[英]Convert Java List to Scala Seq
I need to implement a method that returns a Scala Seq
, in Java.我需要在 Java 中实现一个返回 Scala Seq
的方法。
But I encounter this error:但是我遇到了这个错误:
java.util.ArrayList cannot be cast to scala.collection.Seq
Here is my code so far:到目前为止,这是我的代码:
@Override
public Seq<String> columnNames() {
List<String> a = new ArrayList<String>();
a.add("john");
a.add("mary");
Seq<String> b = (scala.collection.Seq<String>) a;
return b;
}
But scala.collection.JavaConverters
doesn't seem to offer the possibility to convert as a Seq
.但是scala.collection.JavaConverters
似乎没有提供转换为Seq
的可能性。
JavaConverters is what I needed to solve this. JavaConverters 是我解决这个问题所需要的。
import scala.collection.JavaConverters;
public Seq<String> convertListToSeq(List<String> inputList) {
return JavaConverters.asScalaIteratorConverter(inputList.iterator()).asScala().toSeq();
}
JavaConversions
should work. JavaConversions
应该可以工作。 I think, you are looking for something like this: JavaConversions.asScalaBuffer(a).toSeq()
我想,你正在寻找这样的东西: JavaConversions.asScalaBuffer(a).toSeq()
Starting Scala 2.13
, package scala.jdk.javaapi.CollectionConverters
replaces deprecated packages scala.collection.JavaConverters/JavaConversions
:从Scala 2.13
开始,包scala.jdk.javaapi.CollectionConverters
替换了不推荐使用的包scala.collection.JavaConverters/JavaConversions
:
import scala.jdk.javaapi.CollectionConverters;
// List<String> javaList = Arrays.asList("a", "b");
CollectionConverters.asScala(javaList).toSeq();
// Seq[String] = List(a, b)
This worked for me!这对我有用! (Java 8, Spark 2.0.0) (Java 8,火花 2.0.0)
import java.util.ArrayList;
import scala.collection.JavaConverters;
import scala.collection.Seq;
public class Java2Scala
{
public Seq<String> getSeqString(ArrayList<String> list)
{
return JavaConverters.asScalaIterableConverter(list).asScala().toSeq();
}
}
@ Fundhor , the method asScalaIterableConverter
was not showing up in the IDE. @ Fundhor ,方法asScalaIterableConverter
没有出现在 IDE 中。 It may be due to a difference in the versions of Scala.这可能是由于 Scala 的版本不同。 I am using Scala 2.11.我正在使用 Scala 2.11。 Instead, it showed up asScalaIteratorConverter
.相反,它显示为asScalaIteratorConverter
。 I made a slight change to your final snippet and it worked fine for me.我对你的最终片段做了一些小改动,对我来说效果很好。
scala.collection.JavaConverters.asScalaIteratorConverter(columnNames.iterator()).asScala().toSeq()
where columnNames
is a java.util.List
. scala.collection.JavaConverters.asScalaIteratorConverter(columnNames.iterator()).asScala().toSeq()
其中columnNames
是一个java.util.List
。
thanks !谢谢 !
Up to 4 elements, you can simply use the factory method of the Seq class like this :最多 4 个元素,您可以简单地使用 Seq 类的工厂方法,如下所示:
Seq<String> seq1 = new Set.Set1<>("s1").toSeq();
Seq<String> seq2 = new Set.Set2<>("s1", "s2").toSeq();
Seq<String> seq3 = new Set.Set3<>("s1", "s2", "s3").toSeq();
Seq<String> seq4 = new Set.Set4<>("s1", "s2", "s3", "s4").toSeq();
import scala.collection.JavaConverters;
import scala.collection.Seq;
import java.util.ArrayList;
public class Helpers {
public Seq<String> convertListToSeq(ArrayList<String> inputList) {
return JavaConverters.collectionAsScalaIterableConverter(inputList).asScala().toSeq();
}
}
Versions -版本 -
compile 'org.apache.spark:spark-core_2.11:2.3.1'
compile 'org.apache.spark:spark-sql_2.11:2.3.1'
compile group: 'commons-io', name: 'commons-io', version: '2.6'
compile "com.fasterxml.jackson.module:jackson-module-scala_2.11:2.8.8"
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