解析 Seq[Future[Either[A, Seq[B]]]] - Scala Cats

我在尝试解决方法的结果时遇到问题。更具体地说，我有：

def methodA(): Future[Either[Error, Seq[A]]]

在某些时候，我想为列表的每个元素调用此方法并合并结果。 像这样的东西：

val tes: Seq[Future[Either[Error, Seq[A]]]] = relevantRounds.map(round =>
            methodA()
          )

你知道我该如何解决这个问题吗Seq[Future[Either[Error, Seq[A]]]]?

所以我最终想要的是Future[Either[Error, Seq[A]]]包含整个列表结果的序列。

您可能正在寻找

  def combine[A](s: Seq[Future[Either[Error, Seq[A]]]]) = {
    Future.sequence(s)
      .map(x => {
        x.foldRight(Right(Seq()): Either[Error, Seq[A]]) {
          (e, acc) => for (xs <- acc.right; x <- e.right) yield x ++ xs
        }
      }
      )
  }

如果您希望仅在前一个成功时才执行函数，请使用此

  def combine[A](s: Seq[() => Future[Either[Error, Seq[A]]]]): Future[Either[Error, Seq[A]]] =
    combine(Seq(), s)

  def combine[A](acc: Seq[A], s: Seq[() => Future[Either[Error, Seq[A]]]]): Future[Either[Error, Seq[A]]] = s match {
    case x +: Nil =>
      val v = x.apply()
      v.andThen {
        case Success(Right(r)) => Success(Right(acc ++ r))
        case Success(Left(l)) => Success(Left(l))
        case Failure(f) => Failure(f)
      }
    case x +: xs =>
      val v = x.apply()
      v.andThen {
        case Success(Right(r)) => combine(acc ++ r, xs)
        case Success(Left(l)) => Success(Left(l))
        case Failure(f) => Failure(f)
      }
  }