[Haskell-cafe] bit of a noob question

[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])
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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