Question

我尝试将函数edits1应用两次，但在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)

第一个功能是工作，所以错误必须在2'nd函数中，但我无法弄清楚它是什么。我该如何解决？

Answer 1

首先，你应该替换：

edits2 a = nub . f . edits1 a

使用：

edits2 a = nub $ f (edits1 a)

其次，您遇到f (x:xs) = (edits1 x) ++ map (x:) (f xs)：

的问题

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 :)’

根据您的定义，x的类型为String，而xs的类型为[String]

(x:)是一个以[String]为参数的函数，但map会尝试将其应用于[String] String的每个元素。这不起作用。

评论后修改：

edits2功能

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

应替换为此版本：

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

对edits1 a的第一次调用将函数应用于参数。然后它使用折叠第二次调用edits1并连接结果。

应用函数两次编译错误Haskell

1 个答案: