haskell 减半函数

Question

使用库函数，定义一个函数halve :: [a ] → ([a ], [a ]) 将偶数长度列表分成两半。 例如：

> halve [1, 2, 3, 4, 5, 6]
([1, 2, 3], [4, 5, 6])

到目前为止我所拥有的是

halve :: [a] -> ([a],[a])
halve = (\xs -> case xs of
        [] -> ([],[])
        xs -> take ((length xs) `div` 2 ) xs)

这是错误的，因为 xs -> take ((length x) div 2 ) xs 只显示列表的前半部分...请帮助我继续，以便它显示列表的后半部分。

Answer 1

我在Graham Hutton 的Programming in Haskell 中遇到了同样的问题。 我的解决方案是：

halve :: [a] -> ([a], [a]) 
halve xs = 
    ((take s xs), (drop s xs))
    where
        s = (length xs ) `div` 2

曾经给我带来一些麻烦的小事是意识到我需要使用div而不是(/)因为length :: Foldable t => ta -> Int and (/) :: Fractional a => a -> a -> a

Answer 2

感谢您评论一些解决方案。 我解决了……在这里

first_halve :: [a] -> [a]
first_halve = (\xs -> case xs of
            [] -> []
            xs -> take ((length xs) `div` 2 ) xs)

second_halve :: [a] -> [a]
second_halve = (\xs -> case xs of
            [] -> []
            xs -> drop ((length xs) `div` 2 ) xs)

halve :: [a] -> ([a],[a])
halve = (\xs -> case xs of
            [] -> ([],[])
            xs -> (first_halve xs, second_halve xs))

Answer 3

Hm... five years later and I'm probably working through the same text: 
Programming in Haskell 1st edition. Your question gave me the hint I 
needed to solve the problem:
halve :: [a] -> ([a], [a])
halve xs = (take l xs, drop l xs) 
           where l = div (length xs) 2