how to modify the i-th element of a Haskell List?

问题

I'm looking for a way to modify the i-th element of haskell List. let says foobar is such a function, then the following works.

let xs = ["a","b","c","d"]
foobar xs 2 "baba" -- xs = ["a","b","baba","d"]

thanks for any reply!

回答1:

You can do it with splitAt:

Prelude> let xs = ["a","b","c","d"]
Prelude> (\(l,_:r)->l++"baba":r) $ splitAt 2 xs
["a","b","baba","d"]

回答2:

let xs = ["a","b","c","d"]
take 2 xs ++ ["baba"] ++ drop 3 xs

回答3:

change n x = zipWith (\k e -> if k == n then x else e) [0..]

回答4:

A simple function to do it directly:

replaceAt _ _ []     = []
replaceAt 0 x (_:ys) = x:ys
replaceAt n x (y:ys) = y:replaceAt (n - 1) x ys

来源：https://stackoverflow.com/questions/9220022/how-to-modify-the-i-th-element-of-a-haskell-list