I need to implement a method that return a Scala Seq
, in Java.
But I encounter this error:
java.util.ArrayList cannot be cast to scala.collection.Seq
Here is my code so far:
@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
.
JavaConverters is what I needed to solve this.
import scala.collection.JavaConverters;
public Seq<String> convertListToSeq(List<String> inputList) {
return JavaConverters.asScalaIteratorConverter(inputList.iterator()).asScala().toSeq();
}
JavaConversions
should work. I think, you are looking for something like this: JavaConversions.asScalaBuffer(a).toSeq()
@Fundhor, the method asScalaIterableConverter
was not showing up in the IDE. It may be due to a difference in the versions of Scala. I am using Scala 2.11. Instead, it showed up asScalaIteratorConverter
. I made a slight change to your final snippet and it worked fine for me.scala.collection.JavaConverters.asScalaIteratorConverter(columnNames.iterator()).asScala().toSeq()
where columnNames
is a java.util.List
.
thanks !
This worked for me! (Java 8, Spark 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();
}
}
Up to 4 elements, you can simply use the factory method of the Seq class like this :
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();
Starting Scala 2.13
, package scala.jdk.javaapi.CollectionConverters
replaces deprecated packages 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)
来源:https://stackoverflow.com/questions/35988315/convert-java-list-to-scala-seq