There are many ways you can do it. Here are two. The first uses the
Transform List Comp extensions introduced in 6.10.
http://www.haskell.org/ghc/docs/latest/html/users_guide/syntax-extns.html#generalised-list-comprehensions
The second uses more normal Haskell. The second version is probably
not the best 'normal' Haskell implementation though.
{-# LANGUAGE TransformListComp #-}
import Data.Function (on)
import Data.List (groupBy)
import GHC.Exts
test :: (Ord a) => [(a, b)] -> [(a, [b])]
test l = [ (the f, s) | (f,s) <- l , then group by f ]
ex1 = test [('a',1),('a',2),('a',3),('b',1),('b',2)]
test2 :: (Ord a) => [(a, b)] -> [(a, [b])]
test2 l =
map (\grp -> (fst (head grp), map snd grp)) ((groupBy ((==) `on`
fst)) l)
ex2 = test [('a',1),('a',2),('a',3),('b',1),('b',2)]
On Oct 24, 2009, at 5:27 PM, spot135 wrote:
Ok maybe a noob question, but hopefully its an easy one.
This is what I've got so far:
test :: x->[a] -> (b,[b])
test x arrlist = let test1 = x
a = filter (\n -> fst n == test1) arrlist
test2 = map snd a
in (test1, [test2])
so basically I have a list say [(a,1),(a,2),(a,3),(b,1),(b,2)] etc
So I give the function a x value (a or b) in this case and it return
(a,[1,2,3])
which is all gravy
But,
Is there a way that i dont have to supply the a or b ie i call the
function
and it gives me the list
[(a,[1,2,3]),(b,[1,2])...
I presume i need another layer of recursion but I cant figure out
how to do
it.
Any help would be gratefully received :-)
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