如果第一个为空,则切换到不同的IObservable

时间:2013-03-04 19:52:07

标签: c# .net system.reactive

我正在编写一个函数,用于检索有关主题的新闻,并通过IObservable返回值反馈此消息。

但是,我有几个新闻来源。我不想使用Merge将这些来源合并为一个。相反,我想做的是优先订购 -

  1. 调用我的函数时,会查询第一个新闻源(生成代表该函数的IObservable)。
  2. 如果该新闻来源的IObservable完成而未返回任何结果,则会查询下一个新闻来源。
  3. 如果第二个来源完成而没有返回结果,则会查询最终的新闻来源。
  4. 这整个行为被包含在一个可观察的我可以返回给用户。
  5. 这种行为是我可以使用内置的Rx扩展方法完成的,还是我需要实现一个自定义类来处理这个问题?我怎么办?

5 个答案:

答案 0 :(得分:1)

从原始海报编辑:

我接受了这个答案,但把它变成了一个扩展方法 -

/// <summary> Returns the elements of the first sequence, or the values in the second sequence if the first sequence is empty. </summary>
/// <param name="first"> The first sequence. </param>
/// <param name="second"> The second sequence. </param>
/// <typeparam name="T"> The type of elements in the sequence. </typeparam>
/// <returns> The <see cref="IObservable{T}"/> sequence. </returns>
public static IObservable<T> DefaultIfEmpty<T>(this IObservable<T> first, IObservable<T> second)
{
    var signal = new AsyncSubject<Unit>();
    var source1 = first.Do(item => { signal.OnNext(Unit.Default); signal.OnCompleted(); });
    var source2 = second.TakeUntil(signal);

    return source1.Concat(source2); // if source2 is cold, it won't invoke it until source1 is completed
}

原答案:

这可能会成功。

var signal1 = new AsyncSubject<Unit>();
var signal2 = new AsyncSubject<Unit>();
var source1 = a.Do(item => { signal1.onNext(Unit.Default); signal1.onCompleted(); });
var source2 = b.Do(item => { signal2.onNext(Unit.Default); signal2.onCompleted(); })).TakeUntil(signal1);
var source3 = c.TakeUntil(signal2.Merge(signal1));

return Observable.Concat(source1, source2, source3);

编辑:哎呀,第二个来源需要一个单独的信号而第三个不需要发出任何信号。 Edit2:哎呀...类型。我习惯了RxJs :))

P.S。还有更少的RX-y方法可以做到这一点,这可能是一点点打字:

var gotResult = false;
var source1 = a();
var source2 = Observable.Defer(() => return gotResult ? Observable.Empty<T>() : b());
var source3 = Observable.Defer(() => return gotResult ? Observable.Empty<T>() : c());
return Observable.Concat(source1, source2, source3).Do(_ => gotResult = true;);

答案 1 :(得分:1)

另一种方法 - 与其他方法相差甚远,所以我会提出一个新答案:

这里有各种有趣的调试线:

public static IObservable<T> FirstWithValues<T>(this IEnumerable<IObservable<T>> sources)
{
    return Observable.Create<T>(obs =>
    {
        // these are neat - if you set it's .Disposable field, and it already
        // had one in there, it'll auto-dispose it
        SerialDisposable disp = new SerialDisposable();
        // this will trigger our exit condition
        bool hadValues = false;
        // start on the first source (assumed to be in order of importance)
        var sourceWalker = sources.GetEnumerator();
        sourceWalker.MoveNext();

        IObserver<T> checker = null;
        checker = Observer.Create<T>(v => 
            {
                // Hey, we got a value - pass to the "real" observer and note we 
                // got values on the current source
                Console.WriteLine("Got value on source:" + v.ToString());
                hadValues = true;
                obs.OnNext(v);
            },
            ex => {
                // pass any errors immediately back to the real observer
                Console.WriteLine("Error on source, passing to observer");
                obs.OnError(ex);
            },
            () => {
                // A source completed; if it generated any values, we're done;                    
                if(hadValues)
                {
                    Console.WriteLine("Source completed, had values, so ending");
                    obs.OnCompleted();
                }
                // Otherwise, we need to check the next source in line...
                else
                {
                    Console.WriteLine("Source completed, no values, so moving to next source");
                    sourceWalker.MoveNext();
                    disp.Disposable = sourceWalker.Current.Subscribe(checker);
                }
            });
        // kick it off by subscribing our..."walker?" to the first source
        disp.Disposable = sourceWalker.Current.Subscribe(checker);
        return disp.Disposable;
    });
}

用法:

var query = new[]
{
    Observable.Defer(() => GetSource("A")), 
    Observable.Defer(() => GetSource("B")), 
    Observable.Defer(() => GetSource("C")), 
}.FirstWithValues();

输出:

Source A invoked
Got value on source:Article from A
Article from A
Source completed, had values, so ending

Source A invoked
Source completed, no values, so moving to next source
Source B invoked
Got value on source:Article from B
Article from B
Source completed, had values, so ending

Source A invoked
Source completed, no values, so moving to next source
Source B invoked
Source completed, no values, so moving to next source
Source C invoked
Got value on source:Article from C
Article from C
Source completed, had values, so ending

答案 2 :(得分:1)

在我看来,可接受的答案是不可取的,因为它使用SubjectDo,并且在第一个序列不为空时仍订阅第二个序列。如果第二个可观察对象调用任何无关紧要的东西,那么后者可能是一个大问题。我想出了以下解决方案:

public static IObservable<T> SwitchIfEmpty<T>(this IObservable<T> @this, IObservable<T> switchTo)
{
    if (@this == null) throw new ArgumentNullException(nameof(@this));
    if (switchTo == null) throw new ArgumentNullException(nameof(switchTo));
    return Observable.Create<T>(obs =>
    {
        var source = @this.Replay(1);
        var switched = source.Any().SelectMany(any => any ? Observable.Empty<T>() : switchTo);
        return new CompositeDisposable(source.Concat(switched).Subscribe(obs), source.Connect());
    });
}

名称SwitchIfEmptyexisting RxJava implementation一致。 Here正在进行有关将某些RxJava运算符合并到RxNET中的讨论。

我敢肯定,自定义IObservable的实现将比我的实现高效得多。您可以找到一个由ReactiveX成员here编写的akarnokd。也可以在NuGet上使用。

答案 3 :(得分:0)

听起来你可以使用普通的Amb查询。

编辑:基于评论,Amb不会这样做 - 给这个打击:

public static IObservable<T> SwitchIfEmpty<T>(
     this IObservable<T> first, 
     Func<IObservable<T>> second)
{
    return first.IsEmpty().FirstOrDefault() ? second() : first;
}

试验台:

static Random r = new Random();
public IObservable<string> GetSource(string sourceName)
{
    Console.WriteLine("Source {0} invoked", sourceName);
    return r.Next(0, 10) < 5 
        ? Observable.Empty<string>() 
        : Observable.Return("Article from " + sourceName);
}

void Main()
{
    var query = GetSource("A")
        .SwitchIfEmpty(() => GetSource("B"))
        .SwitchIfEmpty(() => GetSource("C"));

    using(query.Subscribe(Console.WriteLine))
    {
        Console.ReadLine();
    }           
}

一些示例运行:

Source A invoked
Article from A

Source A invoked
Source B invoked
Article from B

Source A invoked
Source B invoked
Source C invoked
Article from C

EDITEDIT:

我想你也可以将它概括为:

public static IObservable<T> SwitchIf<T>(
    this IObservable<T> first, 
    Func<IObservable<T>, IObservable<bool>> predicate, 
    Func<IObservable<T>> second)
{
    return predicate(first).FirstOrDefault() 
        ? second() 
        : first;
}

答案 4 :(得分:0)

这是JerKimball的SwitchIfEmpty运算符的非阻塞版本。

/// <summary>Returns the elements of the first sequence, or the elements of the
/// second sequence if the first sequence is empty.</summary>
public static IObservable<T> SwitchIfEmpty<T>(this IObservable<T> first,
    IObservable<T> second)
{
    return Observable.Defer(() =>
    {
        bool isEmpty = true;
        return first
            .Do(_ => isEmpty = false)
            .Concat(Observable.If(() => isEmpty, second));
    });
}

这是同一运算符的版本,它接受多个序列,并返回第一个非空序列的元素:

/// <summary>Returns the elements of the first non-empty sequence.</summary>
public static IObservable<T> SwitchIfEmpty<T>(params IObservable<T>[] sequences)
{
    return Observable.Defer(() =>
    {
        bool isEmpty = true;
        return sequences
            .Select(s => s.Do(_ => isEmpty = false))
            .Select(s => Observable.If(() => isEmpty, s))
            .Concat();
    });
}

Observable.Defer运算符用于防止多个订阅共享相同的bool isEmpty状态(有关此here的更多信息)。