正确处理RxJava中的空Observable

Question

我有一种情况,我正在创建一个包含数据库结果的Observable.然后我将一系列过滤器应用于它们.然后我有一个订阅者正在记录结果.情况可能是没有元素通过过滤器.我的业务逻辑表明这不是错误.但是,当发生这种情况时,我的onError被调用并包含以下异常:java.util.NoSuchElementException: Sequence contains no elements

公认的做法只是检测那种类型的异常并忽略它吗？或者有更好的方法来处理这个问题吗？

版本是1.0.0.

这是一个简单的测试用例,揭示了我所看到的内容.它似乎与在到达地图之前过滤所有事件并减少有关.

  @Test
public void test()
{

    Integer values[] = new Integer[]{1, 2, 3, 4, 5};

    Observable.from(values).filter(new Func1<Integer, Boolean>()
    {
        @Override
        public Boolean call(Integer integer)
        {
            if (integer < 0)
                return true;
            else
                return false;
        }
    }).map(new Func1<Integer, String>()
    {
        @Override
        public String call(Integer integer)
        {
            return String.valueOf(integer);
        }
    }).reduce(new Func2<String, String, String>()
    {
        @Override
        public String call(String s, String s2)
        {
            return s + "," + s2;
        }
    })

            .subscribe(new Action1<String>()
            {
                @Override
                public void call(String s)
                {
                    System.out.println(s);
                }
            });
}

因为我使用的是安全订阅者,所以它最初会抛出一个包含以下异常的OnErrorNotImplementedException:

java.util.NoSuchElementException: Sequence contains no elements
    at rx.internal.operators.OperatorSingle$1.onCompleted(OperatorSingle.java:82)
    at rx.internal.operators.NotificationLite.accept(NotificationLite.java:140)
    at rx.internal.operators.TakeLastQueueProducer.emit(TakeLastQueueProducer.java:73)
    at rx.internal.operators.TakeLastQueueProducer.startEmitting(TakeLastQueueProducer.java:45)
    at rx.internal.operators.OperatorTakeLast$1.onCompleted(OperatorTakeLast.java:59)
    at rx.internal.operators.OperatorScan$2.onCompleted(OperatorScan.java:121)
    at rx.internal.operators.OperatorMap$1.onCompleted(OperatorMap.java:43)
    at rx.internal.operators.OperatorFilter$1.onCompleted(OperatorFilter.java:42)
    at rx.internal.operators.OnSubscribeFromIterable$IterableProducer.request(OnSubscribeFromIterable.java:79)
    at rx.internal.operators.OperatorScan$2$1.request(OperatorScan.java:147)
    at rx.Subscriber.setProducer(Subscriber.java:139)
    at rx.internal.operators.OperatorScan$2.setProducer(OperatorScan.java:139)
    at rx.Subscriber.setProducer(Subscriber.java:133)
    at rx.Subscriber.setProducer(Subscriber.java:133)
    at rx.internal.operators.OnSubscribeFromIterable.call(OnSubscribeFromIterable.java:47)
    at rx.internal.operators.OnSubscribeFromIterable.call(OnSubscribeFromIterable.java:33)
    at rx.Observable$1.call(Observable.java:144)
    at rx.Observable$1.call(Observable.java:136)
    at rx.Observable$1.call(Observable.java:144)
    at rx.Observable$1.call(Observable.java:136)
    at rx.Observable$1.call(Observable.java:144)
    at rx.Observable$1.call(Observable.java:136)
    at rx.Observable$1.call(Observable.java:144)
    at rx.Observable$1.call(Observable.java:136)
    at rx.Observable$1.call(Observable.java:144)
    at rx.Observable$1.call(Observable.java:136)
    at rx.Observable.subscribe(Observable.java:7284)

基于下面@davem的答案,我创建了一个新的测试用例:

@Test
public void testFromBlockingAndSingle()
{

    Integer values[] = new Integer[]{-2, -1, 0, 1, 2, 3, 4, 5};

    List<String> results = Observable.from(values).filter(new Func1<Integer, Boolean>()
    {
        @Override
        public Boolean call(Integer integer)
        {
            if (integer < 0)
                return true;
            else
                return false;
        }
    }).map(new Func1<Integer, String>()
    {
        @Override
        public String call(Integer integer)
        {
            return String.valueOf(integer);
        }
    }).reduce(new Func2<String, String, String>()
    {
        @Override
        public String call(String s, String s2)
        {
            return s + "," + s2;
        }
    }).toList().toBlocking().single();

    System.out.println("Test: " + results + " Size: " + results.size());

}

此测试会导致以下行为:

当输入是:

Integer values[] = new Integer[]{-2, -1, 0, 1, 2, 3, 4, 5};

然后结果(如预期的那样)是:

Test: [-2,-1] Size: 1

当输入是:

Integer values[] = new Integer[]{0, 1, 2, 3, 4, 5};

然后结果是以下堆栈跟踪:

java.util.NoSuchElementException: Sequence contains no elements
at rx.internal.operators.OperatorSingle$1.onCompleted(OperatorSingle.java:82)
at rx.internal.operators.NotificationLite.accept(NotificationLite.java:140)
at rx.internal.operators.TakeLastQueueProducer.emit(TakeLastQueueProducer.java:73)
at rx.internal.operators.TakeLastQueueProducer.startEmitting(TakeLastQueueProducer.java:45)
at rx.internal.operators.OperatorTakeLast$1.onCompleted(OperatorTakeLast.java:59)
at rx.internal.operators.OperatorScan$2.onCompleted(OperatorScan.java:121)
at rx.internal.operators.OperatorMap$1.onCompleted(OperatorMap.java:43)
at rx.internal.operators.OperatorFilter$1.onCompleted(OperatorFilter.java:42)
at rx.internal.operators.OnSubscribeFromIterable$IterableProducer.request(OnSubscribeFromIterable.java:79)
at rx.internal.operators.OperatorScan$2$1.request(OperatorScan.java:147)
at rx.Subscriber.setProducer(Subscriber.java:139)
at rx.internal.operators.OperatorScan$2.setProducer(OperatorScan.java:139)
at rx.Subscriber.setProducer(Subscriber.java:133)
at rx.Subscriber.setProducer(Subscriber.java:133)
at rx.internal.operators.OnSubscribeFromIterable.call(OnSubscribeFromIterable.java:47)
at rx.internal.operators.OnSubscribeFromIterable.call(OnSubscribeFromIterable.java:33)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable$1.call(Observable.java:144)
at rx.Observable$1.call(Observable.java:136)
at rx.Observable.subscribe(Observable.java:7284)
at rx.observables.BlockingObservable.blockForSingle(BlockingObservable.java:441)
at rx.observables.BlockingObservable.single(BlockingObservable.java:340)
at EmptyTest2.test(EmptyTest2.java:19)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:47)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:44)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:271)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:70)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:50)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:238)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:63)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:236)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:53)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:229)
at org.junit.runners.ParentRunner.run(ParentRunner.java:309)
at org.junit.runner.JUnitCore.run(JUnitCore.java:160)
at com.intellij.junit4.JUnit4IdeaTestRunner.startRunnerWithArgs(JUnit4IdeaTestRunner.java:74)
at com.intellij.rt.execution.junit.JUnitStarter.prepareStreamsAndStart(JUnitStarter.java:211)
at com.intellij.rt.execution.junit.JUnitStarter.main(JUnitStarter.java:67)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at com.intellij.rt.execution.application.AppMain.main(AppMain.java:134)

所以似乎问题肯定是使用reduce函数.请参阅以下处理这两种情况的测试用例:

@Test
public void testNoReduce()
{

    Integer values[] = new Integer[]{-2, -1, 0, 1, 2, 3, 4, 5};

    List<String> results = Observable.from(values).filter(new Func1<Integer, Boolean>()
    {
        @Override
        public Boolean call(Integer integer)
        {
            if (integer < 0)
                return true;
            else
                return false;
        }
    }).map(new Func1<Integer, String>()
    {
        @Override
        public String call(Integer integer)
        {
            return String.valueOf(integer);
        }
    }).toList().toBlocking().first();

    Iterator<String> itr = results.iterator();
    StringBuilder b = new StringBuilder();

    while (itr.hasNext())
    {
        b.append(itr.next());

        if (itr.hasNext())
            b.append(",");
    }

    System.out.println("Test NoReduce: " + b);

}

通过以下输入:

Integer values[] = new Integer[]{-2, -1, 0, 1, 2, 3, 4, 5};

我得到了以下预期的结果:

Test NoReduce: -2,-1

并通过以下输入:

Integer values[] = new Integer[]{0, 1, 2, 3, 4, 5};

我得到了以下预期的输出:

Test NoReduce: 

因此,除非我完全误解了某些东西,否则真正处理零元素Observable的唯一方法是在迭代后跟一个map并使用reduce来实现在Observable链之外的reduce逻辑.你们都同意这个说法吗？

---

最终解决方案

这是我实施TomášDvořák和David Motten建议的最终解决方案.我认为这个解决方案是合理的.

@Test
public void testWithToList()
{

    Integer values[] = new Integer[]{-2, -1, 0, 1, 2, 3, 4, 5};

     Observable.from(values).filter(new Func1<Integer, Boolean>()
     {
         @Override
         public Boolean call(Integer integer)
         {
             if (integer < 0)
                 return true;
             else
                 return false;
         }
     }).toList().map(new Func1<List<Integer>, String>()
     {
         @Override
         public String call(List<Integer> integers)
         {
             Iterator<Integer> intItr = integers.iterator();
             StringBuilder b = new StringBuilder();

             while (intItr.hasNext())
             {
                 b.append(intItr.next());

                 if (intItr.hasNext())
                 {
                     b.append(",");
                 }
             }

             return b.toString();
         }
     }).subscribe(new Action1<String>()
     {
         @Override
         public void call(String s)
         {
             System.out.println("With a toList: " + s);
         }
     });

}

以下是给定以下输入时此测试的行为方式.

当给定一个将通过过滤器传递一些值的流时:

Integer values[] = new Integer[]{-2, -1, 0, 1, 2, 3, 4, 5};

结果是:

With a toList: -2,-1

当给定一个没有任何值的流通过过滤器时:

Integer values[] = new Integer[]{0, 1, 2, 3, 4, 5};

结果是:

With a toList: <empty string>

Answer 1

现在更新后,错误非常明显.Reduce在RxJava中将失败,IllegalArgumentException如果它正在减少的可观察量是空的,完全按照规范(http://reactivex.io/documentation/operators/reduce.html).

在函数式编程中,通常有两个泛型运算符将集合聚合为单个值,fold并且reduce.在接受的术语中,fold获取初始累加器值,以及从集合中获取正在运行的累加器和值的函数,并生成另一个累加器值.伪代码中的一个例子:

[1, 2, 3, 4].fold(0, (accumulator, value) => accumulator + value)

将从0开始,最后将1,2,3,4添加到正在运行的累加器,最后得到10,即值的总和.

Reduce非常相似,只是它没有显式获取初始累加器值,它使用第一个值作为初始累加器,然后累积所有剩余值.如果您正在寻找最小值或最大值,这是有道理的.

[1, 2, 3, 4].reduce((accumulator, value) => min(accumulator, value))

看看fold并减少不同的方式,你可能会使用fold聚合值,即使在空集合中也是有意义的(比如,in sum,0有意义),reduce否则(minimum在空集合上没有意义,并且reduce将无法操作这样的收集,在你的情况下抛出一个例外).

您正在进行类似的聚合,使用逗号散布一组字符串以生成单个字符串.那是一个有点困难的情况.它可能对空集合有意义(你可能期望一个空字符串),但另一方面,如果你从一个空的累加器开始,你将在结果中再多一个逗号而不是你期望的.对此的正确解决方案是首先检查集合是否为空,然后返回空集合的回退字符串,或者reduce对非空集合执行.您可能会观察到,通常您实际上并不想在空集合中使用空字符串,但"collection is empty"之类的内容可能更合适,从而进一步向您保证此解决方案是干净的.

顺便说一下,我在这里使用单词集合而不是自由观察,仅用于教育目的.此外,在RxJava,既fold和reduce被称为是相同的,reduce只有有两个版本的方法,以一个只有一个参数,其他两个参数.

至于你的最后一个问题:你不必离开Observable链.正如David Motten建议的那样,只需使用toList()即可.

.filter(...)
.toList()
.map(listOfValues => listOfValues.intersperse(", "))

哪里intersperse可以实现reduce,如果还没有库函数(这是很常见).

collection.intersperse(separator) = 
    if (collection.isEmpty()) 
      ""
    else
      collection.reduce(accumulator, element => accumulator + separator + element)

Answer 2

发生这种情况的原因是您正在使用toBlocking().single()空流.如果您希望流中的值为0或1,则可以toList().toBlocking().single()检查列表中的值(可能为空,但不会引发您获得的异常).