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 :-) -- View this message in context: http://www.nabble.com/bit-of-a-noob-question-tp26043671p26043671.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.