使用文件夹Haskell反转列表

Question

我正在尝试通过使用文件夹反转Haskell中的列表列表。 有一个我想做的例子：

> reverse'' [[1,2,3],[3,4,5]]
[[5,4,3],[3,2,1]]

和我的代码（它不起作用）：

reverse'':: [[a]]->[[a]]
reverse'' x = foldr (\n acum -> acum:(foldr(\m acum1 -> acum1++[m])) [] n) [[]] x

我的IDE在第二个文件夹的开头报告了一个错误。

Answer 1

为了分析您当前的解决方案，我将您的一线代码分解为一些辅助函数，并根据其组成推导了它们的类型：

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [[]] x
    where
        f :: [a] -> a -> [a]
        f n acum = acum : reverse n

        reverse :: [a] -> [a]
        reverse n = foldr append [] n

        append :: a -> [a] -> [a]
        append m acum = acum ++ [m]

当我尝试编译上面的代码时，ghc抱怨reverse''类型签名出现以下错误：

Expected type: [[a]] -> [[a]] -> [[a]]
  Actual type: [[a]] ->  [a]  -> [[a]]

我做了一些挖掘，然后为了使“ reverse''具有类型[[a]] -> [[a]] ，函数f需要具有类型[a] -> [[a]] -> [[a]] 。 但是，当前f类型为[a] -> a -> [a] ，或者在这种情况下为[[a]] -> [a] -> [[a]] 。

以下是正确的类型，但是由于累加器的起始值，所以将一个额外的[]值插入到数组的开头：

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [[]] x
    where
        f :: [a] -> [[a]] -> [[a]]
        f n acum = acum ++ [reverse n]

        reverse :: [a] -> [a]
        reverse n = foldr append [] n

        append :: a -> [a] -> [a]
        append m acum = acum ++ [m]

最终的解决方案

通过将初始累加器值更改为空列表[]而不是空列表[[]]列表，我们得到了一个可行的解决方案：

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [] x
    where
        f :: a -> [a] -> [a]
        f n acum = acum ++ [reverse n]

        reverse :: [a] -> [a]
        reverse n = foldr append [] n

        append :: a -> [a] -> [a]
        append m acum = acum ++ [m]

作为单行

如果您真的希望将此作为单线工作，则为：

reverse'' :: [[a]] -> [[a]]
reverse'' = foldr (\n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]) []

与工作：

append m acum = acum ++ [m]
append = \m acum1 -> acum1 ++ [m]

reverse n = foldr append [] n
reverse = \n -> foldr append [] n
reverse = foldr append []
reverse = foldr (\m acum1 -> acum1 ++ [m]) []

f n acum = acum ++ [reverse n]
f = \n acum -> acum ++ [reverse n]
f = \n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [] x
reverse'' x = foldr (\n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]) [] x
reverse'' = foldr (\n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]) []

附录：一个明确的递归解决方案

这是一个显式递归解决方案（即不使用fold ），提供了map和reverse 。

reverse :: [a] -> [a]
reverse (x:xs) = reverse xs ++ [x]
reverse []     = []

map :: (a -> b) -> [a] -> [b]
map f (x:xs) = f x : map f xs
map _ [] = []

reverse'' :: [[a]] -> [[a]]
reverse'' ls = reverse $ map reverse ls