scala - 在Scalaz中对一系列析取进行分区

Question

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

Answer 1

有一个方法：在 MonadPlus 中separate 定义。这个类型类是 Monad 和 PlusEmpty（广义 Monoid）的组合。所以你需要定义实例Seq：

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 已经是一元的，所以point和bind很容易，empty并且plus是幺半群运算并且Seq是一个自由的幺半群

2) 可折叠[\/]

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)
    }
  }

也很容易，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))

更新

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

还有一个简单的解决方案foldLeft：

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