haskell - 结合功能与列表理解

Question

我有一个简单的问题:

combine :: (a -> b -> c) -> [a] -> [b] -> [c]
combine f (a:as) (b:bs) =  f a b : combine f as bs
combine _  _ _      = [ ]

这是递归的。现在我想使用列表推导来解决同样的问题:

combine f (x:xs) (y:ys) = [ f x y  | x <- (x:xs), y <- (y:ys) ]

但我的问题是元素的组合。我只想结合 x1 y1, x2 y2, xs ys ... 而不是 x1 y1, x1 y2, x1 ys, x2 y1, x2 y2, ......

谢谢!

Answer 1

您需要的是 parallel list compehension .为了能够使用它，您需要为编译器指定一个 ParallelListComp pragma:

{-# LANGUAGE ParallelListComp #-}

combine :: (a -> b -> c) -> [a] -> [b] -> [c]
combine f xs ys = [ f x y | x <- xs | y <- ys ]

编译器将其脱糖到 zipWith 的应用程序中:

combine :: (a -> b -> c) -> [a] -> [b] -> [c]
combine f xs ys = zipWith f xs ys

这实际上是你的功能，所以:

combine :: (a -> b -> c) -> [a] -> [b] -> [c]
combine = zipWith