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

That's how I solved it:

swapElementsAt :: Int -> Int -> [a] -> [a]
swapElementsAt a b list = list1 ++ [list !! b] ++ list2 ++ [list !! a] ++ list3
    where   list1 = take a list;
            list2 = drop (succ a) (take b list);
            list3 = drop (succ b) list

Here I used the convention that position 0 is the first. My function expects a<=b.

What I like most in my program is the line take a list.

Edit: If you want to get more such cool lines, look at this code:

swapElementsAt :: Int -> Int -> [a] -> [a]
swapElementsAt a another list = list1 ++ [list !! another] ++ list2 ++ [list !! a] ++ list3
    where   list1 = take a list;
            list2 = drop (succ a) (take another list);
            list3 = drop (succ another) list