Flatten a list of lists

Question

I have to write a function that flattens a list of lists.

For example flatten [] = [] or flatten [1,2,3,4] = [1,2,3,4] or flatten [[1,2],[3],4,5]] = [1,2,3,4,5]

I'm having trouble with the being able to match the type depending on what is given to the flatten function.

Here's what I have:

data A a = B a | C [a] deriving (Show, Eq, Ord)

flatten::(Show a, Eq a, Ord a)=>A a -> A a
flatten (C []) = (C [])
flatten (C (x:xs) ) = (C flatten x) ++ (C flatten xs)
flatten (B a) = (C [a])

From what I can tell the issue is that the ++ operator is expecting a list for both of its arguments and I'm trying to give it something of type A. I've added the A type so the function can either get a single element or a list of elements.

Does anyone know a different way to do this differently, or explain what I can do to fix the type error?

Answer 1

It's a bit unclear what you are asking for, but flattening a list of list is a standard function called concat in the prelude with type signature [[a]] -> [a].

If you make a data type of nested lists as you have started above, maybe you want to adjust your data type to something like this:

 data Lists a = List [a] | ListOfLists [Lists a]

Then you can flatten these to a list;

 flatten :: Lists a -> [a]
 flatten (List xs) = xs
 flatten (ListOfLists xss) = concatMap flatten xss

As a test,

 > flatten (ListOfLists [List [1,2],List [3],ListOfLists [List [4],List[5]]])
 [1,2,3,4,5]

Answer 2

Firstly, the A type is on the right track but I don't think it's quite correct. You want it to be able to flatten arbitrarily nested lists, so a value of type "A a" should be able to contain values of type "A a":

data A a = B a | C [A a]

Secondly, the type of the function should be slightly different. Instead of returning a value of type "A a", you probably want it to return just a list of a, since by definition the function is always returning a flat list. So the type signature is thus:

flatten :: A a -> [a]

Also note that no typeclass constraints are necessary -- this function is completely generic since it does not look at the list's elements at all.

Here's my implementation:

flatten (B a) = [a]
flatten (C []) = []
flatten (C (x:xs)) = flatten x ++ flatten (C xs)

Answer 3

this one liner will do the job. Although as it was mentioned by Malin the type signature is different:

flatten :: [[a]] -> [a]         
flatten xs = (\z n -> foldr (\x y -> foldr z y x) n xs) (:) []

simple test

frege> li = [[3,4,2],[1,9,9],[5,8]]
frege> flatten li
[3,4,2,1,9,9,5,8]