可能值得考虑使用Sink.fold而不是Sink.reduce：

val possiblyEmpty = Source(Seq(1, 3, 5)).filter(_ % 2 == 0)
val eventualInt = possiblyEmpty.toMat(Sink.fold[Int](0)(_ + _))(Keep.right).run()

如果没有合理的零/单位元素，你可以有一些更一般化的东西：

def reducePossiblyEmpty[T](source: Source[T])(f: (T, T) => T): RunnableGraph[Future[Option[T]]] = {
  val lifted = { (x: Option[T], y: Option[T]) =>
    x.flatMap(a => y.map(f))
  }
  source.map(x => Some(x))
    .concat(Source.single(None))
    .statefulMapConcat[Option[T]] { () =>
      var emptyStream = true
      { x =>
        x match {
          case Some(x) =>
            // element from the given stream
            emptyStream = false
            List(x)

          case None =>
            // given stream completed
            if (emptyStream) {
              List(x)
            } else {
              Nil  // don't emit anything
            }
        }
      }
    }
    .toMat(Sink.reduce[Option[T]](lifted))(Keep.right)
}

如果没有元素，返回的图形将以None完成，或者以约简结果的Some完成。
编辑：您也可以只在Source/Flow上使用orElse，代替上面的.concat.statefulMapConcat：

def reducePossiblyEmpty[T](source: Source[T])(f: (T, T) => T): RunnableGraph[Future[Option[T]]] = {
  val lifted = { (x: Option[T], y: Option[T]) =>
    x.flatMap(a => y.map(f))
  }

  source.map(x => Some(x))
    .orElse(Source.single(None))
    .toMat(Sink.reduce[Option[T]](lifted))(Keep.right)
}

因为你使用的是Sink.reduce[Int]，所以你可以添加一个Source，它保证只有一个0元素，这样Sink.reduce[Int]就可以工作，并产生0。
这里有一个例子

val zero          = Source.single(0)
val possiblyEmpty = Source(List[Int](1, 3, 5)).filter(_ % 2 == 0)
val eventualInt   = zero.merge(possiblyEmpty).toMat(Sink.reduce[Int](_ + _))(Keep.right).run()