Question

I have two lists. One list contains some random data and other list contains the index of first list which needs to be deleted.

For example, let us consider two lists:

let a = [3,4,5,6,6,7,8]
let b = [1,3]

Then, the resultant output should be [3,5,6,7,8]. The number 4 and 6 are deleted since they are on index positions 1 and 3 respectively.

I’m new to Haskell, so finding it difficult to find the solution.

Update: Following code makes it work

import Data.List
dele :: Eq a => [a] -> [Int] -> [a]
dele [] _ = []
dele x [] = x
dele x (y:ys) = dele (delete (x !! y) x) ys

I was just wondering, is there a way to solve it through map/fold way ?

Answer 1

deleteByIndex :: (Enum a, Eq a, Num a) => [a] -> [b] -> [b]
deleteByIndex r = map snd . filter (\(i, _) -> notElem i r) . zip [0..]

[0..] produces an infinite list [0, 1, 2, 3, ...]

zip constructs a list of pairs with the values of this list and your input list in the form [(0,x), (1, y), ...]

filter takes a function a -> Bool. The lambda checks if the index (first element of the pair) is in your input list r.

map snd returns the second element of each pair of the zip list.

zip,filter, map and notElem are documented here