我被rx和特定查询卡住了。 问题:Rx groupby直到条件发生变化
许多单一的更新操作是由连续流产生的。操作可以是插入或删除。我想缓冲这些数据流并在当时执行少量操作,但维护订单非常重要。此外,操作应被缓冲并在序列来完成每X秒
实施例:
在:
insert-insert-insert-delete-delete-insert-delete-delete-delete-delete
输出:
insert(3)-delete(2)-insert(1)-delete(4)
我写了一个简单的应用程序,以测试它,它的作品或多或少,因为我会,但它不尊重传入的插入/删除顺序
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, delete, 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();
}
}
}
}
任何人可以帮助我正确的查询?
感谢
UPDATE 13年3月8日(由JerKimball建议)
以下线的微小变化/补充JerKimball代码打印结果:
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(", {0}", item);
}
}
stringBuilder.Append("]");
Console.WriteLine(stringBuilder);
}
和下面的字符串的操作:
public override string ToString()
{
return string.Format("{0}:{1}", this.Type, this.Seq);
}
顺序被保存,但是:
- 我不知道其他内部认购订阅:这是正确的(这是一个问题,因为我有很久以前,对我而言并不清楚)?
- 我总是有每个列表(即使流产生两个以上的连续值与同类型)的不超过两种元素
'我不知道在另一个订阅订阅:它是正确的< - 你是什么意思?这发生在哪里? (编辑)哦,我看到 - 在你的'打印'...是的,你不想这样做,如果只是因为你现在正在泄漏一个'IDisposable'' – JerKimball 2013-03-08 20:14:20
'Subscribe'有一个重载需要一个'Action' - 使用它;这就是为什么我在我的例子中使用'Console.WriteLine'。如果你改变'Print'的签名来取'IList ',你可以'使用(query.Subscribe(Print))' –
JerKimball
2013-03-08 20:17:12
从Select我得到一个IObservable >>,这就是为什么我用第一次打印(IObservable >操作)。所以我会再次包装第二个操作。订阅(打印)在使用 –
fra
2013-03-08 20:22:01