如何 Maybe-ify 这个递归代码？

Question

我有这段 Haskell 代码可以计算最小二乘回归的成本函数：

cost :: [[Double]] -> [[Double]] -> Double
cost [[]] [[]] = 0
cost ((x:xs):xss) ((y:ys):yss) = (x-y)^2 + cost [xs] [ys] + cost xss yss
cost _ _ = 0

print (cost [[1,2,3],[4,5,6]] [[1,3,3],[4,6,6]]) -- 2.0

这很好用，但现在我想要 return 类型的 Maybe Double。我怎么能那样做？假设最后一个案例 cost _ _ = Nothing。对于此处给出的输入，结果应为 Just 2.0。

Answer 1

为空列表添加一个案例，将所有递归调用转换为绑定，并将 return 应用到每个等式中的最终结果。因此：

cost :: [[Double]] -> [[Double]] -> Maybe Double
cost [] [] = return 0
cost [[]] [[]] = return 0
cost ((x:xs):xss) ((y:ys):yss) = do
    c  <- cost [xs] [ys]
    c' <- cost xss yss
    return ((x-y)^2 + c + c')
cost _ _ = Nothing

但是，案件数量让我担心。你可以考虑写这样的东西，而不是：

import Control.Applicative

costBy :: (a -> a -> Maybe Double) -> [a] -> [a] -> Maybe Double
costBy f [] [] = return 0
costBy f (x:xs) (y:ys) = liftA2 (+) (f x y) (costBy f xs ys)
costBy _ _ _ = Nothing

cost' :: [[Double]] -> [[Double]] -> Maybe Double
cost' = costBy (costBy (\x y -> return ((x-y)^2)))

在这里您不需要 cost [xs] [ys] 恶作剧来重用代码，而是使用多态性在 [Double] 和 [[Double]] 层获得相同的 shape-checking 行为。同样更清楚的是，您已经涵盖了所有感兴趣的案例，因为您一次只处理一层 list-iness。

如何 Maybe-ify 这个递归代码？

How to Maybe-ify this recursive code?

haskell

maybe