haskell - 使用 Maybe 和 Writer 过滤列表并跟踪过滤器命中

Question

我正在使用返回 Maybe 元素的链式函数过滤列表。这部分工作正常。

{-# LANGUAGE TypeSynonymInstances, FlexibleInstances, OverlappingInstances #-}
import Control.Monad
import Control.Monad.Trans.Maybe
import Control.Monad.Writer
import Data.Map (Map, alter, empty, unionWith)

------------------------------------------------

main = do
  let numberList = [1..6]
  let result = filter ((\z -> case z of Just _ -> True; Nothing -> False) . numFilter) numberList
  (putStrLn . show) result

{-
 [2,3,4]
-}

--- Maybe
bigOne :: Int -> Maybe Int
bigOne n | n > 1     = Just n
         | otherwise = Nothing

lessFive :: Int -> Maybe Int
lessFive n | n < 5     = Just n
           | otherwise = Nothing

numFilter :: Int -> Maybe Int
numFilter num = bigOne num
            >>= lessFive

但是我也想统计不同的函数捕获到一个元素的次数。我现在正在使用带有 map 的 Writer 来收集点击量。我尝试将其包装在 MaybeT 中，但这会导致整个过滤器在出现不需要的元素并返回和空列表的情况下失败。

-------------------------------
type FunctionName = String
type Count = Int
type CountMap = Map FunctionName Count

instance Monoid CountMap where
  mempty = empty :: CountMap
  -- default mappend on maps overwrites values with same key,
  -- this increments them
  mappend x y = unionWith (+) x y

{-
  Helper monad to track the filter hits.
-}
type CountWriter = Writer CountMap

incrementCount :: String -> CountMap
incrementCount key = alter addOne key empty

addOne :: Maybe Int -> Maybe Int
addOne Nothing = Just 1
addOne (Just n) = Just (n + 1)

bigOneMW :: Int -> MaybeT CountWriter Int
bigOneMW n | n > 1     = MaybeT $ return (Just n)
           | otherwise = do
                          tell (incrementCount "bigOne")
                          MaybeT $ return Nothing

lessFiveMW :: Int -> MaybeT CountWriter Int
lessFiveMW n | n < 5     = MaybeT $ return (Just n)
             | otherwise = do
                           tell (incrementCount "lessFive")
                           MaybeT $ return Nothing

chainMWBool :: Int -> MaybeT CountWriter Bool
chainMWBool n = do
             a <- bigOneMW n
             b <- lessFiveMW a
             return True

chainerMW :: [Int] -> MaybeT CountWriter [Int]
chainerMW ns = do
               result <- filterM chainMWBool ns
               return result
{-
> runWriter (runMaybeT (chainerMW [1..3]))
(Nothing,fromList [("bigOne",1)])
> runWriter (runMaybeT (chainerMW [2..5]))
(Nothing,fromList [("lessFive",1)])
> runWriter (runMaybeT (chainerMW [2..4]))
(Just [2,3,4],fromList [])
-}

我只是不知道如何让它做我想做的事。我想我正在寻找的类型签名是 [Int] -> CountWriter [Int]，但是当输入是 [1..6] :

([2,3,4], fromList[("bigOne", 1), ("lessFive", 2)])

Answer 1

当你说:

but how to get a result like this when input is [1..6]:

([2,3,4], fromList[("bigOne", 1), ("lessFive", 2)])

换句话说，您需要将列表作为输入并返回列表和 map 作为输出的东西:

newtype Filter a = Filter { runFilter :: [a] -> (CountMap, [a]) }

为什么不直接使用您真正想要的表示对所有过滤器进行编码:

import Data.List (partition)
import qualified Data.Map as M
import Data.Monoid

newtype CountMap = CountMap (M.Map String Int)

instance Show CountMap where
    show (CountMap m) = show m

instance Monoid CountMap where
    mempty = CountMap M.empty
    mappend (CountMap x) (CountMap y) = CountMap (M.unionWith (+) x y)

filterOn :: String -> (a -> Bool) -> Filter a
filterOn str pred = Filter $ \as ->
    let (pass, fail) = partition pred as
    in  (CountMap (M.singleton str (length fail)), pass)

bigOne :: Filter Int
bigOne = filterOn "bigOne" (> 1)

lessFive :: Filter Int
lessFive = filterOn "lessFive" (< 5)

我们遗漏了最后一 block 拼图:如何组合过滤器。好吧，事实证明我们的Filter类型是 Monoid :

instance Monoid (Filter a) where
    mempty = Filter (\as -> (mempty, as))
    mappend (Filter f) (Filter g) = Filter $ \as0 ->
        let (map1, as1) = f as0
            (map2, as2) = g as1
        in  (map1 <> map2, as2)

有经验的读者会认出这只是 State伪装的 monad。

这使得使用 (<>) 组合过滤器变得容易(即 mappend )，我们只需打开我们的 Filter 即可运行它们输入:

ghci> runFilter (bigOne <> lessFive) [1..6]
(fromList [("bigOne",1),("lessFive",2)],[2,3,4])

这表明最佳路径往往是最直接的路径!