我想把一个 List[Either[A, B]]
分成两个列表。
有更好的方法吗?
def lefts[A, B](eithers : List[Either[A, B]]) : List[A] = eithers.collect { case Left(l) => l}
def rights[A, B](eithers : List[Either[A, B]]) : List[B] = eithers.collect { case Right(r) => r}
partitionMap
方法,该方法根据返回 Right
或 Left
的函数对元素进行分区。Right
或 Left
的函数来定义分区,因为我们已经有了 Right
和 Left
。因此,只需要简单地使用 identity
:val (lefts, rights) = List(Right(2), Left("a"), Left("b")).partitionMap(identity)
// lefts: List[String] = List(a, b)
// rights: List[Int] = List(2)
不确定这是否更加整洁,但:
scala> def splitEitherList[A,B](el: List[Either[A,B]]) = {
val (lefts, rights) = el.partition(_.isLeft)
(lefts.map(_.left.get), rights.map(_.right.get))
}
splitEitherList: [A, B](el: List[Either[A,B]])(List[A], List[B])
scala> val el : List[Either[Int, String]] = List(Left(1), Right("Success"), Left(42))
el: List[Either[Int,String]] = List(Left(1), Right(Success), Left(42))
scala> val (leftValues, rightValues) = splitEitherList(el)
leftValues: List[Int] = List(1, 42)
rightValues: List[String] = List("Success")
如果scalaz
是您的依赖之一,我建议直接使用separate
:
import scalaz.std.list._
import scalaz.std.either._
import scalaz.syntax.monadPlus._
val el : List[Either[Int, String]] = List(Left(1), Right("Success"), Left(42))
scala> val (lefts, rights) = el.separate
lefts: List[Int] = List(1, 42)
rights: List[String] = List(Success)
libraryDependencies += "org.scalaz" %% "scalaz-core" % "7.2.2"
,以及 import scalaz.Scalaz._
。更多信息请参见:http://stackoverflow.com/questions/36878459/how-to-turn-a-list-of-eithers-to-a-either-of-lists-using-scalaz-monadplus-separa/36883766#36883766 - David Portabellacats
,你也有 separate
!http://typelevel.org/cats/api/cats/syntax/SeparateOps.html#separate(implicitF:cats.MonadCombine[F],implicitG:cats.Bifoldable[G]):(F[A],F[B]) - Gabriele Petronellaval lefts = list.flatMap(_.left.toOption)
val rights = list.flatMap(_.right.toOption)
您可以使用以下方法:
val (lefts, rights) = eithers.foldRight((List[Int](), List[String]()))((e, p) => e.fold(l => (l :: p._1, p._2), r => (p._1, r :: p._2)))
def splitEitherList[A,B](el: List[Either[A,B]]): (List[A], List[B]) =
(el :\ (List[A](), List[B]()))((e, p) =>
e.fold(l => (l :: p._1, p._2), r => (p._1, r :: p._2)))
val x = List(Left(1), Right(3), Left(2), Left(4), Right(8))
splitEitherList(x) // (List(1, 2, 4), List(3, 8))
el.partition(_.isLeft) match { case (lefts, rights) =>
(lefts.map(_.left.get), rights.map(_.right.get)) }
好的,如果这不必是一行代码,那么它可以变得非常简单。
def split[A,B](eithers : List[Either[A, B]]):(List[A],List[B]) = {
val lefts = scala.collection.mutable.ListBuffer[A]()
val rights = scala.collection.mutable.ListBuffer[B]()
eithers.map {
case Left(l) => lefts += l
case Right(r) => rights += r
}
(lefts.toList, rights.toList)
}
Seq
解决方案。def partition[A, B](seq: Seq[Either[A, B]]): (Seq[A], Seq[B]) = {
seq.foldLeft[(Seq[A], Seq[B])]((Nil, Nil)) { case ((ls, rs), next) =>
next match {
case Left(l) => (ls :+ l, rs)
case Right(r) => (ls, rs :+ r)
}
}
}
类似于 Scalaz 对于 Cats,你也可以使用 separate
import cats.implicits._
val items = List(Right(1), Left("error"), Right(2), Right(3), Left("another error"))
val groupedItems: (List[String], List[Int]) = items.separate
//(List(error, another error),List(1, 2, 3))