替代 Scala 中的线程执行器池

Substitute for thread executor pool in Scala

我的应用程序要求我有多个线程 运行 从不同的 HDFS 节点获取数据。为此,我正在使用线程执行器池和分叉线程。 分叉于:

val pathSuffixList = fileStatuses.getOrElse("FileStatus", List[Any]()).asInstanceOf[List[Map[String, Any]]]
  pathSuffixList.foreach(block => {
    ConsumptionExecutor.execute(new Consumption(webHdfsUri,block))
  })

我的class消费:

class Consumption(webHdfsUri: String, block:Map[String,Any]) extends Runnable {

      override def run(): Unit = {
        val uriSplit = webHdfsUri.split("\?")
        val fileOpenUri = uriSplit(0) + "/" + block.getOrElse("pathSuffix", "").toString + "?op=OPEN"
        val inputStream = new URL(fileOpenUri).openStream()
        val datumReader = new GenericDatumReader[Void]()
        val dataStreamReader = new DataFileStream(inputStream, datumReader)
        //        val schema = dataStreamReader.getSchema()
        val dataIterator = dataStreamReader.iterator()
        while (dataIterator.hasNext) {
          println(" data : " + dataStreamReader.next())
        }
      }

    }

消费执行器:

object ConsumptionExecutor{

  val counter: AtomicLong = new AtomicLong()

  val executionContext: ExecutorService = Executors.newCachedThreadPool(new ThreadFactory {
    def newThread(r: Runnable): Thread = {
      val thread: Thread = new Thread(r)
      thread.setName("ConsumptionExecutor-" + counter.incrementAndGet())
      thread
    }
  })
  executionContext.asInstanceOf[ThreadPoolExecutor].setMaximumPoolSize(200)

  def execute(trigger: Runnable) {
    executionContext.execute(trigger)
  }

}

但是我想使用 Akka streaming/Akka actor,我不需要提供固定的线程池大小,Akka 会处理所有事情。 我对 Akka 以及 Streaming 和 actors 的概念还很陌生。有人可以以示例代码的形式给我任何线索以适合我的用例吗? 提前致谢!

一个想法是创建 ActorPublisher for each HDFS node that you are reading from, and then Merge them in as multiple Sources in a FlowGraph.

的(子类)实例

类似于此伪代码,其中省略了 ActorPublisher 来源的详细信息:

val g = PartialFlowGraph { implicit b =>
  import FlowGraphImplicits._
  val in1 = actorSource1
  val in2 = actorSource2
  // etc.

  val out = UndefinedSink[T]
  val merge = Merge[T]

  in1 ~> merge ~> out
  in2 ~> merge
  // etc.
}

对于一组演员源,这可以通过迭代它们并为每个演员源添加一条边到 merge 来改进,但这给出了想法。