如何将结果 Future[ Option[ T ] ] 组合成 Seq[ T ]

Question

我有一个方法

 def readTree(id: String): Future[Option[CategoryTreeResponse]]

和字符串列表 channels:List[String]。

如何将所有结果迭代组合成一个非Future Sequence？如：

 def readAllTrees(): Seq[CategoryTreeResponse] =  ???

可能没有阻塞。

来自命令式世界，我会这样做：

import scala.concurrent.duration._

def readTrees(): Seq[CategoryTreeResponse] = {
  val list = ListBuffer[CategoryTreeResponse]()
  for (id <- channels) {
    val tree = Await.result(readTree(id), 5.seconds)
    if (tree.isDefined) {
      list += tree.get
    }
  }
  list
}

Answer 1

你可以这样做

def readAllTrees(channels: List[String]): Future[Seq[CategoryTreeResponse]] = {
    Future.sequence(channels.map(readTree(_))).map(_.flatten)
}

我已经更改了 readAllTrees 的签名以接收列表和 return 序列的未来。

如果您想访问生成的序列，您需要等到完成

 Await.result(readAllTrees(channels), Duration.Inf)

但这不是管理期货的好方法，因为它会锁定调用 Await.ready

的线程

Answer 2

Future.sequence 和 Await.result 应该有帮助。不过我同意 Mikel 的观点，最好使用 Future class

的 map/flatMap/foreach 等方法尽可能长时间地保持异步

scala> :paste
// Entering paste mode (ctrl-D to finish)

import scala.concurrent._
import scala.concurrent.duration._
import scala.concurrent.ExecutionContext.Implicits.global

case class CategoryTreeResponse()

val futureResults: List[Future[Option[CategoryTreeResponse]]] = List(
 Future.successful(Option(CategoryTreeResponse())), 
 Future.successful(Option(CategoryTreeResponse())), 
 Future.successful(None)
) 
val futureResult: Future[List[Option[CategoryTreeResponse]]] = Future.sequence(futureResults)
val allResults: List[Option[CategoryTreeResponse]] = Await.result(futureResult, Duration.Inf)
val nonEmptyResults: Seq[CategoryTreeResponse] = allResults.flatMap(_.toSeq)

// Exiting paste mode, now interpreting.

import scala.concurrent._
import scala.concurrent.duration._
import scala.concurrent.ExecutionContext.Implicits.global
defined class CategoryTreeResponse
futureResults: List[scala.concurrent.Future[Option[CategoryTreeResponse]]] = List(Future(Success(Some(CategoryTreeResponse()))), Future(Success(Some(CategoryTreeResponse()))), Future(Success(None)))
futureResult: scala.concurrent.Future[List[Option[CategoryTreeResponse]]] = Future(Success(List(Some(CategoryTreeResponse()), Some(CategoryTreeResponse()), None)))
allResults: List[Option[CategoryTreeResponse]] = List(Some(CategoryTreeResponse()), Some(CategoryTreeResponse()), None)
nonEmptyResults: Seq[CategoryTreeResponse] = List(CategoryTreeResponse(), CategoryTreeResponse())

scala>

如何将结果 Future[ Option[ T ] ] 组合成 Seq[ T ]

how to combine the results Future[ Option[ T ] ] into Seq[ T ]

scala

sequence

future

option