CODEWITHSUNDEEP
scalaziozio-streams
myLittlePony
myLittlePony

Reputation: 1

How can I combine N different ZIO-Streams?

I've been exploring ZIO Streams and ZIO's concurrency model for a while. Recently, I started working on a toy project where I process a large file containing integers. Here's what I've done so far:

  1. Read the file.
  2. Split it into chunks of 10 MB.
  3. Sort each chunk.
  4. Save the sorted chunks to temporary files.

Now, I'm at the next stage: reading all these temporary files into streams, sorting their contents in-memory again, and writing the results to a new file. I'd like to enforce some memory limitations so that I can dive deeper into ZIO's stream framework. You can say I am stuck trying to implement k-way merge, in ZIO.

Here’s the code I have for creating a ZStream from a single file:

private def fileStream(file: Path): Task[Stream[Throwable, Int]] =
  ZIO.succeed {
    ZStream
      .fromPath(file)
      .via(ZPipeline.utf8Decode)
      .via(ZPipeline.splitLines)
      .mapZIO(line =>
        ZIO
          .attempt(line.trim.toInt)
          .orElseFail(new IOException(s"Invalid line in file $file: $line"))
      )
  }

I can create a list of streams using ZIO.foreach: ZIO.foreach(files)(fileStream)

However, I'm stuck on how to combine multiple ZStreams declaratively. My goal is to merge or combine these streams while respecting some memory constraints.

In theory, I was expecting some sort ofZStream#combine method to merge multiple streams, but I haven’t found a clear solution. How can I achieve this in ZIO Streams?

I tried using ZStream#fromIterable and ZStream#fromIteratorZIO constructors, however these are not combining the effects described (which is reading a line) so I don't think they are doing what I am intending to do.

Upvotes: 0

Views: 76

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