我尝试两次应用函数edits1，但是在f的定义中输入“=”时出现解析错误

delete1 :: String -> [String]
delete1 [] = []
delete1 (x:xs) = xs : map (x:) (delete1 xs)

replaced12 :: String -> [String]
replaced12 [] = []
replaced12 (x:xs) = [h c | c <- ['a'..'z']] ++ map (x:) (replaced12 xs)
         where h :: Char -> String
               h c = c:xs
replaced1 :: String -> [String]
replaced1 a = delete a $ replaced12 a

inserted1 :: String -> [String]
inserted1 [] = [] 
inserted1 (x:xs) = [ x:c:xs | c <- ['a'..'z']] ++ map (x:) (inserted1 xs)

transposed1 :: String -> [String]
transposed1 (x:y:xs) = (y:x:xs) : map (x:) (transposed1 (y:xs))
transposed1      xs  = []

edits1 :: String -> [String]
edits1 a = nub $ (delete1 a ++ replaced1 a ++ inserted1 a ++ transposed1 a)


edits2 :: String -> [String]
edits2 a = nub . f . edits1 a
         where f :: [String] -> [String]
               f []     = []
               f (x:xs) = (edits1 x) ++ map (x:) (f xs)

最佳答案

首先，您应该替换:

edits2 a = nub . f . edits1 a

edits2 a = nub $ f (edits1 a)

Couldn't match type ‘[Char]’ with ‘Char’
Expected type: Char
  Actual type: String
In the first argument of ‘(:)’, namely ‘x’
In the first argument of ‘map’, namely ‘(x :)’

edits2 :: String -> [String]
edits2 a = nub . f . edits1 a
         where f :: [String] -> [String]
               f []     = []
               f (x:xs) = (edits1 x) ++ map (x:) (f xs)

edits2 :: String -> [String]
edits2 a = nub $ foldl (\acc x -> acc ++ edits1 x) [] (edits1 a)

https://stackoverflow.com/questions/26951987/