CODEWITHSUNDEEP
haskell
one user
one user

Reputation: 109

Delete from sublists

I'm trying to remove a number (if exists) from sublists, using Higher Order Functions.

removeAux :: Int -> [[Int]] -> (Int -> [Int] -> Bool) -> [[Int]]
removeAux num (l:ls) f 
  | f num l = delete num l  -- ERROR in this line
  |otherwise = removeAux num ls f

removeFromSublists :: Int -> [[Int]] -> [[Int]]
removeFromSublists _ [[]] = [[]]
removeFromSublists num ls = removeAux num ls (elem)

And it's suppose to work like this:

> l = [[0,1,2],[2,3]]

> removeFromSublists 1 l
[[0,2],[2,3]]

> removeFromSublists 0 l  $ removeFromSubLists 1 l
[[2],[2,3]]

> removeFromSublists 2 l
[[0,1],[3]]

But I have two problems that I don't know how to solve. First it will not delete all the occurrences of the number and it should? And I can't understand why the code doesn't work? I keep geting the error:

enter image description here

Can someone help me please?

Upvotes: 2

Views: 114

Answers (1)

eric
eric

Reputation: 1059

It could help to take a type-driven approach here. As in, you want to remove all occurrences of some element from a list of list of elements.

This is, "given an a and a list of list of a's ([[a]]) return a list of list of a's ([[a]]):

removeAll :: Eq a => a -> [[a]] -> [[a]]
removeAll v (list:listOfLists) = ???

We're removing all v from each list in the list (list:listOfLists). Another way of saying this is "for each list in our listOfLists filter out v".

So, we want to perform the same action (filter) for-each value in a list. In Haskell, this translates to mapping the filter action over every list in the listOfLists.

removeAll :: Eq a => a -> [[a]] -> [[a]]
removeAll v xs = map (filter (/= v)) xs

Upvotes: 2

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