Question

使用Scalaz将Seq[A \/ B]分区为(Seq[A], Seq[B])的最佳方法是什么？

Answer 1

MonadPlus中有一个方法：separate defined。这个类型类是Monad和PlusEmpty（广义Monoid）的组合。因此，您需要为Seq：

定义实例

1）MonadPlus [Seq]

implicit val seqmp = new MonadPlus[Seq] {
  def plus[A](a: Seq[A], b: => Seq[A]): Seq[A] = a ++ b
  def empty[A]: Seq[A] = Seq.empty[A]
  def point[A](a: => A): Seq[A] = Seq(a)
  def bind[A, B](fa: Seq[A])(f: (A) => Seq[B]): Seq[B] = fa.flatMap(f)
}

Seq已经是monadic，因此point和bind很容易，empty和plus是幺半群操作，Seq是free monoid

2）Bifoldold [\ /]

implicit val bife = new Bifoldable[\/] {
    def bifoldMap[A, B, M](fa: \/[A, B])(f: (A) => M)(g: (B) => M)(implicit F: Monoid[M]): M = fa match {
      case \/-(r) => g(r)
      case -\/(l) => f(l)
    }

    def bifoldRight[A, B, C](fa: \/[A, B], z: => C)(f: (A, => C) => C)(g: (B, => C) => C): C = fa match {
      case \/-(r) => g(r, z)
      case -\/(l) => f(l, z)
    }
  }

同样简单，标准折叠，但对于具有两个参数的类型构造函数。

现在您可以单独使用：

val seq: Seq[String \/ Int] = List(\/-(10), -\/("wrong"), \/-(22), \/-(1), -\/("exception"))
scala> seq.separate
res2: (Seq[String], Seq[Int]) = (List(wrong, number exception),List(10, 22, 1))

<强>更新

感谢Kenji Yoshida，is有一个Bitraverse [\ /]，所以你只需要MonadPlus。

使用foldLeft的简单解决方案：

seq.foldLeft((Seq.empty[String], Seq.empty[Int])){ case ((as, ai), either) =>
  either match {
    case \/-(r) => (as, ai :+ r)
    case -\/(l) => (as :+ l, ai)
  }
}

在Scalaz中划分一系列析取

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