Remove elements by index in haskell

Question

I\'m new in haskell and I\'m looking for some standard functions to work with lists by indexes.

My exact problem is that i want to remove 3 elements after every 5. If i

Answer 1

Two completely different approaches

1. You can use List.splitAt together with drop:

   import Data.List (splitAt)
   f :: [a] -> [a]
   f [] = []
   f xs = let (h, t) = splitAt 5 xs in h ++ f (drop 3 t)
   

   Now f [1..12] yields [1,2,3,4,5,9,10,11,12]. Note that this function can be expressed more elegantly using uncurry and Control.Arrow.second:

   import Data.List (splitAt)
   import Control.Arrow (second)
   f :: [a] -> [a]
   f [] = []
   f xs = uncurry (++) $ second (f . drop 3) $ splitAt 5 xs
   

   Since we're using Control.Arrow anyway, we can opt to drop splitAt and instead call in the help of Control.Arrow.(&&&), combined with take:

   import Control.Arrow ((&&&))
   f :: [a] -> [a]
   f [] = []
   f xs = uncurry (++) $ (take 5 &&& (f . drop 8)) xs
   

   But now it's clear that an even shorter solution is the following:

   f :: [a] -> [a] 
   f [] = []
   f xs = take 5 xs ++ (f . drop 8) xs
   

   As Chris Lutz notes, this solution can then be generalized as follows:

   nofm :: Int -> Int -> [a] -> [a]
   nofm _ _ [] = []
   nofm n m xs = take n xs ++ (nofm n m . drop m) xs
   

   Now nofm 5 8 yields the required function. Note that a solution with splitAt may still be more efficient!

2. Apply some mathematics using map, snd, filter, mod and zip:

   f :: [a] -> [a]
   f = map snd . filter (\(i, _) -> i `mod` 8 < (5 :: Int)) . zip [0..]
   

   The idea here is that we pair each element in the list with its index, a natural number i. We then remove those elements for which i % 8 > 4. The general version of this solution is:

   nofm :: Int -> Int -> [a] -> [a]
   nofm n m = map snd . filter (\(i, _) -> i `mod` m < n) . zip [0..]