haskell - 可以将分区应用于`a -> IO Bool`吗？

Question

我有以下代码：

import System.Directory
import System.FilePath
import Control.Monad (filterM)

filesAndDirs dir = do
  entries <- getDirectoryContents dir
  let filtered = [dir </> e | e <- entries, e `notElem` [".", ".."]]
  files <- filterM doesFileExist filtered 
  dirs <- filterM doesDirectoryExist filtered 
  return (files, dirs)

我想写的是类似的东西return $ partitionM doesFileExist filtered。有没有办法重复使用或提升partition或者是双重使用filterM的最佳方式？

Answer 1

partitionM在Hayoo上搜索将返回至少 2 个实现该功能的库。这意味着您可以依赖它们或研究它们的来源。

这是此实现的更易读的翻译：

partitionM :: (Monad m) => (a -> m Bool) -> [a] -> m ([a], [a])
partitionM p xs = foldM f ([], []) xs
  where 
    f (a, b) x = do
      flag <- p x
      return $ if flag 
        then (x : a, b) 
        else (a, x : b)

---

关于如何将partition功能提升到的问题partitionM，我提出了以下提升功能的实现：

liftSplitter :: (Monad m) =>
  ((a -> Bool) -> [a] -> ([a], [a])) -> 
  (a -> m Bool) -> [a] -> m ([a], [a])
liftSplitter splitter kleisliPredicate list = do
  predicateResultsAndItems <- sequence $ do
    item <- list
    return $ do
      predicateResult <- kleisliPredicate item
      return (predicateResult, item)
  return $ results $ predicateResultsAndItems
  where
    results [] = ([], [])
    results ((predicateResult, item) : tail) = (a ++ tailA, b ++ tailB)
      where
        (a, b) = splitter (const predicateResult) [item]
        (tailA, tailB) = results tail

您可以使用此功能解除所有类型的功能

(a -> Bool) -> [a] -> ([a], [a])

（即，partition和break）span到

(a -> m Bool) -> [a] -> m ([a], [a])