haskell - Swap two elements in a list by its indices -

is there way swap 2 elements in list if thing know elements position @ occur in list.

to more specific, looking this:

swapelementsat :: int -> int -> [int] -> [int] 

that behave that:

> swapelementsat 1 3 [5,4,3,2,1] -- swap first , third elements [3,4,5,2,1] 

i thought built-in function might exists in haskell wasn't able find it.

haskell doesn't have such function, because little bit un-functional. trying achieve?

anyway, if want it, can implement own version of (maybe there more idiomatic way write this). note assume i < j, trivial extend function correctly handle other cases:

swapelementsat :: int -> int -> [a] -> [a] swapelementsat j xs = let elemi = xs !!                         elemj = xs !! j                         left = take xs                         middle = take (j - - 1) (drop (i + 1) xs)                         right = drop (j + 1) xs                     in  left ++ [elemj] ++ middle ++ [elemi] ++ right