如何在不丢掉RX中的值的情况下放慢Observable的速度？

Question

My scenario: I have a computation that should be run about once a second. After it is run there should be a wait of about 200ms for other stuff to catch up. If the compuation is still running after a second it should be started a second time, but should the program should wait until it is finished and start the next computation 200ms after finishing.

The way I am doing it now:

_refreshFinished = new Subject<bool>();
_autoRefresher = Observable.Interval(TimeSpan.FromMilliseconds(1000))
   .Zip(_refreshFinished, (x,y) => x)
   .Subscribe(x => AutoRefresh(stuff));

The problem with this code is, that i see no way to put in a delay after a computation finished. The Delay method only delays the first element of the observable collection. Usually this behaviour is the right once, since you would have to buffer an endless amount of elements if you wanted to buffer everyone, but since delaying the call to Autorefesh by 200ms delays the output of _refreshFinished by 200ms as well there would be no buffer overhead. Basicly I want an Observable that fires every every MaxTime(some_call,1000ms) then gets delayed by 200ms or even better, some dynamic value. At this point i dont even really care about the values that are running through this, although that might change in the future.

I´m open to any suggestions

Answer 1

Observable.Generate() has a number of overloads which will let you dynamically adjust the time in which the next item is created.

For instance

IScheduler schd = Scheduler.TaskPool;
var timeout = TimeSpan.FromSeconds(1);
var shortDelay = TimeSpan.FromMilliseconds(200);
var longerDelay = TimeSpan.FromMilliseconds(500);
Observable.Generate(schd.Now, 
                    time => true, 
                    time => schd.Now, 
                    time => new object(), // your code here
                    time => schd.Now.Subtract(time) > timeout  ? shortDelay : longerDelay ,
                    schd);

Answer 2

这听起来更像是新异步框架的工作http://msdn.microsoft.com/en-us/vstudio/gg316360

Answer 3

There is a way to do it. Its not the easiest thing ever, since the wait time has to be dynamicly calculated on each value but it works and is pretty generic.

When you use thise code you can just insert the code that should be called in YOURCODE and everything else works automaticly. You code will be basicly be called every Max(yourCodeTime+extraDelay,usualCallTime+extraDelay). This means yourCode wont be called twice at the same time and the app will always have extraDelay of time to do other stuff. If there is some easier/other way to do this i would ove to hear it.

double usualCallTime = 1000;
double extraDealy = 100;
var subject = new Subject<double>();
var subscription =
    sub.TimeInterval()
        .Select(x =>
            {
                var processingTime = x.Interval.TotalMilliseconds - x.Value;
                double timeToWait = 
                     Math.Max(0, usualCallTime - processingTime) + extraDelay;
                return Observable.Timer(TimeSpan.FromMilliseconds(timeToWait))
                    .Select(ignore => timeToWait);
            })
        .Switch()
        .Subscribe(x => {YOURCODE();sub.OnNext(x)});
sub.OnNext(0);

private static void YOURCODE()
{
    // do stuff here
    action.Invoke();
}

Answer 4

If I understand your problem correctly, you have a long-running compute function such as this:

static String compute()
{
    int t = 300 + new Random().Next(1000);
    Console.Write("[{0}...", t);
    Thread.Sleep(t);
    Console.Write("]");
    return Guid.NewGuid().ToString();
}

And you want to call this function at least once per second but without overlapping calls, and with a minimum 200ms recovery time between calls. The code below works for this situation.

I started with a more functional approach (using Scan() and Timestamp() ), more in the style of Rx--because I was looking for a good Rx exercise--but in the end, this non-aggregating approach was just simpler.

static void Main()
{
    TimeSpan period = TimeSpan.FromMilliseconds(1000);
    TimeSpan recovery = TimeSpan.FromMilliseconds(200);

    Observable
        .Repeat(Unit.Default)
        .Select(_ =>
        {
            var s = DateTimeOffset.Now;
            var x = compute();
            var delay = period - (DateTimeOffset.Now - s);
            if (delay < recovery)
                delay = recovery;

            Console.Write("+{0} ", (int)delay.TotalMilliseconds);

            return Observable.Return(x).Delay(delay).First();
        })
        .Subscribe(Console.WriteLine);
}

Here's the output:

[1144...]+200 a7cb5d3d-34b9-4d44-95c9-3e363f518e52
[1183...]+200 359ad966-3be7-4027-8b95-1051e3fb20c2
[831...]+200 f433b4dc-d075-49fe-9c84-b790274982d9
[766...]+219 310c9521-7bee-4acc-bbca-81c706a4632a
[505...]+485 0715abfc-db9b-42e2-9ec7-880d7ff58126
[1244...]+200 30a3002a-924a-4a64-9669-095152906d85
[1284...]+200 e5b1cd79-da73-477c-bca0-0870f4b5c640
[354...]+641 a43c9df5-53e8-4b58-a0df-7561cf4b0483
[1094...]+200 8f25019c-77a0-4507-b05e-c9ab8b34bcc3
[993...]+200 840281bd-c8fd-4627-9324-372636f8dea3

[edit: this sample uses Rx 2.0(RC) 2.0.20612.0]

Answer 5

Suppose you have an existing 'IObservable' , then the following will work

var delay = TimeSpan.FromSeconds(1.0);
var actual = source.Scan(
    new ConcurrentQueue<object>(),
    (q, i) =>
        {
            q.Enqueue(i);
            return q;
        }).CombineLatest(
            Observable.Interval(delay),
            (q, t) =>
                {
                    object item;
                    if (q.TryDequeue(out item))
                    {
                        return item;
                    }

                    return null;
                }).Where(v => v != null);

'actual' is your resultant observable. But keep in mind that the above code has turned that into a Hot observable if it wasn't hot already. So you won't get 'OnCompleted' called.