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Scala 期貨使用 Future.sequence 但返回元組有問題

[英]Scala futures using Future.sequence but having issues with returning a tuple

當我在下面使用Future.sequence時,我在嘗試返回元組時遇到了一些麻煩:

type UserId = Int
  type GameId = Int
  type Score = Double

  case class Game(id: GameId)

  def findGame(id: GameId): Future[Game] = ???
  def findUserGames(id: UserId): Future[Seq[(GameId, Score)]] = ???

  def findGames(id: UserId): Future[Seq[(Game, Score)]] = {
    val userGames = findUserGames(id)

    // ******* I am stuck here
    val games: Future[Seq[Game]] = userGames.flatMap(x => Future.sequence(x.map(y => findGame(y._1))))    
  }

當 Future.sequence 調用只是返回給我一個Seq[Game]時,我如何返回一個(Game, Score)的元組?

另外,如果我們沒有Future.sequence ,我們如何能夠模仿它的作用? 即將List[Future[Game]]轉換成Future[List[Game]]

你可以試試這個:

def findGames(id: UserId): Future[Seq[(Game, Score)]] = {
  findUserGames(id).flatMap(gameIdToScoreSeq =>
    Future.traverse(gameIdToScoreSeq)(gameIdToScore =>
      findGame(gameIdToScore._1).map((_, gameIdToScore._2))
    )
  )
}

只需跟蹤名稱是什么,作為oneliner,您就可以做到

def findGames(id: UserId)(implicit ec: ExecutionContext): Future[Seq[(Game, Score)]] =
  findUserGames(id).flatMap(games =>
    Future.sequence(games.map { case (gameid, score) =>
      findGame(gameid).map(game => game -> score)
    })
  )

你只需要一個 map:

val games: Future[Seq[(Game, Score)]] = userGames.flatMap(x => Future.sequence(x.map(y => findGame(y._1).map(z => (z, y._2)))))

正如你所看到的,它變得有點復雜。 您可以通過使用 for-comprehensions 和模式匹配使您的代碼更具可讀性(可以說是):

val games2: Future[Seq[(Game, Score)]] = for {
  games <- userGames
  result <- Future.sequence(games.map {
    case (gameId, score) => 
      for {
        game <- findGame(gameId)
      } yield (game, score)
  })
} yield result

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