CODEWITHSUNDEEP
listhaskell
JohnnyGat
JohnnyGat

Reputation: 335

Splitting lists inside list haskell

Hi I need to split list by an argument in Haskell. I found function like this

group :: Int -> [a] -> [[a]]
group _ [] = []
group n l
  | n > 0 = (take n l) : (group n (drop n l))
  | otherwise = error "Negative n"

But what if lists that I want to divide are contained by another list?

For example

group 3 [[1,2,3,4,5,6],[2,4,6,8,10,12]]

should return

[[[1,2,3],[4,5,6]],[[2,4,6],[8,10,12]]]

Is there any way to do that ?

Upvotes: 1

Views: 399

Answers (2)

amnn
amnn

Reputation: 3716

You want this:

group n = map (toList . splitAt n)
  where
    toList (x,y) = [x,y]

UPDATE Given that the specification involves repeated taking and dropping here's how I would do it:

import Data.List (unfoldr)
import Data.Maybe (listToMaybe)

group :: Int -> [[a]] -> [[[a]]]
group n = map (unfoldr split)
  where
    split = (\s -> listToMaybe (fst s) >> Just s) . splitAt n

You'll notice that inside here is also the definition of what you have implemented as group. It is unfoldr split.

Upvotes: 0

Manu343726
Manu343726

Reputation: 14174

Just do map over the elements (lists) of that list:

grouplists :: [[a]] -> Int -> [[[a]]]
grouplists input n = map (group n) input

Upvotes: 3

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