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scala - 如何根据另一个 Akka 流的元素聚合一个 Akka 流的元素?

转载 作者:行者123 更新时间:2023-12-01 23:30:56 25 4
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示例场景:将流的字节分组为大小由另一个流(整数)确定的 block 。

def partition[A, B, C](
first:Source[A, NotUsed],
second:Source[B, NotUsed],
aggregate:(Int => Seq[A], B) => C
):Source[C, NotUsed] = ???

val bytes:Source[Byte, NotUsed] = ???
val sizes:Source[Int, NotUsed] = ???

val chunks:Source[ByteString, NotUsed] =
partition(bytes, sizes, (grab, count) => ByteString(grab(count)))

我最初的尝试包括 Flow#scan 的组合和 Flow#prefixAndTail ,但感觉不太对(见下文)。我也看了Framing ,但它似乎不适用于上面的示例场景(也不够通用以适应非字节串流)。我猜我唯一的选择是使用 Graphs (或更一般的 FlowOps#transform ),但我还不够精通 Akka 流来尝试这样做。


这是我到目前为止能够想到的(特定于示例场景):

val chunks:Source[ByteString, NotUsed] = sizes
.scan(bytes prefixAndTail 0) {
(grouped, count) => grouped flatMapConcat {
case (chunk, remainder) => remainder prefixAndTail count
}
}
.flatMapConcat(identity)
.collect { case (chunk, _) if chunk.nonEmpty => ByteString(chunk:_*) }

最佳答案

我认为您可以将处理实现为自定义 GraphStage。舞台将有两个 Inlet 元素。一个获取字节,另一个获取大小。它将有一个 Outlet 元素产生值。

考虑以下输入流。

def randomChars = Iterator.continually(Random.nextPrintableChar())
def randomNumbers = Iterator.continually(math.abs(Random.nextInt() % 50))

val bytes: Source[Char, NotUsed] =
Source.fromIterator(() => randomChars)

val sizes: Source[Int, NotUsed] =
Source.fromIterator(() => randomNumbers).filter(_ != 0)

然后使用描述自定义流处理的信息 (http://doc.akka.io/docs/akka/2.4.2/scala/stream/stream-customize.html),您可以构造 GraphStage

case class ZipFraming() extends GraphStage[FanInShape2[Int, Char, (Int, ByteString)]] {

override def initialAttributes = Attributes.name("ZipFraming")

override val shape: FanInShape2[Int, Char, (Int, ByteString)] =
new FanInShape2[Int, Char, (Int, ByteString)]("ZipFraming")

val inFrameSize: Inlet[Int] = shape.in0
val inElements: Inlet[Char] = shape.in1

def out: Outlet[(Int, ByteString)] = shape.out

override def createLogic(inheritedAttributes: Attributes): GraphStageLogic =
new GraphStageLogic(shape) {
// we will buffer as much as 512 characters from the input
val MaxBufferSize = 512
// the buffer for the received chars
var buffer = Vector.empty[Char]
// the needed number of elements
var needed: Int = -1
// if the downstream is waiting
var isDemanding = false

override def preStart(): Unit = {
pull(inFrameSize)
pull(inElements)
}

setHandler(inElements, new InHandler {
override def onPush(): Unit = {
// we buffer elements as long as we can
if (buffer.size < MaxBufferSize) {
buffer = buffer :+ grab(inElements)
pull(inElements)
}
emit()
}
})

setHandler(inFrameSize, new InHandler {
override def onPush(): Unit = {
needed = grab(inFrameSize)
emit()
}
})

setHandler(out, new OutHandler {
override def onPull(): Unit = {
isDemanding = true
emit()
}
})

def emit(): Unit = {
if (needed > 0 && buffer.length >= needed && isDemanding) {
val (emit, reminder) = buffer.splitAt(needed)
push(out, (needed, ByteString(emit.map(_.toByte).toArray)))
buffer = reminder
needed = -1
isDemanding = false
pull(inFrameSize)
if (!hasBeenPulled(inElements)) pull(inElements)
}
}
}
}

这就是您运行它的方式。

RunnableGraph.fromGraph(GraphDSL.create(bytes, sizes)(Keep.none) { implicit b =>
(bs, ss) =>
import GraphDSL.Implicits._

val zipFraming = b.add(ZipFraming())

ss ~> zipFraming.in0
bs ~> zipFraming.in1

zipFraming.out ~> Sink.foreach[(Int, ByteString)](e => println((e._1, e._2.utf8String)))

ClosedShape
}).run()

关于scala - 如何根据另一个 Akka 流的元素聚合一个 Akka 流的元素?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36109241/

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