如何在 Scala 中的过滤器、映射、平面映射期间轻松地从一种集合类型转换为另一种集合类型？
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How do I easily convert from one collection type to another during a filter, map, flatMap in Scala?
提问by Jean-Philippe Pellet
Suppose I have a List[Int], and I want to call toStringon each element, and get back the result as a Vector[String].
假设我有一个List[Int], 并且我想调用toString每个元素，并将结果作为Vector[String].
What are the various ways to do this in Scala? Is there a solution with a minimal amount of explicit typing? — i.e., I want to specify that I want a Vectorrather than a List, but I'd like the Stringargument to be inferred from the filter function.
在 Scala 中执行此操作的各种方法是什么？是否有最少显式类型的解决方案？— 即，我想指定我想要 aVector而不是 a List，但我希望String从过滤器函数中推断出参数。
Or should I explicitly pass a CanBuildFrominstance? Where do I get these from — for Seqs, Sets and Maps?
或者我应该明确地传递一个CanBuildFrom实例？我从哪里得到这些——对于Seqs、Sets 和Maps？
回答by oxbow_lakes
Using breakOut
使用突破
Use breakOutas the CanBuildFromand let the typer know what you want your result type to be (unfortunately you need to specify String here)
使用breakOutasCanBuildFrom并让打字机知道您希望结果类型是什么（不幸的是，您需要在此处指定 String）
scala> import collection.breakOut
import collection.breakOut
scala> List(1, 2, 3)
res0: List[Int] = List(1, 2, 3)
scala> res0.map(_.toString)(breakOut) : Vector[String]
res2: Vector[String] = Vector(1, 2, 3)
Using to[Collection] (starting with Scala 2.10.0)
使用 to[Collection]（从 Scala 2.10.0 开始）
Scala 2.10.0 introduced an easy way to convert a collection to another collection:
Scala 2.10.0 引入了一种将集合转换为另一个集合的简单方法：
scala> List(1, 2, 3).map(_.toString).to[Vector]
res0: Vector[String] = Vector(1, 2, 3)
Using toIndexedSeq
使用 toIndexedSeq
Alternatively ask for an IndexedSeqexplicitly:
或者IndexedSeq明确要求：
scala> res0.map(_.toString).toIndexedSeq
res4: scala.collection.immutable.IndexedSeq[String] = Vector(1, 2, 3)
If you want to do this without creating the intermediate List, then:
如果您想在不创建中间件的情况下执行此操作List，则：
scala> res0.view.map(_.toString).toIndexedSeq
res5: scala.collection.immutable.IndexedSeq[String] = Vector(1, 2, 3)
Using Natural Transformations
使用自然变换
You could do it (awkwardly, but more generally) using natural transformations
你可以使用自然变换来做到（尴尬，但更普遍）
scala> trait Trans[F[_], G[_]] {
| def f2g[A](f : F[A]) : G[A]
| }
defined trait Trans
Now provide a typeclass instance from the List ~> Vector transformation:
现在从 List ~> Vector 转换中提供一个类型类实例：
scala> implicit val List2Vector = new Trans[List, collection.immutable.Vector] {
| def f2g[A](l : List[A]) : Vector[A] = l.map(identity[A])(collection.breakOut)
| }
List2Vector: java.lang.Object with Trans[List,scala.collection.immutable.Vector] = $anon@56329755
Define a wrapper and an implicit conversion:
定义一个包装器和一个隐式转换：
scala> class Clever[M[_], A](ma : M[A]) { def to[N[_]](implicit t : Trans[M, N]) : N[A] = t.f2g(ma) }
defined class Clever
scala> implicit def ma2clever[M[_], A](ma : M[A]) = new Clever[M, A](ma)
ma2clever: [M[_],A](ma: M[A])Clever[M,A]
Then:
然后：
scala> List(1, 2, 3).map(_.toString).to[Vector]
res4: Vector[java.lang.String] = Vector(1, 2, 3)
