如何 - 如果这个递归代码？

我有这个Haskell代码片段来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,但现在我想要一个返回类型Maybe Double.我怎么能这样做？让我们说最后一个案例cost _ _ = Nothing.结果应该是Just 2.0这里给出的输入.

为空列表添加一个案例,将所有递归调用转换为绑定,并应用于return每个等式中的最终结果.从而:

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,案件数量令我担忧.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]shenanigans来重用代码,而是使用多态来在层[Double]和[[Double]]层上获得相同的形状检查行为.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列表.