我正在学习akka流,并试图使用我写的下面的代码过滤源代码:
import akka.NotUsed;
import akka.actor.ActorSystem;
import akka.stream.javadsl.Sink;
import akka.stream.javadsl.Source;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.concurrent.CompletionStage;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.TimeoutException;
public class sd001 {
private static final ActorSystem system = ActorSystem.create("akkassembly");
private static List<RData> ls = new ArrayList();
private static class RData {
private String id;
public RData(String id){
this.id = id;
}
public List<Integer> getValues(){
return Arrays.asList(1,2,3,4,5);
}
public String getId() {
return this.id;
}
}
final static List<RData> builderFunction() {
try {
ls.add(new RData("1"));
ls.add(new RData("2"));
ls.add(new RData("3"));
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
return ls;
}
private static double calculateAverage(List <Integer> marks) {
return marks.stream()
.mapToDouble(d -> d)
.average()
.orElse(0.0);
}
public static void main(String[] args) throws ExecutionException, InterruptedException, TimeoutException {
final Source<List<RData>, NotUsed> source =
Source.repeat(NotUsed.getInstance()).map(elem -> builderFunction());
source.filter(x -> x.getId().equals("1"));
}
}
实际上,这段代码只是尝试过滤List<RData>
类型的数据结构,但source.filter(x -> x.getId().equals("1"));
行会导致编译器错误:
Cannot resolve method 'getId' in 'List'
通过阅读过滤器的akka文件:https://doc.akka.io/docs/akka/current/stream/operators/Source-or-Flow/filter.html我应该能够过滤列表值?
1条答案
按热度按时间ukqbszuj1#
使用
mapConcat
展平集合似乎可以给予所需的结果: