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

Haskell doesn't have such a function, mainly because it is a little bit un-functional. What are you actually trying to achieve?

You can implement your own version of it (maybe there is a more idiomatic way to write this). Note that I assume that i < j, but it would be trivial to extend the function to correctly handle the other cases:

swapElementsAt :: Int -> Int -> [a] -> [a]
swapElementsAt i j xs = let elemI = xs !! i
                            elemJ = xs !! j
                            left = take i xs
                            middle = take (j - i - 1) (drop (i + 1) xs)
                            right = drop (j + 1) xs
                    in  left ++ [elemJ] ++ middle ++ [elemI] ++ right