Is it possible to create pipes that get all values that have been sent downstream in a certain time period? I'm implementing a server where the protocol allows me to concatenate outgoing packets and compress them together, so I'd like to effectively "empty out" the queue of downstream ByteString
s every 100ms and mappend
them together to then yield on to the next pipe which does the compression.

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So you want to buffer all incoming values for 100ms and then send "one batch" downstream, which contains all the incoming values aggregated? – bennofs Mar 08 '14 at 23:03
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Yes, that is correct. – user3261399 Mar 09 '14 at 00:13
3 Answers
Here's a solution using pipes-concurrency
. You give it any Input
and it will periodically drain the input of all values:
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
This gives you much greater control over how you consume elements from upstream, by selecting the type of Buffer
you use to build the Input
.
If you're new to pipes-concurrency
, you can read the tutorial which explains how to use spawn
, Buffer
and Input
.

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I think I am confused. Doesn't `drainAll` just block indefinitely until the other side of the `Input` is closed? My `Input` would effectively never close, so I need to somehow interrupt the blocking if the delay is up. – user3261399 Mar 11 '14 at 01:08
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@user3261399 It will not block indefinitely. Notice the `recv i <|> return Nothing`. `recv` will `retry` if the buffer is empty but not closed, which wil trigger the second branch of the alternative and return `Nothing` instead. – Gabriella Gonzalez Mar 11 '14 at 05:09
Here is a possible solution. It is based on a Pipe
that tags ByteString
s going downstream with a Bool
, in order to identify ByteStrings
belonging to the same "time bucket".
First, some imports:
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
Here is the tagging Pipe
. It uses StateT
internally:
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
Then we can use functions from the pipes-group
package to group ByteString
s belonging to the same "time bucket" into lazy 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
It seems to batch correctly. This program:
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
Has the output:
1
2
Notice that the contents of a "time bucket" are only yield
ed when the first element of the next bucket arrives. They are not yield
ed automatically each 100ms. This may or may not be a problem for you. It you want to yield
automatically each 100ms, you would need a different solution, possibly based on pipes-concurrency
.
Also, you could consider working directly with the FreeT
-based "effectul lists" provided by pipes-group
. That way you could start compressing the data in a "time bucket" before the bucket is full.

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So unlike Daniel's answer my does not tag the data as it is produced. It just takes at least element from upstream and then continues to aggregate more values in the monoid until the time interval has passed.
This codes uses a list to aggregate, but there are better monoids to aggregate with
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
Depending on cpu load you the output data will be aggregated differently. With at least on element in each chunk.
$ 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]
You might want to look at the source code of pipes-rt it shows one approach to deal with time in pipes.
edit: Thanks to Daniel Díaz Carrete, adapted pipes-parse-1.0 technique to handle upstream termination. A pipes-group solution should be possible using the same technique as well.

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When I run your solution with only one value coming from upstream (like `runEffect $ each [1]`) the value doesn't seem to get printed. – danidiaz Mar 09 '14 at 11:45
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That is a good point. Really a pipe like this needs to know about upstream termination. So either transferring to pipes-parse, pipes-group or using a stream of maybes values is the correct thing to do. – Davorak Mar 09 '14 at 12:10