[Haskell-cafe] streaming package: How to demux a stream properly?
YueCompl
compl.yue at icloud.com
Wed Apr 15 17:42:58 UTC 2020
Not directly answering your question, but off the top of my head is:
why not demux into a Vector (Sink (Of a) m ())
Yet I have no idea if there is actually a Sink construct in any stream lib, but the theory is the Vector should contain reactive actions taking elements, rather than containers to collect elements (even thunks of them demand space).
Just my bit of thoughts, I'm definitely novice at it.
Cheers,
Compl
> On 2020-04-16, at 00:40, ☂Josh Chia (謝任中) <joshchia at gmail.com> wrote:
>
> I have a streaming package question.
>
> Suppose I have a:
> Stream (Of (Int, a)) m ()
>
> I would like to demultiplex it into Vector (Stream (Of a) m ()) using the Int as the index of an item into the Vector of output streams.
>
> How can I do this efficiently (constant memory and linear time)?
>
> Does the following work?
> import qualified Streaming.Prelude as SP
> import qualified Data.Vector as V
>
> type StreamOf a m r = Stream (Of a) m r
>
> demuxStream :: forall a m. MonadIO m
> => Int -> StreamOf (Int, a) m () -> m (Vector (StreamOf a m ()))
> demuxStream numSyms stream =
> let emptyStreams = V.replicate numSyms (pure ())
> processItem v (iD, x) = V.modify (\vm -> VM.modify vm (>> SP.yield x) iD) v
> in SP.fold_ processItem emptyStreams id stream
>
> My guess is that it takes more than constant memory as it goes through the entire input stream before returning.
>
> Josh
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