list Scala：将 Map 映射到元组列表
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Scala: map a Map to list of tuples
提问by user2051561
I tried to use Map.map to convert a map into a List of Tuples. However this fails. I did the following experiments:
我尝试使用 Map.map 将地图转换为元组列表。然而这失败了。我做了以下实验：
val m = Map(("a" -> 1), ("b" -> 2))
//> m : scala.collection.immutable.Map[String,Int] = Map(a -> 1, b -> 2)
val r1 = m.map{ case (k,v) => v} //> r1 : scala.collection.immutable.Iterable[Int] = List(1, 2)
def toTuple[A,B](a:A,b:B) = (a,b) //> toTuple: [A, B](a: A, b: B)(A, B)
//val r2: List[Tuple2[_,_]] = m.map(e => (e._1,e._2))
val r3 = m.map(e => toTuple(e._1,e._2)) //> r3 : scala.collection.immutable.Map[String,Int] = Map(a -> 1, b -> 2)
val r4 = m.toSeq //> r4 : Seq[(String, Int)] = ArrayBuffer((a,1), (b,2))
Notice how a List is generated for single elements (r1) but a Map is produced for tuples (r3). Not even forcing the type worked (r2). Only an explicit call to Seq did it (r4) So my question is, why/how does Map.map "automagically" create a new Map and not a list for example? In fact how is the return type determined (Seq, List, etc.)
请注意如何为单个元素 (r1) 生成 List，但如何为元组 (r3) 生成 Map。甚至不强制类型工作（r2）。只有对 Seq 的显式调用做到了（r4）所以我的问题是，为什么/如何 Map.map“自动”创建一个新的 Map 而不是一个列表？实际上如何确定返回类型（Seq、List 等）
回答by Michael Zajac
A Mapis a collection of tuples already.
AMap已经是元组的集合。
scala> "b" -> 2
res0: (String, Int) = (b,2) // Implicitly converted to a Tuple
When you're mapping a Map, you're mapping the (key, value) pairs that it contains. This can't work, because you're stripping away the keys, and retaining only the values. So what you have is no longer a Map, but a step or two up the collection hierarchy, an Iterable:
映射 a 时Map，就是映射它包含的 (key, value) 对。这是行不通的，因为您正在剥离键，只保留值。所以你所拥有的不再是一个Map，而是集合层次结构的一两步，一个Iterable：
val r1 = m.map{ case (k,v) => v}
Forcing the type cannot work, because a Map[A, B]is not a List[(A, B)]. This is the equivalent of m.map(identity). Notice how you're even accessing ewith tuple accessors:
强制类型不起作用，因为 aMap[A, B]不是 a List[(A, B)]。这相当于m.map(identity). 注意你是如何e使用元组访问器访问的：
val r2: List[Tuple2[_,_]] = m.map(e => (e._1,e._2))
val r3 = m.map(e => toTuple(e._1,e._2))
Here, Seqis more generalized than List:
在这里，Seq比 更普遍List：
val r4 = m.toSeq
The simple solution as stated by @EndeNeu is to just use toList. When you mapa collection, it should return the original collection type if it can. So mapping a Mapshould return another Map, unless the underlying structure has made it no longer a Map(like removing keys entirely) in r1.
@EndeNeu 所述的简单解决方案是使用toList. 当您创建map一个集合时，如果可以，它应该返回原始集合类型。所以映射 aMap应该返回 another Map，除非底层结构使它不再是 a Map（就像完全删除键） in r1。
