Scala：列表[未来]到未来[列表]忽略失败的未来

Question

我正在寻找一种将任意长度的期货列表转换为期货列表的方法。我使用的是Playframework，所以最终，我真正想要的是一个＜code＞未来〔结果〕，但为了让事情更简单，让我们说＜code＞将来〔List[Int]]通常的方法是使用＜code＞Future.sequence（…）

例如，执行以下操作不起作用：

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

我希望能够将1和3从那里拉出来，而不是只得到异常。我尝试将来使用＜code＞。折叠，但这显然只是调用未来。序列在幕后。

提前感谢您的帮助！

Answer 1

我尝试了凯文的答案，我在我的Scala版本（2.11.5）上遇到了一个小故障......我纠正了这一点，并写了一些额外的测试，如果有人感兴趣的话......这是我的版本

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 2

Scala 2.12在＜code＞的未来有了改进。转换，使其自身处于代码较少的anwser中。

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

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

val successes = seq.map(_.collect{case Success(x)=>x})
successes
//res1: Future[Seq[Int]] = Future(Success(List(1)))

val failures = seq.map(_.collect{case Failure(x)=>x})
failures
//res2: Future[Seq[Throwable]] = Future(Success(List(java.lang.Exception)))

Answer 3

诀窍是首先确保没有一个期货失败。.recover 是你在这里的朋友，你可以将它与 map 结合使用，将所有 Future[T] 结果转换为 Future[Try[T]]] 实例，所有这些实例都一定是成功的期货。

注意：您也可以在此处使用Option或任一，但如果您特别想捕获异常，尝试是最干净的方法

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

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

然后在将来使用。顺序与前面一样，为您提供一个未来的列表[Try[T]]]

val futureListOfTrys = Future.sequence(listOfFutureTrys)

然后过滤：

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

如果需要，您甚至可以提取特定的故障：

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