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使用JavaConversions在java和scala集合之间进行隐式转换
EN

Stack Overflow用户
提问于 2016-07-12 05:32:21
回答 2查看 2.5K关注 0票数 0

我使用一个泛型函数将scala Set与scala Map合并。

代码语言:javascript
复制
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)
}

当相同的键发生冲突时,这将处理这种情况。但是,我想用Scala代码中的Java集合来实现它。我研究了一下,偶然发现了JavaConversions。我把它导入并写了这个

代码语言:javascript
复制
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)
}

然而,它说存在一种类型错配。

代码语言:javascript
复制
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)
      ^

JavaConversions不是用来隐式地将util.HashMap转换为mutable.Map吗?我在这里错过了什么?

EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2016-07-12 05:51:22

他们说要尝试JavaConverters,因为JavaConversions是不推荐的。

代码语言:javascript
复制
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]

也许是为了说明为什么它被否决了:

代码语言:javascript
复制
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)) {
                                    ^

注意到for理解会产生一组对。

代码语言:javascript
复制
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,推理不能同时完成转换,然后计算出op类型。

有时,分解表达式有助于推断,但在这里则不然。

代码语言:javascript
复制
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) }
     | }
票数 2
EN

Stack Overflow用户

发布于 2016-07-12 05:49:02

JavaConverter会做你想做的事吗?

代码语言:javascript
复制
scala> import scala.collection.JavaConverters._
import scala.collection.JavaConverters._

scala> val x = (new java.util.HashMap[Int,Int]).asScala
x: scala.collection.mutable.Map[Int,Int] = Map()
票数 7
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/38320708

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