Swap two elements in a list by its indices

Question

Is there any way to swap two elements in a list if the only thing I know about the elements is the position at which they occur in the list.

To be more specific, I a

Answer 1

This is a strange thing to do, but this should work, aside from the off-by-one errors you'll have to fix since I'm writing this on my phone. This version avoids going over the same segments of the list any more times than necessary.

swap' :: Int -> Int -> [a] -> [a]
swap' first second lst = beginning ++ [y] ++ middle ++ [x] ++ end
  where
    (beginning, (x : r)) = splitAt first lst
    (middle, (y : end)) = splitAt (second - first - 1) r

swap x y | x == y = id
         | otherwise = swap' (min x y) (max x y)