在Scalaz中划分一系列析取

Question

什么是分区的最佳方式Seq[A \/ B]为(Seq[A], Seq[B])使用Scalaz？

Answer 1

有一种方法: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)
}

序列已经是一元,所以point和bind容易,empty而且plus是独异的操作和Seq是免费的幺

2)双折[\ /]

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

由于吉田健二,有是一个Bitraverse [\ /],所以你只需要MonadPlus.

一个简单的解决方案使用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)
  }
}