Convert Java List to Scala Seq

user_4708077 0
Viewed 68350

I need to implement a method that returns 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.

seq scala-java-interop scala java
7 Answers

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)
user_92971440

@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 !

user_33593100 
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 -

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"
user_16473200
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