使用JavaConversions在Java和Scala集合之间进行隐式转换

  • Z时代
  • 2024-01-10
  • 分类：问答

我使用通用函数合并了scala Set的scalaMap

def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] =
(Map[A, B]() /: (for (m <- ms; kv <- m) yield kv))
{
  (a, kv) =>
  a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv)
}

当存在相同的键冲突时，可以处理这种情况。但是，我想使用Scala

Code中的Java集合来实现。我研究了一下，发现了JavaConversions。我导入并编写了这个

def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] =
(new util.HashMap[A, B] /: (for (m <- ms; kv <- m) yield kv))
{
  case (a, kv) =>
    a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv)
}

但是，它说类型不匹配

Error:(67, 11) type mismatch;
found   : scala.collection.mutable.Map[A,B]
required: java.util.HashMap[A,B]
    a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv)
      ^

是不是JavaConversions用来隐式转换util.HashMap为mutable.Map？我在这里想念什么？

回答：

他们说尝试使用JavaConverters，因为不推荐使用JavaConversions。

scala> import collection.JavaConverters._
import collection.JavaConverters._
scala> def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] =
     | (new java.util.HashMap[A, B] /: (for (m <- ms; kv <- m) yield kv)) {
     | case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
<console>:16: error: value contains is not a member of java.util.HashMap[A,B]
       case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
                                  ^
<console>:16: error: java.util.HashMap[A,B] does not take parameters
       case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
                                                               ^
<console>:16: error: type mismatch;
 found   : (A, B)
 required: String
       case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
                                                                                    ^
<console>:15: error: type mismatch;
 found   : java.util.HashMap[A,B]
 required: Map[A,B]
       (new java.util.HashMap[A, B] /: (for (m <- ms; kv <- m) yield kv)) {
                                    ^
scala> def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] =
     | (new java.util.HashMap[A, B].asScala /: (for (m <- ms; kv <- m) yield kv)) {
     | case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
<console>:15: error: type mismatch;
 found   : scala.collection.mutable.Map[A,B]
 required: scala.collection.immutable.Map[A,B]
       (new java.util.HashMap[A, B].asScala /: (for (m <- ms; kv <- m) yield kv)) {
                                            ^
scala> def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] =
     | (new java.util.HashMap[A, B].asScala.toMap /: (for (m <- ms; kv <- m) yield kv)) {
     | case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
mergeMaps: [A, B](ms: Set[Map[A,B]])(f: (B, B) => B)Map[A,B]

也许要说明为什么不推荐使用它：

scala> def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] =
     | (new java.util.HashMap[A, B] /: (for (m <- ms; kv <- m) yield kv)) {
     | case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
<console>:19: error: type mismatch;
 found   : scala.collection.mutable.Map[A,B]
 required: java.util.HashMap[A,B]
       case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
                         ^
<console>:18: error: type mismatch;
 found   : java.util.HashMap[A,B]
 required: Map[A,B]
       (new java.util.HashMap[A, B] /: (for (m <- ms; kv <- m) yield kv)) {
                                    ^

注意for理解会产生一对对。

scala> def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B) = for (m <- ms; kv <- m) yield kv
mergeMaps: [A, B](ms: Set[Map[A,B]])(f: (B, B) => B)scala.collection.immutable.Set[(A, B)]

显然，推论既无法完成转换又无法找出操作类型。

有时将表达式分开会有助于推理，但不能在此进行。

scala> def mergeMaps[A, B](ms: Set[Map[A, B]])(f: (B, B) => B): Map[A, B] = {
     | val ss = for (m <- ms; kv <- m) yield kv
     | (new java.util.HashMap[A, B] /: ss) {
     | case (a, kv) => a + (if (a.contains(kv._1)) kv._1 -> f(a(kv._1), kv._2) else kv) }
     | }

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