在Scalaz中對分隔序列進行分區

Question

使用Scalaz將Seq[A \/ B]分區爲(Seq[A], Seq[B])的最佳方法是什麼？

Answer 1

在MonadPlus中有一種方法：separatedefined。此類型類型是Monad與PlusEmpty（廣義Monoid）的組合。所以，你需要爲Seq定義實例：

1）MonadPlus [序列]

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) 
} 

序列已經單子，所以point和bind容易，empty和plus是幺操作和Seq是free monoid

2）Bifoldable [\ /]

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，標準摺疊，但對於具有兩個參數的類型構造函數。

現在你可以使用不同的：

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

感謝Kenji Yoshida，有is一個Bitraverse [\ /]，所以你只需要MonadPlus。

而且使用foldLeft一個簡單的解決方案：

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