RxJava:如何使代码异步工作

Question

如果我对异步工作的结论有错，请与我合作，并给出如何使其异步工作的任何建议。

此代码在网格单元中执行计算的设置结果。我假设异步工作应该按部分显示网格单元。因此，如果我们有8个核心作为结果，我们可以看到显示了8个单元格，在一段时间内，显示了另外8个，以此类推(以防我设置了一些时间延迟(。但就目前而言，显示结果的单元格变成了一个接一个。

型号：

public class ListRepository implements ListRepositoryInterface {
private final Integer insertValue = 1000000;
private HashMap<String, BaseUnit> unitMap = new HashMap<>();
@Inject
public ListRepository() {}
public PublishSubject<BaseUnit> exec(int inputNumber) {
PublishSubject<BaseUnit> subject = PublishSubject.create();
Observable<BaseUnit> observable = getListObservable(inputNumber)
.subscribeOn(Schedulers.computation())
.flatMap(resultList ->
Observable.fromIterable(resultList)
.flatMap(listElem ->
Observable.fromArray(ListOperationName.values())
.map(operationElem -> {
ListUnit unit = new ListUnit(operationElem, listElem, 0);
calculate(unit, listElem);
unitMap.put(unit.getViewId(), unit);
return unit;
})
)
);
observable.subscribe(subject);
return subject;
}

private Observable<ArrayList<List<Integer>>> getListObservable(int inputNumber) {
return Observable.fromCallable(() -> {
ArrayList<List<Integer>> list = new ArrayList<>();
Integer[] populatedArray = new Integer[inputNumber];
Arrays.fill(populatedArray, insertValue);
list.add(new ArrayList<>(Arrays.asList(populatedArray)));
list.add(new LinkedList<>(Arrays.asList(populatedArray)));
list.add(new CopyOnWriteArrayList<>(Arrays.asList(populatedArray)));
return list;
});
}

private void calculate(ListUnit unit, List<Integer> list) {
try {
TimeUnit.MILLISECONDS.sleep(50);
} catch (InterruptedException e) {
e.printStackTrace();
}
double start = getTime();
//noinspection SynchronizationOnLocalVariableOrMethodParameter
synchronized (list) {
switch (unit.getOperationName()) {
case ADD_FIRST:
list.add(0, insertValue);
break;
case ADD_MID:
list.add(list.size() / 2, insertValue);
break;
case ADD_LAST:
list.add(insertValue);
break;
case SEARCH:
//noinspection unused
boolean contains = list.contains(insertValue);
break;
case RM_FIRST:
list.remove(0);
break;
case RM_MID:
list.remove(list.size() / 2);
break;
case RM_LAST:
list.remove(list.size() - 1);
break;
}
}
unit.setTime(getTime() - start);
}
private double getTime() {
return System.nanoTime();
}
public HashMap<String, BaseUnit> getUnitMap() {
return unitMap;
}
}

演示者：

public void calculate(int inputNumber) {
fragment.showAllProgressBars();
PublishSubject<BaseUnit> subject = repository.exec(inputNumber);
Disposable disposable = subject.observeOn(AndroidSchedulers.mainThread())
.subscribe(unit -> {
fragment.setCellText(unit.getViewId(), unit.getTimeString());
}, Throwable::printStackTrace);
}

UPD：现在我做了测试示例，并试图正确订阅主题。如果我使用subject.onNext((，它是异步工作的，但我认为这是错误的，因为它没有检查subject.hasComplete((和get"真"；。参见评论"；TODO"；在下面

private void run() {
Log.d("APP", "INIT");
PublishSubject<String> subject = exec(1000000);
subject.observeOn(AndroidSchedulers.mainThread())
.subscribe(unit -> {
Log.d("STRING RESULT = ", unit);
if (subject.hasComplete()) {
//TODO: this condition should work if we use observable.subscribe(subject), not subject.onNext
Log.d("SUBJECT", "COMPLETED");
}
}, Throwable::printStackTrace);
}
private int insertValue = 1000000;
public PublishSubject<String> exec(int inputNumber) {
PublishSubject<String> subject = PublishSubject.create();
getListObservable(inputNumber)
.flatMap(resultList -> getOperationsObservable()
.flatMap(operationElem -> getResultListObservable(resultList)
.map(listElem ->
calculate(operationElem, listElem)
//TODO: should be smth like this, i.e. calculate.subscribe(subject)
//                        .subscribe(subject)
.subscribe(subject::onNext)
)
)
).subscribe();
return subject;
}

private Observable<ArrayList<List<Integer>>> getListObservable(int inputNumber) {
return Observable.fromCallable(() -> {
ArrayList<List<Integer>> list = new ArrayList<>();
Integer[] populatedArray = new Integer[inputNumber];
Arrays.fill(populatedArray, insertValue);
list.add(new ArrayList<>(Arrays.asList(populatedArray)));
list.add(new LinkedList<>(Arrays.asList(populatedArray)));
list.add(new CopyOnWriteArrayList<>(Arrays.asList(populatedArray)));
return list;
}).subscribeOn(Schedulers.computation());
}

private Observable<String> calculate(ListOperationName operationName, List<Integer> list) {
return Observable.fromCallable(() -> {
try {
TimeUnit.MILLISECONDS.sleep(100);
} catch (InterruptedException e) {
e.printStackTrace();
}
double start = getTime();
synchronized (list) {
switch (operationName) {
case ADD_FIRST:
list.add(0, insertValue);
break;
case ADD_MID:
list.add(list.size() / 2, insertValue);
break;
case ADD_LAST:
list.add(insertValue);
break;
case SEARCH:
//noinspection unused
boolean contains = list.contains(insertValue);
break;
case RM_FIRST:
list.remove(0);
break;
case RM_MID:
list.remove(list.size() / 2);
break;
case RM_LAST:
list.remove(list.size() - 1);
break;
}
}
return Double.toString(getTime() - start);
}).subscribeOn(Schedulers.newThread());
}
private Observable<ListOperationName> getOperationsObservable() {
return Observable.fromArray(ListOperationName.values());
}
private Observable<List<Integer>> getResultListObservable(ArrayList<List<Integer>> list) {
return Observable.fromIterable(list);
}
private double getTime() {
return System.nanoTime();
}
public enum ListOperationName {
ADD_FIRST,
ADD_MID,
ADD_LAST,
SEARCH,
RM_FIRST,
RM_MID,
RM_LAST;
}

等级：

dependencies {
implementation 'io.reactivex.rxjava3:rxandroid:3.0.0'
implementation 'io.reactivex.rxjava3:rxjava:3.0.4'
}

Answer 1

如果我在最近的示例中从calculate(...)方法中删除.subscribeOn(Schedulers.newThread());，并对exec(...)进行以下更改，我似乎得到了所需的并行度。

public Observable<String> exec( int inputNumber )
{
return getListObservable( inputNumber )
.flatMap( resultList -> getOperationsObservable()
.flatMap( operationElem -> getResultListObservable( resultList )
.flatMap( listElem -> Observable.just( listElem )
.subscribeOn( Schedulers.computation() )
.flatMap( __ -> calculate( operationElem, listElem )))));
}

我删除了PublishSubject——它似乎是多余的，你可以订阅生成的Observable，不是吗？