Question

我正在寻找一种方法将Futures的任意长度列表转换为List of Future。我正在使用Playframework，所以最终，我真正想要的是Future[Result]，但为了简单起见，我们只需说Future[List[Int]]这样做的正常方法是使用Future.sequence(...)但是有一个扭曲...我给出的列表通常有大约10-20个期货，并且其中一个期货失败并不常见（它们正在制作外部Web服务请求）。在其中一个失败的情况下，我不想重试所有这些，而是希望能够获得那些成功的并返回那些。

例如，执行以下操作无效

import scala.concurrent._
import scala.concurrent.ExecutionContext.Implicits.global
import scala.util.Success
import scala.util.Failure

val listOfFutures = Future.successful(1) :: Future.failed(new Exception("Failure")) :: 
                    Future.successful(3) :: Nil

val futureOfList = Future.sequence(listOfFutures)

futureOfList onComplete {
  case Success(x) => println("Success!!! " + x)
  case Failure(ex) => println("Failed !!! " + ex)
}

scala> Failed !!! java.lang.Exception: Failure

我不想得到唯一的例外，我希望能够将1和3拉出来。我尝试使用Future.fold，但显然只是在幕后调用Future.sequence。

提前感谢您的帮助！

Answer 1

诀窍是首先确保没有一个未来失败。 .recover在这里是您的朋友，您可以将其与map结合使用，将所有Future[T]结果转换为Future[Try[T]]]个实例，所有这些都确定是成功的期货。

注意：您也可以使用Option或Either，但如果您特别想要捕获异常，则Try是最简洁的方法

def futureToFutureTry[T](f: Future[T]): Future[Try[T]] =
  f.map(Success(_)).recover(x => Failure(x))

val listOfFutures = ...
val listOfFutureTrys = listOfFutures.map(futureToFutureTry(_))

然后像以前一样使用Future.sequence，为您提供Future[List[Try[T]]]

val futureListOfTrys = Future.sequence(listOfFutureTrys)

然后过滤：

val futureListOfSuccesses = futureListOfTrys.map(_.filter(_.isSuccess))

如果您需要，您甚至可以提取特定的故障：

val futureListOfFailures = futureListOfTrys.map(_.filter(_.isFailure))

Answer 2

我试过凯文的答案，我在我的Scala版本（2.11.5）上遇到了一个小故障...我纠正了这个问题，并且如果有人有兴趣的话还写了一些额外的测试......这是我的版本＆gt;

implicit class FutureCompanionOps(val f: Future.type) extends AnyVal {

    /** Given a list of futures `fs`, returns the future holding the list of Try's of the futures from `fs`.
      * The returned future is completed only once all of the futures in `fs` have been completed.
      */
    def allAsTrys[T](fItems: /* future items */ List[Future[T]]): Future[List[Try[T]]] = {
      val listOfFutureTrys: List[Future[Try[T]]] = fItems.map(futureToFutureTry)
      Future.sequence(listOfFutureTrys)
    }

    def futureToFutureTry[T](f: Future[T]): Future[Try[T]] = {
      f.map(Success(_)) .recover({case x => Failure(x)})
    }

    def allFailedAsTrys[T](fItems: /* future items */ List[Future[T]]): Future[List[Try[T]]] = {
      allAsTrys(fItems).map(_.filter(_.isFailure))
    }

    def allSucceededAsTrys[T](fItems: /* future items */ List[Future[T]]): Future[List[Try[T]]] = {
      allAsTrys(fItems).map(_.filter(_.isSuccess))
    }
}


// Tests... 



  // allAsTrys tests
  //
  test("futureToFutureTry returns Success if no exception") {
    val future =  Future.futureToFutureTry(Future{"mouse"})
    Thread.sleep(0, 100)
    val futureValue = future.value
    assert(futureValue == Some(Success(Success("mouse"))))
  }
  test("futureToFutureTry returns Failure if exception thrown") {
    val future =  Future.futureToFutureTry(Future{throw new IllegalStateException("bad news")})
    Thread.sleep(5)            // need to sleep a LOT longer to get Exception from failure case... interesting.....
    val futureValue = future.value

    assertResult(true) {
      futureValue match {
        case Some(Success(Failure(error: IllegalStateException)))  => true
      }
    }
  }
  test("Future.allAsTrys returns Nil given Nil list as input") {
    val future =  Future.allAsTrys(Nil)
    assert ( Await.result(future, 100 nanosecond).isEmpty )
  }
  test("Future.allAsTrys returns successful item even if preceded by failing item") {
    val future1 =  Future{throw new IllegalStateException("bad news")}
    var future2 = Future{"dog"}

    val futureListOfTrys =  Future.allAsTrys(List(future1,future2))
    val listOfTrys =  Await.result(futureListOfTrys, 10 milli)
    System.out.println("successItem:" + listOfTrys);

    assert(listOfTrys(0).failed.get.getMessage.contains("bad news"))
    assert(listOfTrys(1) == Success("dog"))
  }
  test("Future.allAsTrys returns successful item even if followed by failing item") {
    var future1 = Future{"dog"}
    val future2 =  Future{throw new IllegalStateException("bad news")}

    val futureListOfTrys =  Future.allAsTrys(List(future1,future2))
    val listOfTrys =  Await.result(futureListOfTrys,  10 milli)
    System.out.println("successItem:" + listOfTrys);

    assert(listOfTrys(1).failed.get.getMessage.contains("bad news"))
    assert(listOfTrys(0) == Success("dog"))
  }
  test("Future.allFailedAsTrys returns the failed item and only that item") {
    var future1 = Future{"dog"}
    val future2 =  Future{throw new IllegalStateException("bad news")}

    val futureListOfTrys =  Future.allFailedAsTrys(List(future1,future2))
    val listOfTrys =  Await.result(futureListOfTrys,  10 milli)
    assert(listOfTrys(0).failed.get.getMessage.contains("bad news"))
    assert(listOfTrys.size == 1)
  }
  test("Future.allSucceededAsTrys returns the succeeded item and only that item") {
    var future1 = Future{"dog"}
    val future2 =  Future{throw new IllegalStateException("bad news")}

    val futureListOfTrys =  Future.allSucceededAsTrys(List(future1,future2))
    val listOfTrys =  Await.result(futureListOfTrys,  10 milli)
    assert(listOfTrys(0) == Success("dog"))
    assert(listOfTrys.size == 1)
  }

Answer 3

我刚刚遇到了这个问题并提供了另一种解决方案：

def allSuccessful[A, M[X] <: TraversableOnce[X]](in: M[Future[A]])
                                                (implicit cbf: CanBuildFrom[M[Future[A]], A, M[A]], 
                                                 executor: ExecutionContext): Future[M[A]] = {
    in.foldLeft(Future.successful(cbf(in))) {
      (fr, fa) ⇒ (for (r ← fr; a ← fa) yield r += a) fallbackTo fr
    } map (_.result())
}

这里的想法是，在折叠中你正在等待列表中的下一个元素完成（使用for-comprehension语法），如果下一个元素失败，你只需要回到你已经拥有的东西。

Answer 4

Scala 2.12在Future.transform上有一个改进，它适用于代码较少的anwser。

val futures = Seq(Future{1},Future{throw new Exception})

val seq = Future.sequence(futures.map(_.transform(Success(_)))) // instead of map and recover

val successes = seq.map(_.collect{case Success(x)=>x})
//successes: Future[Seq[Int]] = Future(Success(List(1)))

val failures = seq.map(_.collect{case Failure(x)=>x})
//failures: Future[Seq[Throwable]] = Future(Success(List(java.lang.Exception)))

Answer 5

您可以使用选项轻松包装未来结果，然后展平列表：

def futureToFutureOption[T](f: Future[T]): Future[Option[T]] =
    f.map(Some(_)).recover {
      case e => None
    }
val listOfFutureOptions = listOfFutures.map(futureToFutureOption(_))

val futureListOfOptions = Future.sequence(listOfFutureOptions)

val futureListOfSuccesses = futureListOfOptions.flatten

Answer 6

您还可以在不同的列表中收集成功和失败的结果：

def safeSequence[A](futures: List[Future[A]]): Future[(List[Throwable], List[A])] = {
  futures.foldLeft(Future.successful((List.empty[Throwable], List.empty[A]))) { (flist, future) =>
    flist.flatMap { case (elist, alist) =>
      future
        .map { success => (elist, alist :+ success) }
        .recover { case error: Throwable => (elist :+ error, alist) }
    }
  }
}

Scala：将[未来]列入未来[List]，忽略失败的未来

6 个答案: