Am Sonntag 25 Oktober 2009 00:27:50 schrieb spot135:

Ok maybe a noob question, but hopefully its an easy one.

This is what I've got so far:

test :: x->[a] -> (b,[b])

That can't be. The implementation below has type Eq a => a -> [(a,b)] -> (a,[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 :-)

If the type of the elements of your list belongs to Ord, you should take a look at sort, sortBy (if only the first component belongs to Ord) and groupBy from Data.List. With these functions you can do it pretty easily. If the types don't belong to Ord but just to Eq, a recursion using partition (also from Data.List) works well (but slower).

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#ge... 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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