[Haskell-cafe] bit of a noob question
spot135
aca08sas at shef.ac.uk
Sat Oct 24 18:27:50 EDT 2009
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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