[英]Limiting pipes based on time?
是否可以創建管道以獲取在特定時間段內已向下游發送的所有值? 我正在實現一個服務器,該協議允許我連接傳出的數據包並將它們壓縮在一起,因此我想每隔100ms有效地“清空”下游ByteString
的隊列, mappend
它們mappend
在一起,然后產生下一個壓縮的管道。
這是使用pipes-concurrency
的解決方案。 您給它任何Input
,它將定期消耗所有值的輸入:
import Control.Applicative ((<|>))
import Control.Concurrent (threadDelay)
import Data.Foldable (forM_)
import Pipes
import Pipes.Concurrent
drainAll :: Input a -> STM (Maybe [a])
drainAll i = do
ma <- recv i
case ma of
Nothing -> return Nothing
Just a -> loop (a:)
where
loop diffAs = do
ma <- recv i <|> return Nothing
case ma of
Nothing -> return (Just (diffAs []))
Just a -> loop (diffAs . (a:))
bucketsEvery :: Int -> Input a -> Producer [a] IO ()
bucketsEvery microseconds i = loop
where
loop = do
lift $ threadDelay microseconds
ma <- lift $ atomically $ drainAll i
forM_ ma $ \a -> do
yield a
loop
通過選擇用於構建Input
的Buffer
的類型,可以更好地控制從上游消耗元素的方式。
如果您不pipes-concurrency
,則可以閱讀該教程 , 該教程說明了如何使用spawn
, Buffer
和Input
。
這是一個可能的解決方案。 它基於Pipe
,該Pipe
使用Bool
標記向下游的ByteString
,以便標識屬於同一“時間段”的ByteStrings
。
首先,一些進口:
import Data.AdditiveGroup
import qualified Data.ByteString as B
import qualified Data.ByteString.Lazy as BL
import qualified Data.ByteString.Lazy.Builder as BB
import Data.Thyme.Clock
import Data.Thyme.Clock.POSIX
import Control.Monad.State.Strict
import Control.Lens (view)
import Control.Concurrent (threadDelay)
import Pipes
import Pipes.Lift
import qualified Pipes.Prelude as P
import qualified Pipes.Group as PG
這是標記Pipe
。 它StateT
內部使用StateT
:
tagger :: Pipe B.ByteString (B.ByteString,Bool) IO ()
tagger = do
startTime <- liftIO getPOSIXTime
evalStateP (startTime,False) $ forever $ do
b <- await
currentTime <- liftIO getPOSIXTime
-- (POSIXTime,Bool) inner state
(baseTime,tag) <- get
if (currentTime ^-^ baseTime > timeLimit)
then let tag' = not tag in
yield (b,tag') >> put (currentTime, tag')
else yield $ (b,tag)
where
timeLimit = fromSeconds 0.1
然后,我們可以使用的功能從pipes-group
包組ByteString
小號屬於相同的“時間桶”到懶惰ByteString
S:
batch :: Producer B.ByteString IO () -> Producer BL.ByteString IO ()
batch producer = PG.folds (<>) mempty BB.toLazyByteString
. PG.maps (flip for $ yield . BB.byteString . fst)
. view (PG.groupsBy $ \t1 t2-> snd t1 == snd t2)
$ producer >-> tagger
它似乎正確批處理。 該程序:
main :: IO ()
main = do
count <- P.length $ batch (yield "boo" >> yield "baa")
putStrLn $ show count
count <- P.length $ batch (yield "boo" >> yield "baa"
>> liftIO (threadDelay 200000) >> yield "ddd")
putStrLn $ show count
具有輸出:
1
2
請注意,僅當下一個存儲桶的第一個元素到達時,才會yield
“時間存儲桶”的內容。 他們不是yield
自動編每100毫秒。 這可能對您來說不是問題。 您想每100ms自動yield
一次,可能需要基於pipes-concurrency
的不同解決方案。
另外,您可以考慮直接使用pipes-group
提供的基於FreeT
的“效果列表”。 這樣,您可以在存儲桶裝滿之前開始在“時間存儲桶”中壓縮數據。
因此,與Daniel的回答不同,我不會在生成數據時標記數據。 它僅從上游獲取至少一個元素,然后繼續在monoid中聚合更多的值,直到時間間隔過去為止。
該代碼使用列表進行匯總,但可以使用更好的monoid進行匯總
import Pipes
import qualified Pipes.Prelude as P
import Data.Time.Clock
import Data.Time.Calendar
import Data.Time.Format
import Data.Monoid
import Control.Monad
-- taken from pipes-rt
doubleToNomDiffTime :: Double -> NominalDiffTime
doubleToNomDiffTime x =
let d0 = ModifiedJulianDay 0
t0 = UTCTime d0 (picosecondsToDiffTime 0)
t1 = UTCTime d0 (picosecondsToDiffTime $ floor (x/1e-12))
in diffUTCTime t1 t0
-- Adapted from from pipes-parse-1.0
wrap
:: Monad m =>
Producer a m r -> Producer (Maybe a) m r
wrap p = do
p >-> P.map Just
forever $ yield Nothing
yieldAggregateOverTime
:: (Monoid y, -- monoid dependance so we can do aggregation
MonadIO m -- to beable to get the current time the
-- base monad must have access to IO
) =>
(t -> y) -- Change element from upstream to monoid
-> Double -- Time in seconds to aggregate over
-> Pipe (Maybe t) y m ()
yieldAggregateOverTime wrap period = do
t0 <- liftIO getCurrentTime
loop mempty (dtUTC `addUTCTime` t0)
where
dtUTC = doubleToNomDiffTime period
loop m ts = do
t <- liftIO getCurrentTime
v0 <- await -- await at least one element
case v0 of
Nothing -> yield m
Just v -> do
if t > ts
then do
yield (m <> wrap v)
loop mempty (dtUTC `addUTCTime` ts)
else do
loop (m <> wrap v) ts
main = do
runEffect $ wrap (each [1..]) >-> yieldAggregateOverTime (\x -> [x]) (0.0001)
>-> P.take 10 >-> P.print
根據您的CPU負載,輸出數據的聚合方式會有所不同。 每個塊中至少有一個元素。
$ ghc Main.hs -O2
$ ./Main
[1,2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
$ ./Main
[1,2]
[3]
[4]
[5]
[6,7,8,9,10]
[11,12,13,14,15,16,17,18]
[19,20,21,22,23,24,25,26]
[27,28,29,30,31,32,33,34]
[35,36,37,38,39,40,41,42]
[43,44,45,46,47,48,49,50]
$ ./Main
[1,2,3,4,5,6]
[7]
[8]
[9,10,11,12,13,14,15,16,17,18,19,20]
[21,22,23,24,25,26,27,28,29,30,31,32,33]
[34,35,36,37,38,39,40,41,42,43,44]
[45,46,47,48,49,50,51,52,53,54,55]
[56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72]
[73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88]
[89,90,91,92,93,94,95,96,97,98,99,100,101,102,103]
$ ./Main
[1,2,3,4,5,6,7]
[8]
[9]
[10,11,12,13,14,15,16,17,18]
[19,20,21,22,23,24,25,26,27]
[28,29,30,31,32,33,34,35,36,37]
[38,39,40,41,42,43,44,45,46]
[47,48,49,50]
[51,52,53,54,55,56,57]
[58,59,60,61,62,63,64,65,66]
您可能想看一下pipes-rt的源代碼,它顯示了一種處理管道時間的方法。
編輯:感謝DanielDíazCarrete,他采用了pipes-parse-1.0技術來處理上游終端。 管道組解決方案也應該使用相同的技術。
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