将 Java 列表转换为 Scala Seq
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35988315/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Convert Java List to Scala Seq
提问by Fundhor
I need to implement a method that returns a Scala Seq, in Java.
我需要Seq在 Java 中实现一个返回 Scala 的方法。
But I encounter this error:
但是我遇到了这个错误：
java.util.ArrayList cannot be cast to scala.collection.Seq
Here is my code so far:
到目前为止，这是我的代码：
@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.JavaConvertersdoesn't seem to offer the possibility to convert as a Seq.
但scala.collection.JavaConverters似乎没有提供转换为Seq.
采纳答案by Fundhor
JavaConverters is what I needed to solve this.
JavaConverters 是我解决这个问题所需要的。
import scala.collection.JavaConverters;
public Seq<String> convertListToSeq(List<String> inputList) {
return JavaConverters.asScalaIteratorConverter(inputList.iterator()).asScala().toSeq();
}
回答by Dima
JavaConversionsshould work. I think, you are looking for something like this: JavaConversions.asScalaBuffer(a).toSeq()
JavaConversions应该管用。我想，你正在寻找这样的东西：JavaConversions.asScalaBuffer(a).toSeq()
回答by Neeleshkumar S
@Fundhor, the method asScalaIterableConverterwas 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 columnNamesis a java.util.List.
@ Fundhor，该方法asScalaIterableConverter未出现在 IDE 中。这可能是由于 Scala 的版本不同。我正在使用 Scala 2.11。相反，它出现了asScalaIteratorConverter。我对你的最终片段做了一些小改动，对我来说效果很好。scala.collection.JavaConverters.asScalaIteratorConverter(columnNames.iterator()).asScala().toSeq()哪里columnNames是java.util.List.
thanks !
谢谢 ！
回答by loicmathieu
Up to 4 elements, you can simply use the factory method of the Seq class like this :
最多 4 个元素，您可以简单地使用 Seq 类的工厂方法，如下所示：
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();
回答by Nikhil
This worked for me! (Java 8, Spark 2.0.0)
这对我有用！（Java 8，火花 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();
}
}
回答by Xavier Guihot
Starting Scala 2.13, package scala.jdk.javaapi.CollectionConvertersreplaces deprecated packages scala.collection.JavaConverters/JavaConversions:
开始Scala 2.13，包scala.jdk.javaapi.CollectionConverters替换不推荐使用的包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)
回答by sapy
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 -
版本 -
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.Hymanson.module:Hymanson-module-scala_2.11:2.8.8"
