Change Java List to Scala Seq?

Question

I have the following list from my configuration:

val markets = Configuration.getStringList("markets");

To create a sequence out of it I write this code:

JavaConverters.asScalaIteratorConverter(markets.iterator()).asScala.toSeq

I wish I could do it in a less verbose way, such as:

markets.toSeq

And then from that list I get the sequence. I will have more configuration in the near future; is there a solution that provides this kind of simplicity?

I want a sequence regardless of the configuration library I am using. I don't want to have the stated verbose solution with the JavaConverters.

com.typesafe.config, it is config-1.3.1.jar it helps me to get values from my configuration files — juan garcia
– juan garcia, Commented Mar 15, 2018 at 10:26
the question is already there stackoverflow.com/questions/17913215/… — Anahcolus
– Anahcolus, Commented Mar 15, 2018 at 10:27
Possible duplicate of How to get a list with the Typesafe config library — Anahcolus
– Anahcolus, Commented Mar 15, 2018 at 10:27
Nop I want a Sequence out of it... I don't need a list, and I want as simple as defined by @Evgeny — juan garcia
– juan garcia, Commented Mar 15, 2018 at 10:28

Jeffrey Chung · Accepted Answer · 2018-03-15 10:39:14Z

6

JavaConversions is deprecated since Scala 2.12.0. Use JavaConverters; you can import scala.collection.JavaConverters._ to make it less verbose:

import scala.collection.JavaConverters._

val javaList = java.util.Arrays.asList("one", "two")
val scalaSeq = javaList.asScala.toSeq

answered Mar 15, 2018 at 10:39

Jeffrey Chung

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vindev · Accepted Answer · 2018-03-15 10:21:51Z

1

Yes. Just import implicit conversions:

import java.util

import scala.collection.JavaConversions._

val jlist = new util.ArrayList[String]()
jlist.toSeq

edited Mar 15, 2018 at 10:21

vindev

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answered Mar 15, 2018 at 10:18

Evgeny

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