scala - "Flattening"Scala 和 Haskell 中的列表

Question

给定一个 List[Option[Int]] :

scala> list
res8: List[Option[Int]] = List(Some(1), Some(2), None)

我可以得到 List(1,2) ，即提取 list通过 flatMap和 flatten :

scala> list.flatten
res9: List[Int] = List(1, 2)

scala> list.flatMap(x => x)
res10: List[Int] = List(1, 2)

鉴于以下 [Maybe Int]在 Haskell 中，我该如何执行上述操作？

我尝试了以下失败:

import Control.Monad

maybeToList :: Maybe a -> [b]
maybeToList Just x  = [x]
maybeToList Nothing = []

flatten' :: [Maybe a] -> [a]
flatten' xs = xs >>= (\y -> y >>= maybeToList)

Answer 1

您可以使用 catMaybes :

import Data.Maybe
catMaybes xs

如果你想使用 >>=你需要一个函数 Maybe a -> [a] .这是 maybeToList :

xs >>= maybeToList

正如评论指出的那样，您可以转换任何 Foldable到列表中，以便您可以制作 flatten'更一般的:

flatten' :: Foldable f => [f a] -> [a]
flatten' xs = xs >>= toList