c# – Rx groupby直到条件改变

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我坚持使用rx和特定的查询.
问题:

Many single update operations are produced by continuous stream. The operations can be insert or delete. I want to buffer those streams and perform few operations at the time,but it is really important to preserve the order. Additionally,operations should be buffered and done in sequences every X seconds

例:

在:

insert-insert-insert-delete-delete-insert-delete-delete-delete-delete

日期:

insert(3)-delete(2)-insert(1)-delete(4)

我写了一个简单的应用程序来测试它,它或多或少地工作,但它不尊重传入插入/删除的顺序

namespace RxTests
{
using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
using System.Reactive.Concurrency;
using System.Reactive.Linq;
using System.Reactive.Subjects;
using System.Text;
using System.Threading;

internal class Program
{
    private static readonly Random Random = new Random();

    private static readonly CancellationTokenSource ProducerStopped = new CancellationTokenSource();

    private static readonly ISubject<UpdateOperation> operations = new Subject<UpdateOperation>();

    private static void Main(string[] args)
    {
        Console.WriteLine("Starting production");
        var producerScheduler = new EventLoopScheduler();
        var consumerScheduler = new EventLoopScheduler();
        var producer =
            Observable.Interval(TimeSpan.FromSeconds(2))
                      .SubscribeOn(producerScheduler)
                      .Subscribe(Produce,WriteProductionCompleted);
        var consumer =
            operations.ObserveOn(producerScheduler)
                      .GroupBy(operation => operation.Delete)
                      .SelectMany(observable => observable.Buffer(TimeSpan.FromSeconds(8),50))
                      .SubscribeOn(consumerScheduler)
                      .Subscribe(WriteUpdateOperations);
        Console.WriteLine("Type any key to stop");
        Console.ReadKey();
        consumer.Dispose();
        producer.Dispose();
    }

    private static void Produce(long time)
    {
        var delete = Random.NextDouble() < 0.5;
        Console.WriteLine("Produce {0},{1} at {2}",time + 1,delete,time);
        var idString = (time + 1).ToString(CultureInfo.InvariantCulture);
        var id = time + 1;
        operations.OnNext(
            new UpdateOperation(id,idString,time.ToString(CultureInfo.InvariantCulture)));
    }

    private static void WriteProductionCompleted()
    {
        Console.WriteLine("Production completed");
        ProducerStopped.Cancel();
    }

    private static void WriteUpdateOperation(UpdateOperation updateOperation)
    {
        Console.WriteLine("Consuming {0}",updateOperation);
    }

    private static void WriteUpdateOperations(IList<UpdateOperation> updateOperation)
    {
        foreach (var operation in updateOperation)
        {
            WriteUpdateOperation(operation);
        }
    }

    private class UpdateOperation
    {
        public UpdateOperation(long id,bool delete,params string[] changes)
        {
            this.Id = id;
            this.Delete = delete;
            this.Changes = new List<string>(changes ?? Enumerable.Empty<string>());
        }

        public bool Delete { get; set; }

        public long Id { get; private set; }

        public IList<string> Changes { get; private set; }

        public override string ToString()
        {
            var stringBuilder = new StringBuilder("{UpdateOperation ");
            stringBuilder.AppendFormat("Id: {0},Delete: {1},Changes: [",this.Id,this.Delete);
            if (this.Changes.Count > 0)
            {
                stringBuilder.Append(this.Changes.First());
                foreach (var change in this.Changes.Skip(1))
                {
                    stringBuilder.AppendFormat(",{0}",change);
                }
            }

            stringBuilder.Append("]}");
            return stringBuilder.ToString();
        }
    }
}

}

任何人都可以帮助我正确的查询

谢谢

更新08.03.13(JerKimball的建议)

以下几行是对JerKimball代码的小改动/补充,用于打印结果:

using(query.Subscribe(Print))
{
    Console.ReadLine();
    producer.Dispose();        
}

使用以下打印方法

private static void Print(IObservable<IList<Operation>> operations)
{
    operations.Subscribe(Print);
}

private static void Print(IList<Operation> operations)
{
    var stringBuilder = new StringBuilder("[");
    if (operations.Count > 0)
    {
        stringBuilder.Append(operations.First());
        foreach (var item in operations.Skip(1))
        {
            stringBuilder.AppendFormat(",item);
        }
    }

    stringBuilder.Append("]");
    Console.WriteLine(stringBuilder);
 }

以及要操作的字符串:

public override string ToString()
{
    return string.Format("{0}:{1}",this.Type,this.Seq);
}

订单保留,但是:

>我不确定在另一个订阅订阅:它是否正确(这是我很久以前的一个问题,我从来都不清楚)?
>我在每个列表上总是不超过两个元素(即使流生成两个以上具有相同类型的连续值)

解决方法

让我们尝试一种新方法(因此新答案):

首先,让我们定义一个扩展方法,该方法将根据键“折叠”项目列表,同时保留顺序:

public static class Ext
{
    public static IEnumerable<List<T>> ToRuns<T,TKey>(
            this IEnumerable<T> source,Func<T,TKey> keySelector) 
    {
        using (var enumerator = source.GetEnumerator()) 
        {
            if (!enumerator.MoveNext())
                yield break;

            var currentSet = new List<T>();

            // inspect the first item
            var lastKey = keySelector(enumerator.Current);
            currentSet.Add(enumerator.Current);

            while (enumerator.MoveNext()) 
            {
                var newKey = keySelector(enumerator.Current);
                if (!Equals(newKey,lastKey)) 
                {
                    // A difference == new run; return what we've got thus far
                    yield return currentSet;
                    lastKey = newKey;
                    currentSet = new List<T>();
                }
                currentSet.Add(enumerator.Current);
            }

            // Return the last run.
            yield return currentSet;

            // and clean up
            currentSet = new List<T>();
            lastKey = default(TKey);
        }
    }
}

相当简单 – 给定IEnumerable< T>,将返回List< List< T>>每个子列表将具有相同的密钥.

现在,喂它并使用它:

var rnd = new Random();
var fakeSource = new Subject<Operation>();
var producer = Observable
    .Interval(TimeSpan.FromMilliseconds(1000))
    .Subscribe(i => 
        {
            var op = new Operation();
            op.Type = rnd.NextDouble() < 0.5 ? "insert" : "delete";
            fakeSource.OnNext(op);
        });    

var singleSource = fakeSource
    .Publish().RefCount();

var query = singleSource
    // change this value to alter your "look at" time window
    .Buffer(TimeSpan.FromSeconds(5))    
    .Select(buff => buff.ToRuns(op => op.Type).Where(run => run.Count > 0));

using(query.Subscribe(batch => 
{
    foreach(var item in batch)
    {
        Console.WriteLine("{0}({1})",item.First().Type,item.Count);
    }
}))
{
    Console.ReadLine();
    producer.Dispose();     
}

给它一个旋转 – 这是我在典型的运行中看到的:

insert(4)
delete(2)
insert(1)
delete(1)
insert(1)
insert(1)
delete(1)
insert(1)
delete(2)
delete(2)
insert(2)
delete(1)
insert(1)
delete(2)
insert(2)

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