Scala：从一种类型的集合到另一种类型的收益

Scala: Yielding from one type of collection to another

提问by Zecrates

Concerning the yieldcommand in Scala and the following example:

关于Scala 中的yield命令和以下示例：

val values = Set(1, 2, 3)
val results = for {v <- values} yield (v * 2)

Can anyone explain how Scala knows which type of collection to yield into? I know it is based on values, but how would I go about writing code that replicates yield?

谁能解释 Scala 如何知道要生成哪种类型的集合？我知道它基于值，但是我将如何编写复制yield 的代码？

Is there any way for me to change the type of the collection to yield into? In the example I want resultsto be of type Listinstead of Set.

有什么方法可以让我改变要产生的集合的类型吗？在示例中，我希望结果的类型为List而不是Set。

Failing this, what is the best way to convert from one collection to another? I know about _:*, but as a Setis not a Seqthis does not work. The best I could find thus far is val listResults = List() ++ results.

Ps. I know the example does not following the recommended functional way (which would be to use map), but it is just an example.

如果做不到这一点，从一个集合转换到另一个集合的最佳方法是什么？我知道 _:*，但是因为Set不是Seq，所以这是行不通的。迄今为止我能找到的最好的是val listResults = List() ++ results。

附言。我知道该示例没有遵循推荐的功能方式（即使用map），但这只是一个示例。

回答by missingfaktor

The forcomprehensions are translated by compiler to map/flatMap/filtercalls using thisscheme.

的for内涵是由编译器来编译map/ flatMap/filter使用调用这个方案。

This excellent answer by Danielanswers your first question.

Daniel 的这个出色回答回答了您的第一个问题。

To change the type of result collection, you can use collection.breakout(also explained in the post I linked above.)

要更改结果集合的类型，您可以使用collection.breakout（也在我上面链接的帖子中进行了解释。）

scala> val xs = Set(1, 2, 3)
xs: scala.collection.immutable.Set[Int] = Set(1, 2, 3)

scala> val ys: List[Int] = (for(x <- xs) yield 2 * x)(collection.breakOut)
ys: List[Int] = List(2, 4, 6)

You can convert a Setto a Listusing one of following ways:

您可以使用以下方式之一将 a 转换Set为 a List：

scala> List.empty[Int] ++ xs
res0: List[Int] = List(1, 2, 3)

scala> xs.toList
res1: List[Int] = List(1, 2, 3)

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Recommended read:The Architecture of Scala Collections

推荐阅读：Scala 集合的架构

回答by Kim Stebel

If you use map/flatmap/filterinstead of for comprehensions, you can use scala.collection.breakOutto create a different type of collection:

如果使用map/ flatmap/filter而不是为解析，你可以使用scala.collection.breakOut来创建不同类型的集合：

scala> val result:List[Int] = values.map(2*)(scala.collection.breakOut)
result: List[Int] = List(2, 4, 6)

If you wanted to build your own collection classes (which is the closest thing to "replicating yield" that makes any sense to me), you should have a look at this tutorial.

如果你想构建你自己的集合类（这是最接近“复制产量”的东西，对我来说很有意义），你应该看看这个教程。

回答by user3613516

Try this:

试试这个：

val values = Set(1, 2, 3)
val results = for {v <- values} yield (v * 2).toList