Converting a Java collection into a Scala collection

Question

Related to Stack Overflow question Scala equivalent of new HashSet(Collection) , how do I convert a Java collection (java.util.List say) into a Scala collection List?

I am actually trying to convert a Java API call to Spring's SimpleJdbcTemplate, which returns a java.util.List<T>, into a Scala immutable HashSet. So for example:

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

This seems to work. Criticism is welcome!

import scala.collection.immutable.Set
import scala.collection.jcl.Buffer 
val s: scala.collection.Set[String] =
                      Set(Buffer(javaApi.query( ... ) ) : _ *)

score 128 · Answer 1 · edited May 23 '17 at 12:25

128

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.

edited May 23 '17 at 12:25

Community

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answered Oct 19 '10 at 14:49

robinst

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Looks like now `JavaConversions` has some implicits making the `toSet` call not necessary. – Rajish Aug 22 '12 at 08:08
@Rajish I think it's better if the conversion is explicit (see http://stackoverflow.com/questions/8301947/what-is-the-difference-between-javaconverters-and-javaconversions-in-scala) – krookedking Mar 17 '14 at 10:37
you are right, its better to use import scala.collection.JavaConverters._ – Abhishek Sengupta May 31 '20 at 16:41

score 62 · Answer 2 · edited May 23 '17 at 12:02

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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

edited May 23 '17 at 12:02

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answered Jun 15 '11 at 11:11

Ben James

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score 25 · Answer 3 · edited May 23 '17 at 11:54

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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.

edited May 23 '17 at 11:54

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answered Aug 18 '14 at 08:52

stempler

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Why do you say the JavaConversions and JavaConverters are deprecated? In the current documentation 2.11.8 both are not marked as being deprecated and they also do not have a reference to 'decorateAll' or 'wrapAll' stating that is the new and preferred usage. – Joost den Boer Aug 24 '16 at 12:22
@JoostdenBoer Seems there were some deprecated decorators [there](https://github.com/scala/scala/blob/2.10.x/src/library/scala/collection/JavaConverters.scala) in Scala 2.10, though they seem to have been removed again with Scala 2.11. – stempler Aug 25 '16 at 12:07

score 16 · Accepted Answer · edited Sep 08 '13 at 19:10

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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.

edited Sep 08 '13 at 19:10

Erik Kaplun

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answered Mar 24 '09 at 07:51

Jorge Ortiz

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This suggestion doesn't work where I want to keep the type information – oxbow_lakes Mar 24 '09 at 08:50

score 13 · Answer 5 · answered May 12 '10 at 18:47

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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

answered May 12 '10 at 18:47

Surya Suravarapu

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That library is the greatest thing ever. Really really works well. – Michael Neale Oct 09 '10 at 00:16
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Revisiting this after couple of years, Scala's JavaConverters is the way to go. – Surya Suravarapu Jun 30 '12 at 14:32

Xavier Guihot · Answer 6 · 2022-04-15T05:57:51.163

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")

score 3 · Answer 7 · edited Oct 14 '10 at 08:30

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val array = java.util.Arrays.asList("one","two","three").toArray

val list = array.toList.map(_.asInstanceOf[String])

edited Oct 14 '10 at 08:30

oxbow_lakes

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answered Oct 08 '10 at 07:26

jamesqiu

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this is a great answer and solved my problem. Most of the methods mentioned on the internet just convert the (eg) java.util.List[java.lang.Long] to scala.collection.immutable.List[java.lang.Long]. So this method worked for me to convert it to scala.collection.immutable.List[scala.Long] – ajkl Dec 21 '14 at 05:01

score 2 · Answer 8 · answered Mar 23 '10 at 19:43

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

score 1 · Answer 9 · answered Mar 07 '14 at 05:38

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Another simple way to solve this problem:

import collection.convert.wrapAll._

answered Mar 07 '14 at 05:38

aleksandr_hramcov

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score 1 · Answer 10 · answered Mar 23 '09 at 18:50

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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)

answered Mar 23 '09 at 18:50

Fabian Steeg

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This isn't great because my java.util.List is coming back out of a Java API as a parametrized list (so my call to the API yields a java.util.List) - I'm trying to turn this into a scala immutable HashSet – oxbow_lakes Mar 23 '09 at 18:55

Converting a Java collection into a Scala collection

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