Lists of lists of lists - haskell

Question

What is a good way to represent the type LoL a, being a list of lists
of ... of a? The nesting level is arbitrary, but uniform over all
elements of the outer list.
The case I have in mind is to apply a grouping on the members of a
list, and then to apply a next grouping on each subgroup, and so on. It
is not known up front how many groupings one will have to apply. Hence:
rGroupBy :: [(a -> a -> Bool)] -> [a] -> [...[a]...]
Extra brownie points for the type signature of rGroupBy ;-)
Example:
Suppose deweyGroup i groups the elements based on the i-th number
rGroupBy [deweyGroup 1, deweyGroup 2]
["1.1", "1.2.1", "1.2.2", "2.1", "2.2", "3"]
gives:
[ [ [ "1.1" ], [ "1.2.1", "1.2.2" ] ],
[ [ "2.1" ], [ "2.2" ] ],
[ [ "3" ] ]
]
Postscript
One day later, we have 4 excellent and complementary solutions. I'm very pleased with the answers; thank you all.

Answer 1

Another way to enforce the constraint that all branches have equal depth is to use a nested datatype:
data LoL a = One [a] | Many (LoL [a])
mapLoL :: ([a] -> [b]) -> LoL a -> LoL b
mapLoL f (One xs) = One (f xs)
mapLoL f (Many l) = Many $ mapLoL (map f) l
rGroupBy :: [a -> a -> Bool] -> [a] -> LoL a
rGroupBy [] xs = One xs
rGroupBy (f:fs) xs = Many $ mapLoL (groupBy f) $ rGroupBy fs xs
Expanding the definition of LoL, we see that informally,
LoL a = [a] | [[a]] | [[[a]]] | ...
Then we can say, for example:
ghci> rGroupBy [(==) `on` fst, (==) `on` (fst . snd)] [ (i,(j,k)) | i<-[1..3], j<-[1..3], k<-[1..3]]
to get back
Many (Many (One [[[(1,(1,1)),(1,(1,2)),(1,(1,3))]],[[(1,(2,1)),(1,(2,2)),(1,(2,3)), ...

Answer 2

What you actually have is a tree. Try representing it with a recursive data structure:
data LoL a = SoL [a] | MoL [LoL a] deriving (Eq, Show)
rGroupBy :: [(a -> a -> Bool)] -> [a] -> LoL a
rGroupBy (f:fs) = MoL . map (rGroupBy fs) . groupBy f
rGroupBy [] = SoL
deweyGroup :: Int -> String -> String -> Bool
deweyGroup i a b = a!!idx == b!!idx where idx = 2*(i-1)
rGroupBy [deweyGroup 1, deweyGroup 2] ["1.1", "1.2.1", "1.2.2", "2.1", "2.2", "3.0"] gives:
MoL [MoL [SoL ["1.1"],
SoL ["1.2.1","1.2.2"]],
MoL [SoL ["2.1"],
SoL ["2.2"]],
MoL [SoL ["3.0"]]
]

Answer 3

If you want to enforce uniform depth, there is a (fairly) standard trick to do that involving polymorphic recursion. What we'll do is have a spine of "deeper" constructors telling how deeply nested the list is, then a final "here" constructor with the deeply-nested list:
data GroupList a = Deeper (GroupList [a]) | Here a deriving (Eq, Ord, Show, Read)
Actually, the type as defined has one aesthetic choice that you may wish to vary in your code: the Here constructor takes a single a and not a list of as. The consequences of this choice are sort of scattered through the rest of this answer.
Here's an example of a value of this type exhibiting lists-of-lists; it has two Deeper constructors corresponding to the depth-two nesting that it has:
> :t Deeper (Deeper (Here [[1,2,3], []]))
Num a => GroupList a
Here's see a few sample functions.
instance Functor GroupList where
fmap f (Here a ) = Here (f a)
fmap f (Deeper as) = Deeper (fmap (fmap f) as)
-- the inner fmap is at []-type
-- this type signature is not optional
flatten :: GroupList [a] -> GroupList a
flatten (Here a ) = Deeper (Here a)
flatten (Deeper as) = Deeper (flatten as)
singleGrouping :: (a -> a -> Bool) -> GroupList [a] -> GroupList [a]
singleGrouping f = flatten . fmap (groupBy f)
rGroupBy :: [a -> a -> Bool] -> [a] -> GroupList [a]
rGroupBy fs xs = foldr singleGrouping (Here xs) fs

Answer 4

I believe the following example should be close to what you had in mind. First we declare type-level natural numbers. Then we define vectors, which carry their length as a phantom type (see Fixed-length vectors in Haskell, Part 1: Using GADTs). And then we define a structure for nested lists of lists of ... which carries the depth as a phantom type. Finally we can define correctly typed rGroupBy.
{-# LANGUAGE GADTs #-}
{-# LANGUAGE EmptyDataDecls #-}
import Data.List (groupBy)
data Zero
data Succ n
data Vec n a where
Nil :: Vec Zero a
Cons :: a -> Vec n a -> Vec (Succ n) a
data LList n a where
Singleton :: a -> LList Zero a
SuccList :: [LList n a] -> LList (Succ n) a
-- Not very efficient, but enough for this example.
instance Show a => Show (LList n a) where
showsPrec _ (Singleton x) = shows x
showsPrec _ (SuccList lls) = shows lls
rGroupBy :: Vec n (a -> a -> Bool) -> [a] -> LList (Succ n) a
rGroupBy Nil
= SuccList . map Singleton
rGroupBy (Cons f fs)
= SuccList . map (rGroupBy fs) . groupBy f
-- TEST ------------------------------------------------------------
main = do
let input = ["1.1", "1.2.1", "1.2.2", "2.1", "2.2", "3"]
-- don't split anything
print $ rGroupBy Nil input
-- split on 2 levels
print $ rGroupBy (Cons (deweyGroup 1)
(Cons (deweyGroup 2) Nil))
input
where
deweyGroup :: Int -> String -> String -> Bool
deweyGroup i a b = a!!idx == b!!idx where idx = 2*(i-1)

Answer 5

As a type-hackery exercise it is possible to implement this with standard lists.
All we need is an arbitrary depth groupStringsBy function:
{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, FlexibleContexts,
UndecidableInstances, IncoherentInstances,
TypeFamilies, ScopedTypeVariables #-}
import Data.List
import Data.Function
class StringGroupable a b where
groupStringBy :: Pred -> a -> b
instance (StringGroupable a b, r ~ [b]) => StringGroupable [a] r where
groupStringBy f = map (groupStringBy f)
instance (r ~ [[String]]) => StringGroupable [String] r where
groupStringBy p = groupBy p
Which works like this:
*Main> let lst = ["11","11","22","1","2"]
*Main> groupStringBy ((==) `on` length) lst
[["11","11","22"],["1","2"]]
*Main> groupStringBy (==) . groupStringBy ((==) `on` length) $ lst
[[["11","11"],["22"]],[["1"],["2"]]]
So we can use this function directly (although it has to be put in reverse order):
inp = ["1.1", "1.2.1", "1.2.2", "2.1", "2.2", "3"]
deweyGroup :: Int -> String -> String -> Bool
deweyGroup i a b = a!!idx == b!!idx where idx = 2*(i-1)
-- gives: [[["1.1"],["1.2.1","1.2.2"]],[["2.1"],["2.2"]],[["3"]]]
test1 = groupStringBy (deweyGroup 2) . groupStringBy (deweyGroup 1) $ inp
But if you want to use your original sample, we can hack it too.
First we need a variable argument function which pipelines all the arguments but the last one in reverse order via . and then applies the resulting function to the last argument:
class App a b c r where
app :: (a -> b) -> c -> r
instance (b ~ c, App a d n r1, r ~ (n -> r1)) => App a b (c -> d) r where
app c f = \n -> app (f . c) n
instance (a ~ c, r ~ b) => App a b c r where
app c a = c a
Works like this:
*Main> app not not not True
False
*Main> app (+3) (*2) 2
10
Then expand it with a custom rule for our predicate type type Pred = String -> String -> Bool:
type Pred = String -> String -> Bool
instance (StringGroupable b c, App a c n r1, r ~ (n -> r1)) => App a b Pred r where
app c p = app ((groupStringBy p :: b -> c) . c)
And finally wrap it in rGroupBy (supplying id function to be the first in the pipeline):
rGroupBy :: (App [String] [String] Pred r) => Pred -> r
rGroupBy p = app (id :: [String] -> [String]) p
Now it should work for any number of grouping predicates of type Pred producing the list of the depth equal to the number of supplied predicates:
-- gives: [["1.1","1.2.1","1.2.2"],["2.1","2.2"],["3"]]
test2 = rGroupBy (deweyGroup 1) inp
-- gives: [[["1.1"],["1.2.1","1.2.2"]],[["2.1"],["2.2"]],[["3"]]]
test3 = rGroupBy (deweyGroup 1) (deweyGroup 2) inp
-- gives: [[[["1.1"]],[["1.2.1","1.2.2"]]],[[["2.1"]],[["2.2"]]],[[["3"]]]]
test4 = rGroupBy (deweyGroup 1) (deweyGroup 2) (deweyGroup 1) inp
So it is possible (and probably can be simplified) but as always with this sort of hackery is not recommended to be used for anything but the exercise.