在 java 中将 java.util.HashMap 转换为 scala.collection.immutable.Map
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Convert java.util.HashMap to scala.collection.immutable.Map in java
提问by user1590420
I'm using some Scala library from my Java code. And I have a problem with collections. I need to pass scala.collection.immutable.Mapas a parameter of a method. I can convert or build immutable.Mapfrom my Java code but I do not know how to do it. Suggestions?
我正在使用我的 Java 代码中的一些 Scala 库。我在收藏方面有问题。我需要scala.collection.immutable.Map作为方法的参数传递。我可以immutable.Map从我的 Java 代码转换或构建，但我不知道如何去做。建议？
回答by Travis Brown
It's entirely possible to use JavaConvertersin Java code—there are just a couple of additional hoops to jump through:
它完全可以用JavaConverters在 Java 代码中——只需要跳过几个额外的环节：
import java.util.HashMap;
import scala.Predef;
import scala.Tuple2;
import scala.collection.JavaConverters;
import scala.collection.immutable.Map;
public class ToScalaExample {
public static <A, B> Map<A, B> toScalaMap(HashMap<A, B> m) {
return JavaConverters.mapAsScalaMapConverter(m).asScala().toMap(
Predef.<Tuple2<A, B>>conforms()
);
}
public static HashMap<String, String> example() {
HashMap<String, String> m = new HashMap<String, String>();
m.put("a", "A");
m.put("b", "B");
m.put("c", "C");
return m;
}
}
We can show that this works from the Scala REPL:
我们可以证明这在 Scala REPL 中是有效的：
scala> val jm: java.util.HashMap[String, String] = ToScalaExample.example
jm: java.util.HashMap[String,String] = {b=B, c=C, a=A}
scala> val sm: Map[String, String] = ToScalaExample.toScalaMap(jm)
sm: Map[String,String] = Map(b -> B, c -> C, a -> A)
But of course you could just as easily call these methods from Java code.
当然，您也可以从 Java 代码中轻松调用这些方法。
回答by Kohei Nozaki
My solution for Java 1.7 and Scala 2.11:
我的 Java 1.7 和 Scala 2.11 解决方案：
@SuppressWarnings("unchecked")
private static <K, V> scala.collection.immutable.Map<K, V> toScalaImmutableMap(java.util.Map<K, V> javaMap) {
final java.util.List<scala.Tuple2<K, V>> list = new java.util.ArrayList<>(javaMap.size());
for (final java.util.Map.Entry<K, V> entry : javaMap.entrySet()) {
list.add(scala.Tuple2.apply(entry.getKey(), entry.getValue()));
}
final scala.collection.Seq<Tuple2<K, V>> seq = scala.collection.JavaConverters.asScalaBufferConverter(list).asScala().toSeq();
return (scala.collection.immutable.Map<K, V>) scala.collection.immutable.Map$.MODULE$.apply(seq);
}
回答by zella
This worked for me with java 1.8 and scala 2.12:
这对我来说适用于 java 1.8 和 Scala 2.12：
public static <K, V> scala.collection.immutable.Map<K, V> toScalaImmutableMap(java.util.Map<K, V> jmap) {
List<Tuple2<K, V>> tuples = jmap.entrySet()
.stream()
.map(e -> Tuple2.apply(e.getKey(), e.getValue()))
.collect(Collectors.toList());
Seq<Tuple2<K, V>> scalaSeq = JavaConverters.asScalaBuffer(tuples).toSeq();
return (Map<K, V>) Map$.MODULE$.apply(scalaSeq);
}
回答by Brian Agnew
Can you provide an additional API call that takes/provides a java.util.Map converted using JavaConverters?
您能否提供一个额外的 API 调用来获取/提供使用JavaConverters转换的 java.util.Map ？
class Example {
import scala.collection.JavaConverters._
def fromMap(m:Map[...]) = ...
// generics etc. elided
def fromJava(m:java.util.Map) = {
fromMap(m.asScala.toMap)
}
}
You may wish to extract the conversion and provide a decorator (especially as I note you're working to a Scala library). Note dhg's comment re. immutability.
您可能希望提取转换并提供装饰器（尤其是我注意到您正在使用 Scala 库）。请注意dhg的评论。不变性。
