Question

你将如何为Data.Set实现catMaybes？

我想出了：

import qualified Data.Set as Set
import qualified Data.Maybe as Maybe
setCatMaybes a = Set.map Maybe.fromJust . Set.delete Nothing $ a

fnord = Set.fromList [Nothing, Just 41, Just 43, Just 47]

然后我得到以下

setCatMaybes fnord == fromList [41,43,47]

Answer 1

我认为你已经拥有的解决方案可能是最好的解决方案。按照John的解决方案，这里有一个相当短的解决方案：

setCatMaybes :: Ord a => Set.Set (Maybe a) -> Set.Set a
setCatMaybes s = Set.fromAscList [x | Just x <- Set.toAscList s]

或者这是一个更长的，可能更快：

setCatMaybes2 :: Ord a => Set.Set (Maybe a) -> Set.Set a
setCatMaybes2 s
  | Set.null s = Set.empty
  | otherwise  = Set.mapMonotonic Maybe.fromJust $ case Set.deleteFindMin s of
                   (Nothing, s') ->  s'
                   _ -> s

Answer 2

由于Set (Maybe a)是一种奇怪的类型，只有在应用f :: a -> Maybe b之后出现，为什么不一石二鸟并为mapMaybe创建Data.Set ？

import qualified Data.Set
import qualified Data.Maybe

mapMaybe :: Ord b => (a -> Maybe b) -> Data.Set.Set a -> Data.Set.Set b
mapMaybe f = Data.Set.fromList . Data.Maybe.mapMaybe f . Data.Set.toList

通过这种方式，奇怪的Set (Maybe a)永远不会存在。

Answer 3

这个怎么样：

setCatMaybes = Set.fromList . catMaybes

这只需要遍历列表一次，因为Set.fromList是一个很好的消费者。

Haskell：catMaybe for Data.Set？

3 个答案: