Scala 阿克卡流配料

学习阿卡河。我有一个记录流，每个时间单位有很多，已经从Slick按时间排序，我想通过检测时间步长的变化将它们批处理到时间组中进行处理

范例

如果传入流是

Record(1, "a")
Record(1, "k")
Record(1, "k")
Record(1, "a")
Record(2, "r")
Record(2, "o")
Record(2, "c")
Record(2, "k")
Record(2, "s")
Record(3, "!")
...

我想把它转换成

Batch(1, Seq("a","k","k","a"))
Batch(2, Seq("r","o","c","k","s"))
Batch(3, Seq("!"))
...

到目前为止，我只发现了按固定数量的记录进行分组，或者拆分为多个子流，但从我的角度来看，我不需要多个子流

更新：我发现了，但它看起来更关注背压，而不仅仅是一直在批处理。

是Akka Streams库中的multitool

val records =
  Source(List(
    Record(1, "a"),
    Record(1, "k"),
    Record(1, "k"),
    Record(1, "a"),
    Record(2, "r"),
    Record(2, "o"),
    Record(2, "c"),
    Record(2, "k"),
    Record(2, "s"),
    Record(3, "!")
  ))
  .concat(Source.single(Record(0, "notused"))) // needed to print the last element

records
  .statefulMapConcat { () =>
    var currentTime = 0
    var payloads: Seq[String] = Nil

    record =>
      if (record.time == currentTime) {
        payloads = payloads :+ record.payload
        Nil
      } else {
        val previousState = (currentTime, payloads)
        currentTime = record.time
        payloads = Seq(record.payload)
        List(previousState)
      }
  }
  .runForeach(println)

运行上述命令将打印以下内容：

(0,List())
(1,List(a, k, k, a))
(2,List(r, o, c, k, s))
(3,List(!))

您可以调整示例以打印批处理对象

(0,List())
(1,List(a, k, k, a))
(2,List(r, o, c, k, s))
(3,List(!))