haskell - 折叠在绳子上

Question

我想折叠一个字符串，以便 @ 前面出现的任何零都被替换为“k”。因此“a.@0.1.2.0”变为“a.@k.1.2.0”。我怎么做？到目前为止我的尝试是

test = foldl(\x acc-> if((last acc) == "@" && x == "0" then acc ++ "k" else acc ++ x)) "" "a.@0.1.2.0"

但是，它不起作用。 Foldl 需要字符串列表，而我提供的只是一个字符串。我该如何克服这个问题？

Answer 1

听取chi的建议，

rep "" = ""
rep ('@' : '0' : xs) = "@k" ++ rep xs
rep (x : xs) = x : rep xs

如果我们想变得更优秀，

rep = snd . mapAccumL go False
  where
    go True '0' = (False, 'k')
    go _ '@' = (True, '@')
    go _ x = (False, x)

甚至

rep = snd . mapAccumL go 'x'
  where
    go '@' '0' = ('0', 'k')
    go _ y = (y, y)

将 foldr 与第二种方法一起使用(只是因为它更短；第一种方法也可以正常工作，并且允许泛化)，

rep xs = foldr go (const "") xs 'x'
  where
    go '0' r '@' = 'k' : r '0'
    go x r _ = x : r x

使用zipWith(这更难以概括):

rep xs = zipWith go ('x' : xs) xs where
  go '@' '0' = 'k'
  go _ x = x