Haskell equivalent to Scala's groupBy - haskell

Question

Scala has a function groupBy on lists that accepts a function for extracting keys from list items, and returns another list where the items are tuples consisting of the key and the list of items producing that key. In other words, something like this:
List(1,2,3,4,5,6,7,8,9).groupBy(_ % 2)
// List((0, List(2,4,6,8)), (1, List(1,3,5,7,9)))
(Actually, it looks like in current versions it provides a Map instead, but that's not important). C# has an even more useful version that lets you map the values at the same time (very useful if, say, your key function is just extracting part of a tuple).
Haskell has a groupBy, but it's somewhat different - it groups runs of things according to some comparison function.
Before I go and write it, is there an equivalent of Scala's groupBy in Haskell? Hoogle doesn't have anything for what I'd expect the signature to look like (below), but I may have just got it wrong.
Eq b => (a -> b) -> [a] -> [(b,[a])]

Answer 1

You can write the function yourself rather easily, but you need to place an Ord or Hashable constraint on the result of the classifier function if you want an efficient solution. Example:
import Control.Arrow ((&&&))
import Data.List
import Data.Function
myGroupBy :: (Ord b) => (a -> b) -> [a] -> [(b, [a])]
myGroupBy f = map (f . head &&& id)
. groupBy ((==) `on` f)
. sortBy (compare `on` f)
> myGroupBy (`mod` 2) [1..9]
[(0,[2,4,6,8]),(1,[1,3,5,7,9])]
You can also use a hash map like Data.HashMap.Strict instead of sorting for expected linear time.

Answer 2

Specifically, the following should work:
scalaGroupBy f = groupBy ((==) `on` f) . sortBy (comparing f)
modulo that this doesn't get you the result of f in each group, but if you really need it you can always post-process with
map (\xs -> (f (head xs), xs)) . scalaGroupBy f

Answer 3

This isn't a function in the List library.
You can write it as the composition of sortBy and groupBy.

Answer 4

Putting a trace in f reveals that, with #Niklas solution, f is evaluated 3 times for each element on any list of length 2 or more. I took the liberty of modifying it so that f is applied to each element only once. It's not clear however whether the cost of creating and destroying tuples is less than the cost of evaluating f multiple times (since f can be arbitrary).
import Control.Arrow ((&&&))
import Data.List
import Data.Function
myGroupBy' :: (Ord b) => (a -> b) -> [a] -> [(b, [a])]
myGroupBy' f = map (fst . head &&& map snd)
. groupBy ((==) `on` fst)
. sortBy (compare `on` fst)
. map (f &&& id)

Answer 5

This solution will break and group by on (f x), regardless wether it is sorted or not
f = (`mod` (2::Int))
list = [1,3,4,6,8,9] :: [Int]
myGroupBy :: Eq t => (b -> t) -> [b] -> [(t, [b])]
myGroupBy f (z:zs) = reverse $ foldl (g f) [(f z,[z])] zs
where
-- folding function
g f ((tx, xs):previous) y = if (tx == ty)
then (tx, y:xs):previous
else (ty, [y]):(tx, reverse xs):previous
where ty = f y
main = print $ myGroupBy f list
result:
[(1,[1,3]),(0,[4,6,8]),(1,[9])]

Answer 6

Since Scala groupBy returns an immutable HashMap, which does not require ordering, the corresponding Haskell implementation should return a HashMap as well.
import qualified Data.HashMap.Strict as M
scalaGroupBy :: (Eq k, Hashable k) => (v -> k) -> [v] -> M.HashMap k [v]
scalaGroupBy f l = M.fromListWith (++) [ (f a, [a]) | a <- l]

Answer 7

We can also use the SQL-like then group by syntax in list comprehension, which requires TransformListComp language extension.
Since Scala groupBy returns a Map, we can call fromDistinctAscList to convert the list comprehension to a Map.
$ stack repl --package containers
Prelude> :set -XTransformListComp
Prelude> import Data.Map.Strict ( fromDistinctAscList, Map )
Prelude Data.Map.Strict> import GHC.Exts ( groupWith, the )
Prelude Data.Map.Strict GHC.Exts> :{
Prelude Data.Map.Strict GHC.Exts| scalaGroupBy f l =
Prelude Data.Map.Strict GHC.Exts| fromDistinctAscList
Prelude Data.Map.Strict GHC.Exts| [ (the key, value)
Prelude Data.Map.Strict GHC.Exts| | value <- l
Prelude Data.Map.Strict GHC.Exts| , let key = f value
Prelude Data.Map.Strict GHC.Exts| , then group by key using groupWith
Prelude Data.Map.Strict GHC.Exts| ]
Prelude Data.Map.Strict GHC.Exts| :}
Prelude Data.Map.Strict GHC.Exts> :type scalaGroupBy
scalaGroupBy :: Ord b => (t -> b) -> [t] -> Map b [t]
Prelude Data.Map.Strict GHC.Exts> scalaGroupBy (`mod` 2) [1, 2, 3, 4, 5, 6, 7, 8, 9]
fromList [(0,[2,4,6,8]),(1,[1,3,5,7,9])]
The only difference from Scala groupBy is that the above implementation returns a sorted map instead of a hash map. For implementation that returns a hash map, see my other answer at https://stackoverflow.com/a/64204797/955091.