将多个异步 Observable<List> 减少为一个 Observable<List>

Question

所以这就是我正在尝试做的事情：
对于每个事件，我想调用一些 Service.getList() 10 次，然后将 10 个列表合并为一个。

现在我尝试了这两种方法，它们都在 UT 中工作但在真实应用程序中失败（我猜我在真实应用程序中给定异步 http 调用时我没有正确执行 reduce 操作）。对于这两种情况，我都看不到 reduce()、onNext() 和 onError() 之后的日志，所以我猜测 reduce() 操作没有完成。

尝试 #1：

public Observable<List<Event>> getEventsForLocation(Location location) {
List<Observable<List<Event>>> obs = new ArrayList<>();
for (Venue v : location.getVenues()) {
     obs.add(getEventsForVenue(v)); //does one http call, returns Observable<List<Event>> 
}
return Observable.concat(Observable.from(obs))
            .reduce((List<Event>) new ArrayList<Event>(), (events, events2) -> {
                events.addAll(events2);
                return events;
            })
            .doOnNext(events -> Log.d("reduce ", events.toString()))
            .doOnError(throwable -> Log.e("reduce error", throwable.toString()));}

尝试#2：

public Observable<List<Event>> getEventsForLocation(Location location) {
return Observable
        .from(location.getVenues())
        .flatMap(venue -> getEventsForVenue(venue)) //does one http call, returns Observable<List<Event>> 
        .reduce((List<Event>) new ArrayList<Event>(), (events, events2) -> {
            events.addAll(events2);
            return events;
        })
       .doOnNext(events -> Log.d("service", "total events " + events.toString()))
       .doOnError(t -> Log.e("service", "total events error2 " + t.toString()));}

两种方法都通过的 UT：

    @Test
    public void getEventsForLocation() {
        Location loc = new Location("test", newArrayList(new Venue("v1", "url1"),new Venue("v2", "url2")));

        when(httpGateway.downloadWebPage(Mockito.anyString())).thenReturn(
                Observable.just(readResource("eventsForVenue1.html")),
                Observable.just(readResource("eventsForVenue2.html"))
        );

        TestSubscriber<List<Event>> probe = new TestSubscriber<>();
        service.getEventsForLocation(loc).subscribe(probe);

        probe.assertNoErrors();

        //assert the next event containts contents of all lists
        List<Event> events = probe.getOnNextEvents().get(0);

        //first list
        Assert.assertEquals("Unexpected title", "event1", events.get(0).getName());
        Assert.assertEquals("Unexpected artist", "artist1", events.get(0).getArtist());
        //second list
        Assert.assertEquals("Unexpected title", "event2", events.get(1).getName());
        Assert.assertEquals("Unexpected artist", "artist2", events.get(1).getArtist());
    }

更新

这是更完整的代码，带有调度程序。

Observable
                .just(loc)
                .subscribeOn(AndroidSchedulers.mainThread())
                .observeOn(Schedulers.io())
                .flatMap(location -> service.getEventsForLocation(location))
                .observeOn(AndroidSchedulers.mainThread())
                .subscribe(getObserver();

Answer 1

您可以使用 Collect 来解决问题。但是在我合并多个项目的这种情况下，通常我更愿意使用 Scan。对于每个新项目，它都会为您提供最后处理的项目。因此，您可以将每个项目附加到之前的项目。

看看弹珠图是否适合你的情况 http://reactivex.io/documentation/operators/scan.html

尝试从这个简单的示例开始，然后尝试将其应用到您的代码中

/**
 * apply this function for every item against the previous emitted item from the source.
 *  Emitted:
            0
            1
            3
            6
            10
            15
 */
@Test
public void scanObservable() {
    Integer[] numbers = {0, 1, 2, 3, 4, 5};

    Observable.from(numbers)
              .scan((lastItemEmitted, newItem) -> (lastItemEmitted + newItem))
              .subscribe(System.out::println);
}

Answer 2

我终于找到了一个解决方案，使用 zip() 但我不喜欢它。
问题应该可以通过 flatMap/reduce

的组合来解决

public Observable<List<Event>> getEventsForLocation(Location location) {
        List<Observable<List<Event>>> venues = new ArrayList<>();
        for (Venue v : location.getVenues()) {
            venues.add(Observable.just(v).flatMap(venue -> getEventsForVenue(venue)));
        }
        return Observable.zip(venues, new FuncN<List<Event>>() {
            @Override
            public List<Event> call(Object... args) {
                List<Event> allEvents = new ArrayList<Event>();
                for (Object o : args) {
                    List<Event> le = (List<Event>) o;
                    allEvents.addAll(le);
                }
                return allEvents;
            }
        });

Answer 3

如果你不关心列表中的最后顺序，你可以直接使用from + flatMap + flatMapIterable + toList:

Observable.from(location.getVenues())
.flatMap(venue -> getEventsForVenue(venue))
.flatMapIterable(list -> list)
.toList();

如果顺序很重要并且您想在 "parallel" 中执行 getEventsForVenue，您可以将 flatMap 替换为 concatMapEager:

Observable.from(location.getVenues())
.concatMapEager(venue -> getEventsForVenue(venue))
.concatMapIterable(list -> list)
.toList();

将多个异步 Observable<List> 减少为一个 Observable<List>

Reduce many async Observable<List> into one Observable<List>

java

multithreading

reactive-programming

rx-java

rx-android