Update 'x'th element of list - Haskell [duplicate]

问题

Possible Duplicate:
Replace individual list elements in Haskell?

I have managed to make some progress in this part of my assignment but have attached part of the code below that I have made:

module Grid where

data State = On | Off deriving (Eq, Show)

next :: State -> State
next On = Off
next Off = On

type Row = [State]


updateRow :: Row -> Int -> Row
updateRow  (r:rs) x 
    | x == 0     = next r:rs
--  | otherwise     = ........????

As shown in the last line just above, I have managed to get updateRow to work for when x = 0, as shown below (with the 0th element inverted).

*Grid> updateRow [Off,Off,Off,Off] 0
[On,Off,Off,Off]
*Grid>

It all comes unstuck however when I try inverting other elements of this list. I can't seem to 'genralise' a formula in this function.

I also MUST follow THIS type convention:

updateRow :: Row -> Int -> Row

Thanks in advance.

回答1:

Something like that:

module Grid where

data State = On | Off deriving (Eq, Show)

next :: State -> State
next On = Off
next Off = On

type Row = [State]


updateRow :: Row -> Int -> Row
updateRow  (r:rs) x 
    | x == 0     = next r:rs
    | otherwise  = r : (updateRow rs (x-1))
updateRow [] x = []

回答2:

Use the function provided for you in the previous question you asked. It works for lists of any type and I think it does what you want to do here.

回答3:

How about a general update function?

update i a as = map repl $ zip as [0..] where
   repl (a',i') | i == i' = a
                | otherwise = a'

I guess there are more performant versions, but this one is easy to understand and good enough for short lists. It replaces the ith element (if there is any) in as with a.

回答4:

Actually the idea for the otherwise part of your updateRow function is similar to the what you have in the replace of your other question.

The idea is: if x is not zero, then you want to skip the element r (which is at position zero) and call updateRow on rs at a position that is x - something (where something takes into account the one position that you just skipped).

I hope this helps

来源：https://stackoverflow.com/questions/5987945/update-xth-element-of-list-haskell

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