <$> circleCoords <*> [c]
circleDraggeds = snd $ mapAccumL (\a b -> if a then (a, not a)
else (b,b)) False bools
What's happening part;
(,) <$> circleCoords <*> [c]
circleCoords is a list of coordinates like [c0,c1,c2] and we fmap (the infix version (<$>) here) (,) function to it and it becomes an applicative of coordinates like [(c0,),(c1,),(c2,)]. Then we apply it to [c] aka [(x,y)] to turn it into [(c0,c),(c1,c),(c2,c)].
fmap checkMouseOverlaps $ toAbove
obviously yields to
[checkMouseOverlaps (c0,c), checkMouseOverlaps (c1,c), checkMouseOverlaps (c2,c)]
which is bools :: [Bool].
The the rest follows the logic explained at the top of my answer.
circleDraggeds = snd $ mapAccumL (\a b -> if a then (a, not a)
else (b,b)) False bools
This can be solve directly with recursion. Similar to chi's solution but without function composition
mapUntil :: (a -> Bool) -> [a] -> [Bool]
mapUntil _ [] = []
mapUntil f (x:xs) =
let b = f x -- calculate f x
in if b -- if true
then b : map (const False) xs -- prepend to the solution and map False to the rest of the list (b is True)
else b : mapUntil f xs -- keep applying mapUntil (b is False)
>>> mapUntil (>5) [1,2,3,4,5,6,7,8,9]
[False,False,False,False,False,True,False,False,False]
Map the condition over the list, then zip the result with the False prefix of the result concatenated with a True followed by an infinite list of Falses:
{-# LANGUAGE BlockArguments, ApplicativeDo, ViewPatterns #-}
import Control.Applicative (ZipList(..))
f :: (a -> Bool) -> [a] -> [Bool]
f cond (map cond -> bs) = getZipList do
r <- ZipList $ takeWhile not bs ++ [True] ++ repeat False
_ <- ZipList $ bs
pure r
or, equivalently:
f' :: (a -> Bool) -> [a] -> [Bool]
f' cond (map cond -> bs) = zipWith const (takeWhile not bs ++ [True] ++ repeat False) bs
I have to split the given list into non-empty sub-lists each of which
is either in strictly ascending order, in strictly descending order, or contains all equal elements. For example, [5,6,7,2,1,1,1] should become [[5,6,7],[2,1],[1,1]].
Here is what I have done so far:
splitSort :: Ord a => [a] -> [[a]]
splitSort ns = foldr k [] ns
where
k a [] = [[a]]
k a ns'#(y:ys) | a <= head y = (a:y):ys
| otherwise = [a]:ns'
I think I am quite close but when I use it it outputs [[5,6,7],[2],[1,1,1]] instead of [[5,6,7],[2,1],[1,1]].
Here is a kinda ugly solution, with three reverse in one line of code :).
addElement :: Ord a => a -> [[a]] -> [[a]]
addElement a [] = [[a]]
addElement a (x:xss) = case x of
(x1:x2:xs)
| any (check a x1 x2) [(==),(<),(>)] -> (a:x1:x2:xs):xss
| otherwise -> [a]:(x:xss)
_ -> (a:x):xss
where
check x1 x2 x3 op = (x1 `op` x2) && (x2 `op` x3)
splitSort xs = reverse $ map reverse $ foldr addElement [] (reverse xs)
You can possibly get rid of all the reversing if you modify addElement a bit.
EDIT:
Here is a less reversing version (even works for infinite lists):
splitSort2 [] = []
splitSort2 [x] = [[x]]
splitSort2 (x:y:xys) = (x:y:map snd here):splitSort2 (map snd later)
where
(here,later) = span ((==c) . uncurry compare) (zip (y:xys) xys)
c = compare x y
EDIT 2:
Finally, here is a solution based on a single decorating/undecorating, that avoids comparing any two values more than once and is probably a lot more efficient.
splitSort xs = go (decorate xs) where
decorate :: Ord a => [a] -> [(Ordering,a)]
decorate xs = zipWith (\x y -> (compare x y,y)) (undefined:xs) xs
go :: [(Ordering,a)] -> [[a]]
go ((_,x):(c,y):xys) = let (here, later) = span ((==c) . fst) xys in
(x : y : map snd here) : go later
go xs = map (return . snd) xs -- Deal with both base cases
Every ordered prefix is already in some order, and you don't care in which, as long as it is the longest:
import Data.List (group, unfoldr)
foo :: Ord t => [t] -> [[t]]
foo = unfoldr f
where
f [] = Nothing
f [x] = Just ([x], [])
f xs = Just $ splitAt (length g + 1) xs
where
(g : _) = group $ zipWith compare xs (tail xs)
length can be fused in to make the splitAt count in unary essentially, and thus not be as strict (unnecessarily, as Jonas Duregård rightly commented):
....
f xs = Just $ foldr c z g xs
where
(g : _) = group $ zipWith compare xs (tail xs)
c _ r (x:xs) = let { (a,b) = r xs } in (x:a, b)
z (x:xs) = ([x], xs)
The initial try turned out to be lengthy probably inefficient but i will keep it striked for the sake of integrity with the comments. You best just skip to the end for the answer.
Nice question... but turns out to be a little hard candy. My approach is in segments, those of each i will explain;
import Data.List (groupBy)
splitSort :: Ord a => [a] -> [[a]]
splitSort (x:xs) = (:) <$> (x :) . head <*> tail $ interim
where
pattern = zipWith compare <$> init <*> tail
tuples = zipWith (,) <$> tail <*> pattern
groups = groupBy (\p c -> snd p == snd c) . tuples $ (x:xs)
interim = groups >>= return . map fst
*Main> splitSort [5,6,7,2,1,1,1]
[[5,6,7],[2,1],[1,1]]
The pattern function (zipWith compare <$> init <*> tail) is of type Ord a => [a] -> [Ordering] when fed with [5,6,7,2,1,1,1] compares the init of it by the tail of it by zipWith. So the result would be [LT,LT,GT,GT,EQ,EQ]. This is the pattern we need.
The tuples function will take the tail of our list and will tuple up it's elements with the corresponding elements from the result of pattern. So we will end up with something like [(6,LT),(7,LT),(2,GT),(1,GT),(1,EQ),(1,EQ)].
The groups function utilizes Data.List.groupBy over the second items of the tuples and generates the required sublists such as [[(6,LT),(7,LT)],[(2,GT),(1,GT)],[(1,EQ),(1,EQ)]]
Interim is where we monadically get rid of the Ordering type values and tuples. The result of interim is [[6,7],[2,1],[1,1]].
Finally at the main function body (:) <$> (x :) . head <*> tail $ interim appends the first item of our list (x) to the sublist at head (it has to be there whatever the case) and gloriously present the solution.
Edit: So investigating the [0,1,0,1] resulting [[0,1],[0],[1]] problem that #Jonas Duregård discovered, we can conclude that in the result there shall be no sub lists with a length of 1 except for the last one when singled out. I mean for an input like [0,1,0,1,0,1,0] the above code produces [[0,1],[0],[1],[0],[1],[0]] while it should [[0,1],[0,1],[0,1],[0]]. So I believe adding a squeeze function at the very last stage should correct the logic.
import Data.List (groupBy)
splitSort :: Ord a => [a] -> [[a]]
splitSort [] = []
splitSort [x] = [[x]]
splitSort (x:xs) = squeeze $ (:) <$> (x :) . head <*> tail $ interim
where
pattern = zipWith compare <$> init <*> tail
tuples = zipWith (,) <$> tail <*> pattern
groups = groupBy (\p c -> snd p == snd c) $ tuples (x:xs)
interim = groups >>= return . map fst
squeeze [] = []
squeeze [y] = [y]
squeeze ([n]:[m]:ys) = [n,m] : squeeze ys
squeeze ([n]:(m1:m2:ms):ys) | compare n m1 == compare m1 m2 = (n:m1:m2:ms) : squeeze ys
| otherwise = [n] : (m1:m2:ms) : squeeze ys
squeeze (y:ys) = y : squeeze s
*Main> splitSort [0,1, 0, 1, 0, 1, 0]
[[0,1],[0,1],[0,1],[0]]
*Main> splitSort [5,6,7,2,1,1,1]
[[5,6,7],[2,1],[1,1]]
*Main> splitSort [0,0,1,0,-1]
[[0,0],[1,0,-1]]
Yes; as you will also agree the code has turned out to be a little too lengthy and possibly not so efficient.
The Answer: I have to trust the back of my head when it keeps telling me i am not on the right track. Sometimes, like in this case, the problem reduces down to a single if then else instruction, much simpler than i had initially anticipated.
runner :: Ord a => Maybe Ordering -> [a] -> [[a]]
runner _ [] = []
runner _ [p] = [[p]]
runner mo (p:q:rs) = let mo' = Just (compare p q)
(s:ss) = runner mo' (q:rs)
in if mo == mo' || mo == Nothing then (p:s):ss
else [p] : runner Nothing (q:rs)
splitSort :: Ord a => [a] -> [[a]]
splitSort = runner Nothing
My test cases
*Main> splitSort [0,1, 0, 1, 0, 1, 0]
[[0,1],[0,1],[0,1],[0]]
*Main> splitSort [5,6,7,2,1,1,1]
[[5,6,7],[2,1],[1,1]]
*Main> splitSort [0,0,1,0,-1]
[[0,0],[1,0,-1]]
*Main> splitSort [1,2,3,5,2,0,0,0,-1,-1,0]
[[1,2,3,5],[2,0],[0,0],[-1,-1],[0]]
For this solution I am making the assumption that you want the "longest rally". By that I mean:
splitSort [0, 1, 0, 1] = [[0,1], [0,1]] -- This is OK
splitSort [0, 1, 0, 1] = [[0,1], [0], [1]] -- This is not OK despite of fitting your requirements
Essentially, There are two pieces:
Firstly, split the list in two parts: (a, b). Part a is the longest rally considering the order of the two first elements. Part b is the rest of the list.
Secondly, apply splitSort on b and put all list into one list of list
Taking the longest rally is surprisingly messy but straight. Given the list x:y:xs: by construction x and y will belong to the rally. The elements in xs belonging to the rally depends on whether or not they follow the Ordering of x and y. To check this point, you zip every element with the Ordering is has compared against its previous element and split the list when the Ordering changes. (edge cases are pattern matched) In code:
import Data.List
import Data.Function
-- This function split the list in two (Longest Rally, Rest of the list)
splitSort' :: Ord a => [a] -> ([a], [a])
splitSort' [] = ([], [])
splitSort' (x:[]) = ([x],[])
splitSort' l#(x:y:xs) = case span ( (o ==) . snd) $ zip (y:xs) relativeOrder of
(f, s) -> (x:map fst f, map fst s)
where relativeOrder = zipWith compare (y:xs) l
o = compare y x
-- This applies the previous recursively
splitSort :: Ord a => [a] -> [[a]]
splitSort [] = []
splitSort (x:[]) = [[x]]
splitSort (x:y:[]) = [[x,y]]
splitSort l#(x:y:xs) = fst sl:splitSort (snd sl)
where sl = splitSort' l
I wonder whether this question can be solve using foldr if splits and groups a list from
[5,6,7,2,1,1,1]
to
[[5,6,7],[2,1],[1,1]]
instead of
[[5,6,7],[2],[1,1,1]]
The problem is in each step of foldr, we only know the sorted sub-list on right-hand side and a number to be processed. e.g. after read [1,1] of [5,6,7,2,1,1,1] and next step, we have
1, [[1, 1]]
There are no enough information to determine whether make a new group of 1 or group 1 to [[1,1]]
And therefore, we may construct required sorted sub-lists by reading elements of list from left to right, and why foldl to be used. Here is a solution without optimization of speed.
EDIT:
As the problems that #Jonas Duregård pointed out on comment, some redundant code has been removed, and beware that it is not a efficient solution.
splitSort::Ord a=>[a]->[[a]]
splitSort numList = foldl step [] numList
where step [] n = [[n]]
step sublists n = groupSublist (init sublists) (last sublists) n
groupSublist sublists [n1] n2 = sublists ++ [[n1, n2]]
groupSublist sublists sortedList#(n1:n2:ns) n3
| isEqual n1 n2 = groupIf (isEqual n2 n3) sortedList n3
| isAscen n1 n2 = groupIfNull isAscen sortedList n3
| isDesce n1 n2 = groupIfNull isDesce sortedList n3
| otherwise = mkNewGroup sortedList n3
where groupIfNull check sublist#(n1:n2:ns) n3
| null ns = groupIf (check n2 n3) [n1, n2] n3
| otherwise = groupIf (check (last ns) n3) sublist n3
groupIf isGroup | isGroup = addToGroup
| otherwise = mkNewGroup
addToGroup gp n = sublists ++ [(gp ++ [n])]
mkNewGroup gp n = sublists ++ [gp] ++ [[n]]
isEqual x y = x == y
isAscen x y = x < y
isDesce x y = x > y
My initial thought looks like:
ordruns :: Ord a => [a] -> [[a]]
ordruns = foldr extend []
where
extend a [ ] = [ [a] ]
extend a ( [b] : runs) = [a,b] : runs
extend a (run#(b:c:etc) : runs)
| compare a b == compare b c = (a:run) : runs
| otherwise = [a] : run : runs
This eagerly fills from the right, while maintaining the Ordering in all neighbouring pairs for each sublist. Thus only the first result can end up with a single item in it.
The thought process is this: an Ordering describes the three types of subsequence we're looking for: ascending LT, equal EQ or descending GT. Keeping it the same every time we add on another item means it will match throughout the subsequence. So we know we need to start a new run whenever the Ordering does not match. Furthermore, it's impossible to compare 0 or 1 items, so every run we create contains at least 1 and if there's only 1 we do add the new item.
We could add more rules, such as a preference for filling left or right. A reasonable optimization is to store the ordering for a sequence instead of comparing the leading two items twice per item. And we could also use more expressive types. I also think this version is inefficient (and inapplicable to infinite lists) due to the way it collects from the right; that was mostly so I could use cons (:) to build the lists.
Second thought: I could collect the lists from the left using plain recursion.
ordruns :: Ord a => [a] -> [[a]]
ordruns [] = []
ordruns [a] = [[a]]
ordruns (a1:a2:as) = run:runs
where
runs = ordruns rest
order = compare a1 a2
run = a1:a2:runcontinuation
(runcontinuation, rest) = collectrun a2 order as
collectrun _ _ [] = ([], [])
collectrun last order (a:as)
| order == compare last a =
let (more,rest) = collectrun a order as
in (a:more, rest)
| otherwise = ([], a:as)
More exercises. What if we build the list of comparisons just once, for use in grouping?
import Data.List
ordruns3 [] = []
ordruns3 [a] = [[a]]
ordruns3 xs = unfoldr collectrun marked
where
pairOrder = zipWith compare xs (tail xs)
marked = zip (head pairOrder : pairOrder) xs
collectrun [] = Nothing
collectrun ((o,x):xs) = Just (x:map snd markedgroup, rest)
where (markedgroup, rest) = span ((o==).fst) xs
And then there's the part where there's a groupBy :: (a -> a -> Bool) -> [a] -> [[a]] but no groupOn :: Eq b => (a -> b) -> [a] -> [[a]]. We can use a wrapper type to handle that.
import Data.List
data Grouped t = Grouped Ordering t
instance Eq (Grouped t) where
(Grouped o1 _) == (Grouped o2 _) = o1 == o2
ordruns4 [] = []
ordruns4 [a] = [[a]]
ordruns4 xs = unmarked
where
pairOrder = zipWith compare xs (tail xs)
marked = group $ zipWith Grouped (head pairOrder : pairOrder) xs
unmarked = map (map (\(Grouped _ t) -> t)) marked
Of course, the wrapper type's test can be converted into a function to use groupBy instead:
import Data.List
ordruns5 [] = []
ordruns5 [a] = [[a]]
ordruns5 xs = map (map snd) marked
where
pairOrder = zipWith compare xs (tail xs)
marked = groupBy (\a b -> fst a == fst b) $
zip (head pairOrder : pairOrder) xs
These marking versions arrive at the same decoration concept Jonas Duregård applied.
Suppose I have two lists A and B of the same length. I want to keep elements in A which are greater than corresponding elements in B. Let A=[1,5,8], B=[2,4,9], the result should be [5] because 1<2, 5>4, 8<9.
I come up with a solution. Let C=zip A B, then filter C, finally get result by taking fst of each element in C. It's not so elegant. Is there a simpler way?
Code:
map fst (filter (\ x-> (fst x) > (snd x)) (zip a b))
Your described solution looks fine to me.
An alternative which is not necessarily better:
import Data.Maybe
import Control.Monad
catMaybes $ zipWith (\a b -> guard (a>b) >> return a) list1 list2
According to the desugaring of monad comprehensions this should also work
{-# LANGUAGE MonadComprehensions #-}
[ a | ( a <- list1 | b <- list2 ), a > b ]
... but in practice it does not. It is a pity because I find it quite elegant.
I wonder whether I got it wrong or it is a GHC bug.
I was working on something similar and as a newbie this is the best I came up with:
filterGreaterThan xs ys = do (x,y) <- zip xs ys
guard (x > y)
return x
This solution is easier to reason about than the others. The do notation really shines here.
I'm not sure how your code looks but the following function look quite elegant to me:
greater :: Ord a => [a] -> [a] -> [a]
greater xs = map fst . filter ((>) <$> fst <*> snd) . zip xs
example :: [Int]
example = greater [1,5,8] [2,4,9] -- result is [5]
This pattern is well known in the Lisp community as the decorate-process-undecorate pattern.
A recursive approach, not so elegant as (any) of the other approaches, this relies on no explicit zipping and we get the result in one pass,
greater :: Ord a => [a] -> [a] -> [a]
greater [] [] = []
greater (x:xs) (y:ys)
| x > y = x : greater xs ys
| otherwise = greater xs ys
If you want to generalize this idea nicely, I would recommend looking to mapMaybe:
mapMaybe
:: (a -> Maybe b)
-> [a] -> [b]
Applying that idea to zipWith yields
zipWithMaybe
:: (a -> b -> Maybe c)
-> [a] -> [b] -> [c]
zipWithMaybe f xs ys =
[c | Just c <- zipWith f xs ys]
Now you can write your function
keepGreater :: Ord a => [a] -> [a] -> [a]
keepGreater = zipWithMaybe $
\x y -> x <$ guard (x > y)
Is it really worth the trouble? For lists, probably not. But something like this turns out to be useful in the context of merges for Data.Map.
Pretty similar to #chi's solution with Lists concant:
concat $ zipWith (\a b -> last $ []:[[a] | a > b]) as bs
I want to take a list (or a string) and split it into sub-lists of N elements. How do I do it in Haskell?
Example:
mysteryFunction 2 "abcdefgh"
["ab", "cd", "ef", "gh"]
cabal update
cabal install split
And then use chunksOf from Data.List.Split
Here's one option:
partition :: Int -> [a] -> [[a]]
partition _ [] = []
partition n xs = (take n xs) : (partition n (drop n xs))
And here's a tail recursive version of that function:
partition :: Int -> [a] -> [[a]]
partition n xs = partition' n xs []
where
partition' _ [] acc = reverse acc
partition' n xs acc = partition' n (drop n xs) ((take n xs) : acc)
You could use:
mysteryFunction :: Int -> [a] -> [[a]]
mysteryFunction n list = unfoldr takeList list
where takeList [] = Nothing
takeList l = Just $ splitAt n l
or alternatively:
mysteryFunction :: Int -> [a] -> [[a]]
mysteryFunction n list = unfoldr (\l -> if null l then Nothing else Just $ splitAt n l) list
Note this puts any remaining elements in the last list, for example
mysteryFunction 2 "abcdefg" = ["ab", "cd", "ef", "g"]
import Data.List
import Data.Function
mysteryFunction n = map (map snd) . groupBy ((==) `on` fst) . zip ([0..] >>= replicate n)
... just kidding...
mysteryFunction x "" = []
mysteryFunction x s = take x s : mysteryFunction x (drop x s)
Probably not the elegant solution you had in mind.
There's already
Prelude Data.List> :t either
either :: (a -> c) -> (b -> c) -> Either a b -> c
and
Prelude Data.List> :t maybe
maybe :: b -> (a -> b) -> Maybe a -> b
so there really should be
list :: t -> ([a] -> t) -> [a] -> t
list n _ [] = n
list _ c xs = c xs
as well. With it,
import Data.List (unfoldr)
g n = unfoldr $ list Nothing (Just . splitAt n)
without it,
g n = takeWhile (not.null) . unfoldr (Just . splitAt n)
A fancy answer.
In the answers above you have to use splitAt, which is recursive, too. Let's see how we can build a recursive solution from scratch.
Functor L(X)=1+A*X can map X into a 1 or split it into a pair of A and X, and has List(A) as its minimal fixed point: List(A) can be mapped into 1+A*List(A) and back using a isomorphism; in other words, we have one way to decompose a non-empty list, and only one way to represent a empty list.
Functor F(X)=List(A)+A*X is similar, but the tail of the list is no longer a empty list - "1" - so the functor is able to extract a value A or turn X into a list of As. Then List(A) is its fixed point (but no longer the minimal fixed point), the functor can represent any given list as a List, or as a pair of a element and a list. In effect, any coalgebra can "stop" decomposing the list "at will".
{-# LANGUAGE DeriveFunctor #-}
import Data.Functor.Foldable
data N a x = Z [a] | S a x deriving (Functor)
(which is the same as adding the following trivial instance):
instance Functor (N a) where
fmap f (Z xs) = Z xs
fmap f (S x y) = S x $ f y
Consider the definition of hylomorphism:
hylo :: (f b -> b) -> (c -> f c) -> c -> b
hylo psi phi = psi . fmap (hylo psi phi) . phi
Given a seed value, it uses phi to produce f c, to which fmap applies hylo psi phi recursively, and psi then extracts b from the fmapped structure f b.
A hylomorphism for the pair of (co)algebras for this functor is a splitAt:
splitAt :: Int -> [a] -> ([a],[a])
splitAt n xs = hylo psi phi (n, xs) where
phi (n, []) = Z []
phi (0, xs) = Z xs
phi (n, (x:xs)) = S x (n-1, xs)
This coalgebra extracts a head, as long as there is a head to extract and the counter of extracted elements is not zero. This is because of how the functor was defined: as long as phi produces S x y, hylo will feed y into phi as the next seed; once Z xs is produced, functor no longer applies hylo psi phi to it, and the recursion stops.
At the same time hylo will re-map the structure into a pair of lists:
psi (Z ys) = ([], ys)
psi (S h (t, b)) = (h:t, b)
So now we know how splitAt works. We can extend that to splitList using apomorphism:
splitList :: Int -> [a] -> [[a]]
splitList n xs = apo (hylo psi phi) (n, xs) where
phi (n, []) = Z []
phi (0, xs) = Z xs
phi (n, (x:xs)) = S x (n-1, xs)
psi (Z []) = Cons [] $ Left []
psi (Z ys) = Cons [] $ Right (n, ys)
psi (S h (Cons t b)) = Cons (h:t) b
This time the re-mapping is fitted for use with apomorphism: as long as it is Right, apomorphism will keep using hylo psi phi to produce the next element of the list; if it is Left, it produces the rest of the list in one step (in this case, just finishes off the list with []).