我的目标可能最容易用大理石图来解释。我有两个 Observable，xs 和 ys。我想返回一个名为 rs 的 Observable。
xs --x---x---x---x-------x---x---x-
\ \ \
ys ----y-----------y---y---y-------
| | |
rs ----x-----------x-------x-------
y y y
所以，我需要类似于 CombineLatest 的东西，除了它应该只在 xs 后跟 ys 时触发。该模式之外的其他 xs 或 ys 不应触发输出，应丢弃。
CombineLatest、Zip 或 And/Then/When 不做我需要的事情，我找不到任何方法来指定更复杂的连接结构。
对于此类问题，提供测试场景非常有用。你已经给出了你想要的一个很棒的小大理石图，然后将它转换为测试用例真的很容易。然后其他论坛读者可以获取测试代码，实施他们的想法并知道他们是否满足您的要求。
[Test]
public void Should_only_get_latest_value_from_Y_but_not_while_x_produes()
{
// 11111111112222222222333
// 12345678901234567890123456789012
//xs --x---x---x---x-------x---x---x-
// \ \ \
//ys ----y-----------y---y---y-------
// | | |
//rs ----x-----------x-------x-------
// y y y
var testScheduler = new TestScheduler();
// 11111111112222222222333
// 12345678901234567890123456789012
//xs --x---x---x---x-------x---x---x-
var xs = testScheduler.CreateColdObservable(
new Recorded<Notification<char>>(3, Notification.CreateOnNext('1')),
new Recorded<Notification<char>>(7, Notification.CreateOnNext('2')),
new Recorded<Notification<char>>(10, Notification.CreateOnNext('3')),
new Recorded<Notification<char>>(15, Notification.CreateOnNext('4')),
new Recorded<Notification<char>>(23, Notification.CreateOnNext('5')),
new Recorded<Notification<char>>(27, Notification.CreateOnNext('6')),
new Recorded<Notification<char>>(31, Notification.CreateOnNext('7')));
// 11111111112222222222333
// 12345678901234567890123456789012
//ys ----y-----------y---y---y-------
var ys = testScheduler.CreateColdObservable(
new Recorded<Notification<char>>(5, Notification.CreateOnNext('A')),
new Recorded<Notification<char>>(17, Notification.CreateOnNext('B')),
new Recorded<Notification<char>>(21, Notification.CreateOnNext('C')),
new Recorded<Notification<char>>(25, Notification.CreateOnNext('D')));
//Expected :
//Tick x y
//5 1 A
//17 4 B
//25 5 D
var expected = new[]
{
new Recorded<Notification<Tuple<char, char>>>(5, Notification.CreateOnNext(Tuple.Create('1', 'A'))),
new Recorded<Notification<Tuple<char, char>>>(17, Notification.CreateOnNext(Tuple.Create('4', 'B'))),
new Recorded<Notification<Tuple<char, char>>>(25, Notification.CreateOnNext(Tuple.Create('5', 'D')))
};
var observer = testScheduler.CreateObserver<Tuple<char, char>>();
//Passes HOT, fails Cold. Doesn't meet the requirements due to Timeout anyway.
//xs.Select(x => ys.Take(1)
// .Timeout(TimeSpan.FromSeconds(0.5),
// Observable.Empty<char>())
// .Select(y => Tuple.Create(x, y))
// )
// .Switch()
// .Subscribe(observer);
//Passes HOT. Passes Cold
xs.Join(ys,
_ => xs.Merge(ys),
_ => Observable.Empty<Unit>(),
Tuple.Create)
.Subscribe(observer);
testScheduler.Start();
//You may want to Console.WriteLine out the two collections to validate for yourself.
CollectionAssert.AreEqual(expected, observer.Messages);
}
PS 顺便说一句，Merge 很好地使用了答案。
I ended up using Join.
var rs = xs.Join(ys,
_ => xs.Merge(ys),
_ => Observable.Empty<Unit>(),
Tuple.Create);
This is a good article that explains how Join works: http://blogs.microsoft.co.il/blogs/bnaya/archive/2012/04/04/rx-join.aspx
To state the diagram another way, each x needs to take the first y without an additional x before the y. Those first requirement suggests Take(1). At that point, you will have an IObservable<IObservable<...>>. The second requirement and the type generated by the first two points to Switch. Putting it all together, you should be able to match the diagram with:
var rs = xs.Select(x => ys.Take(1)
.Select(y => Tuple.Create(x, y))
)
.Switch()