I have a list with this type: [(a, [a])]. For example:

[("a",["x","y"]),("x",["a","y"]),("z",[]),("y",["a","x"])]

I would like to filter this list. The filtering should work like this:

1. start at first item, traverse items from first tuple
2. remove keys that are in the first tuple (so remove "x", "y")

End result should become:

[("a",["x","y"]),("z",[])] -- the "x" and "y" are removed because they live in "a"

The idea is to get a list and then to apply Map.delete from import qualified Data.Map as Map on that list. However, I'm stuck on the things when Maybe is introduced. This is my code:

import System.IO
import Data.Maybe as Maybe
import qualified Data.Map as Map


list = [("a",["x","y"]),("x",["a","y"]),("z",[]),("y",["a","x"])]
    
main :: IO ()
main = do
    let myMap = Map.fromList list
    let myList = funcA list myMap 
    print myList
    
funcA (x:xs) myMap
   | f == Nothing = Nothing
--   | otherwise = f
--   | otherwise = funcC f myMap
   | otherwise = Maybe.maybeToList f
    where a = fst x
          f = funcB a myMap

funcB key myMap = Map.lookup key myMap

funcC portion myMap = portion

Can you help me out?

-------------- edit

After answering the code will look like this:

import System.IO
import Data.Maybe as Maybe
import qualified Data.Map as Map


list = [("a",["x","y"]),("x",["a","y"]),("z",[]),("y",["a","x"])]
    
main :: IO ()
main = do
    let myList = removeKeys list
    print myList


removeKeys :: Eq a => [(a, [a])] -> [(a, [a])]
removeKeys [] = []
removeKeys (kv@(_, vs): kvs) = kv : removeKeys (filter ((`notElem` vs). fst) kvs)

CodePudding user response：

You can make a recursive function that filters the rest of the list with the items in the second item of the 2-tuples, so:

removeKeys :: Eq a => [(a, [a])] -> [(a, [a])]
removeKeys [] = []
removeKeys (kv@(_, vs): kvs) = kv : removeKeys (filter ((`notElem` vs). fst) kvs)

The items in the values that are in the key-value pairs of removed items will not be considered. So if "x" for example had ("x", ["a", "y", "z"]), it will not remove "z", since it was in an item that was already removed.