Convert Java List to Scala Seq

Question

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.

Possible duplicate of [Converting a Java collection into a Scala collection](http://stackoverflow.com/questions/674713/converting-a-java-collection-into-a-scala-collection) — Tzach Zohar, Mar 14 '16 at 13:04
You're asking for trouble if you try to do anything with Scala collections in Java. Write a bit of Scala code which does the conversion using scala.collection.JavaConverters. — Luigi Plinge, Mar 14 '16 at 13:15
@TzachZohar most of the answer are done is scala. My code is Java, it's part of a huge program, I can't externalize this treatment... — Fundhor, Mar 14 '16 at 13:20
As @LuigiPlinge mentioned this is not a good idea! You'll probably have a lot of problems trying to invoke method on this sequence, god forbid you try invoking anything that requires implicits — Mateusz Dymczyk, Mar 14 '16 at 13:31
"they can result in unexpected behavior and performance" in the docs doesn't sound very convincing ... especially, since the very next sentence suggests `JavaConverters` that "offer the _same_ conversions". :) — Dima, May 16 '17 at 13:04

Fundhor · Accepted Answer · 2018-10-26T07:28:03.187

51

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();
}

edited Oct 26 '18 at 07:28

answered Mar 14 '16 at 13:29

Fundhor

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Slightly shorter version: `JavaConverters.asScalaIterableConverter(a).asScala().toSeq();` – Adrien Brunelat Mar 12 '18 at 13:23
`JavaConverters.collectionAsScalaIterableConverter(list).asScala().toSeq()` was the proper approach for me with Scala 2.12.6. There is also a `JavaConverters.iterableAsScalaIterableConverter` method if the source is an iterable instead of a collection. – Per Lundberg May 16 '19 at 08:55
asScalaIteratorConverter (java.util.Iterator) in JavaConverters cannot be applied to (scala.collection.Iterator) reason: no instance(s) of type variable(s) A exist so that Iterator conforms to Iterator – sapy Jan 16 '20 at 16:12

score 24 · Answer 2 · answered Mar 14 '16 at 13:48

24

JavaConversions should work. I think, you are looking for something like this: JavaConversions.asScalaBuffer(a).toSeq()

answered Mar 14 '16 at 13:48

Dima

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`JavaConversions` is deprecated and now it should be written as `JavaConverters.asScalaBuffer(a)` – kap May 16 '17 at 12:09
_Should_ be? Nah ... Recommended by some unknown person who thought it was a better idea is more like it :) – Dima May 16 '17 at 12:54
1

Probably. I just followed the hint in the Scala API. I do not know either of the persons who was writing and/or deprecating it ;-) – kap May 16 '17 at 12:59

Xavier Guihot · Answer 3 · 2022-01-03T10:30:04.063

16

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)

edited Jan 03 '22 at 10:30

answered Mar 29 '19 at 21:39

Xavier Guihot

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score 6 · Answer 4 · answered Jul 25 '17 at 13:14

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();
        }

}

score 5 · Answer 5 · edited May 23 '17 at 11:47

@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 !

score 5 · Answer 6 · answered Mar 13 '17 at 15:17

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();

score 4 · Answer 7 · answered Jan 16 '20 at 16:17

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"

Convert Java List to Scala Seq

7 Answers7

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