haskell 減半函數

Question

使用庫函數，定義一個函數halve :: [a ] → ([a ], [a ]) 將偶數長度列表分成兩半。 例如：

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

到目前為止我所擁有的是

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

這是錯誤的，因為 xs -> take ((length x) div 2 ) xs 只顯示列表的前半部分...請幫助我繼續，以便它顯示列表的后半部分。

Answer 1

我在Graham Hutton 的Programming in Haskell 中遇到了同樣的問題。 我的解決方案是：

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

曾經給我帶來一些麻煩的小事是意識到我需要使用div而不是(/)因為length :: Foldable t => ta -> Int and (/) :: Fractional a => a -> a -> a

Answer 2

感謝您評論一些解決方案。 我解決了……在這里

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