将 Java 集合转换为 Scala 集合

与 Stack Overflow 问题 Scala 相当于新的 HashSet (Collection)相关，我如何将一个 Java 集合(比如说 java.util.List)转换成一个 Scala 集合 List？

我实际上正在尝试将 JavaAPI 调用转换为返回 java.util.List<T>的 春天的SimpleJdbcTemplate，转换为 Scala 不可变的 HashSet。例如:

val l: java.util.List[String] = javaApi.query( ... )
val s: HashSet[String] = //make a set from l

这似乎行得通，欢迎批评！

import scala.collection.immutable.Set
import scala.collection.jcl.Buffer
val s: scala.collection.Set[String] =
Set(Buffer(javaApi.query( ... ) ) : _ *)
82027 次浏览
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You could convert the Java collection to an array and then create a Scala list from that:

val array = java.util.Arrays.asList("one","two","three").toArray
val list = List.fromArray(array)
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最佳答案

Your last suggestion works, but you can also avoid using jcl.Buffer:

Set(javaApi.query(...).toArray: _*)

Note that scala.collection.immutable.Set is made available by default thanks to Predef.scala.

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You can add the type information in the toArray call to make the Set be parameterized:

 val s = Set(javaApi.query(....).toArray(new Array[String](0)) : _*)

This might be preferable as the collections package is going through a major rework for Scala 2.8 and the scala.collection.jcl package is going away

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You may also want to explore this excellent library: scalaj-collection that has two-way conversion between Java and Scala collections. In your case, to convert a java.util.List to Scala List you can do this:

val list = new java.util.ArrayList[java.lang.String]
list.add("A")
list.add("B")
list.asScala
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val array = java.util.Arrays.asList("one","two","three").toArray


val list = array.toList.map(_.asInstanceOf[String])
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For future reference: With Scala 2.8, it could be done like this:

import scala.collection.JavaConversions._
val list = new java.util.ArrayList[String]()
list.add("test")
val set = list.toSet

set is a scala.collection.immutable.Set[String] after this.

Also see Ben James' answer for a more explicit way (using JavaConverters), which seems to be recommended now.

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If you want to be more explicit than the JavaConversions demonstrated in robinst's answer, you can use JavaConverters:

import scala.collection.JavaConverters._
val l = new java.util.ArrayList[java.lang.String]
val s = l.asScala.toSet
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Another simple way to solve this problem:

import collection.convert.wrapAll._
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JavaConversions (robinst's answer) and JavaConverters (Ben James's answer) have been deprecated with Scala 2.10.

Instead of JavaConversions use:

import scala.collection.convert.wrapAll._

as suggested by aleksandr_hramcov.

Instead of JavaConverters use:

import scala.collection.convert.decorateAll._

For both there is also the possibility to only import the conversions/converters to Java or Scala respectively, e.g.:

import scala.collection.convert.wrapAsScala._

Update: The statement above that JavaConversions and JavaConverters were deprecated seems to be wrong. There were some deprecated properties in Scala 2.10, which resulted in deprecation warnings when importing them. So the alternate imports here seem to be only aliases. Though I still prefer them, as IMHO the names are more appropriate.

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Starting Scala 2.13, package scala.jdk.CollectionConverters replaces packages scala.collection.JavaConverters/JavaConversions._:

import scala.jdk.CollectionConverters._


// val javaList: java.util.List[String] = java.util.Arrays.asList("one","two","three")
javaList.asScala
// collection.mutable.Buffer[String] = Buffer("one", "two", "three")
javaList.asScala.toSet
// collection.immutable.Set[String] = Set("one", "two", "three")

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