观测资料的可扩展组合

时间:2018-09-17 08:38:33

标签: observable reactive rx.net combinelatest

我想将List的IObservables存储在容器中,并订阅这些可观察项的组合以检索合并的List。然后,我希望能够添加更多的Observable而不需要续订订阅,而仍然可以获得新结果。理想情况下,在向商店添加新的可观察物时也应触发。以下代码应详细说明:

using System;
using System.Collections.Generic;
using System.Reactive;
using System.Reactive.Linq;
using System.Reactive.Subjects;
using System.Linq;

namespace dynamic_combine
{
    class ObservableStuff
    {
        private List<IObservable<List<String>>> _listOfObservables = new List<IObservable<List<String>>>();

        public ObservableStuff() { }

        public void AddObservable(IObservable<List<String>> obs)
        {
            _listOfObservables.Add(obs);
        }

        public IObservable<IList<String>> GetCombinedObservable()
        {
            return Observable.CombineLatest(_listOfObservables)
                .Select((all) =>
                {
                    List<String> mergedList = new List<String>();
                    foreach(var list in all)
                    {
                        mergedList = mergedList.Concat(list).ToList();
                    }
                    return mergedList;
                });
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            ObservableStuff Stuff = new ObservableStuff();
            BehaviorSubject<List<String>> A = new BehaviorSubject<List<String>>(new List<String>() { "a", "b", "c" });
            BehaviorSubject<List<String>> B = new BehaviorSubject<List<String>>(new List<String>() { "d", "e", "f" });
            BehaviorSubject<List<String>> C = new BehaviorSubject<List<String>>(new List<String>() { "x", "y", "z" });

            Stuff.AddObservable(A.AsObservable());
            Stuff.AddObservable(B.AsObservable());

            Stuff.GetCombinedObservable().Subscribe((x) =>
            {
                Console.WriteLine(String.Join(",", x));
            });

            // Initial Output: a,b,c,d,e,f

            A.OnNext(new List<String>() { "1", "2", "3", "4", "5" });
            // Output: 1,2,3,4,5,d,e,f

            B.OnNext(new List<String>() { "6", "7", "8", "9", "0" });
            // Output: 1,2,3,4,5,6,7,8,9,0

            Stuff.AddObservable(C.AsObservable());
            // Wishful Output: 1,2,3,4,5,6,7,8,9,0,x,y,z

            C.OnNext(new List<String>() { "y", "e", "a", "h" });
            // Wishful Output: 1,2,3,4,5,6,7,8,9,0,y,e,a,h

            Console.WriteLine("Press the any key...");
            Console.ReadKey();
        }
    }
}

尽管示例在C#中,但最终需要在rxCpp中实现。同样,在Rx的其他变体中实现也将是一件令人不安的事情。

我已经建立了一个存储库来检查代码,并且可能会将其扩展到其他语言: https://gitlab.com/dwaldorf/rx-examples

BR, 丹尼尔

1 个答案:

答案 0 :(得分:1)

首先,要进行一些更改,因为您的代码不那么容易阅读。 GetCombinedObservable可以重写为:

public IObservable<IList<String>> GetCombinedObservable()
{
    return Observable.CombineLatest(_listOfObservables)
        .Select(l => l.SelectMany(s => s).ToList());
}

您的问题可以归结为两件事:您希望_listOfObservables是动态的,这意味着将其从List<IObservable<T>>更改为IObservable<IObservable<T>>。但是,这样做的问题在于CombineLatest不支持IObservable<IObservable<T>>,因此我们必须创建一个。

这给我们带来了这个有趣的,丑陋的小功能(使用nuget包System.Collections.Immutable):

public static class X
{
    public static IObservable<List<T>> DynamicCombineLatest<T>(this IObservable<IObservable<T>> source)
    {
        return source
            .SelectMany((o, i) => o.Select(item => (observableIndex: i, item: item)))
            .Scan(ImmutableDictionary<int, T>.Empty, (state, t) => state.SetItem(t.observableIndex, t.item))
            .Select(dict => dict.OrderBy(kvp => kvp.Key).Select(kvp => kvp.Value).ToList());
    }
}

现在我们可以更新您的课程:

class ObservableStuff
{
    private ReplaySubject<IObservable<List<String>>> _subject = new ReplaySubject<IObservable<List<String>>>(int.MaxValue);

    public ObservableStuff() { }

    public void AddObservable(IObservable<List<String>> obs)
    {
        _subject.OnNext(obs);
    }

    public IObservable<IList<String>> GetCombinedObservable()
    {
        return _subject
            .DynamicCombineLatest()
            .Select(l => l.SelectMany(s => s).ToList());
    }
}