haskell函数优化

Question

下面的函数计算元音和辅音的数量：

    '''
    countVowelConsonant :: [Char] -> Int
    countVowelConsonant cs = count v c cs
        where
            v = 0
            c = 0
        count :: Int -> Int -> [Char] -> Int
        count v c [] = comp v c 
            where 
                comp v c
                    | v == c = 0
                    | v > c = v
                    | otherwise = c
        count v c (a:cs)
        | or [a == 'a', a == 'e', a == 'i', a == 'o', a == 'u'] = count (v+1) c cs
        | otherwise = count v (c+1) cs

    main :: IO()
    main = do
    print $ countVowelConsonant ['a','l','l']
    '''

如何进一步优化，看起来有点乱

Answer 1

代码片段如下：

countVowelConsonant :: [Char] -> Int
    countVowelConsonant cs = count v c cs
        where
            v = 0
            c = 0

看起来很奇怪：在Haskell中，变量是不可变的，所以你不能递增或递减

v

或

c

，所以定义一个变量

v = 0

，然后将它传递给函数是非常奇怪的。

但我们并不真正需要这个。我们只需要实现一个元音检查功能即可：

isVowel :: Char -> Bool
isVowel 'a' = True
isVowel 'e' = True
isVowel 'i' = True
isVowel 'o' = True
isVowel 'u' = True
isVowel 'A' = True
isVowel 'E' = True
isVowel 'I' = True
isVowel 'O' = True
isVowel 'U' = True
isVowel _ = False

然后我们计算元音和辅音的数量：

countBoth :: (a -> Bool) -> [a] -> (Int, Int)
countBoth pred = go 0 0
  where
    go t f [] = (t, f)
    go t f (x : xs)
      | pred x = t `seq` go (t + 1) f xs
      | otherwise = f `seq` go t (f + 1) xs

然后我们可以对结果进行后处理，例如：

countVowelConsonant :: String -> Int
countVowelConsonant cs = go (countBoth isVowel cs)
  where
    go (v, c)
      | v == c = 0
      | otherwise = max v c