问题
I've got two event streams. One is from an inductance loop, the other is an IP camera. Cars will drive over the loop and then hit the camera. I want to combine them if the events are within N milliseconds of each other (car will always hit the loop first), but I also want the unmatched events from each stream (either hardware can fail) all merged into a single stream. Something like this:
---> (only unmatched a's, None)
/ \
stream_a (loop) \
\ \
--> (a, b) ---------------------------> (Maybe a, Maybe b)
/ /
stream_b (camera) /
\ /
--> (None, only unmatched b's)
Now certainly I can hack my way around by doing the good ole Subject anti-pattern:
unmatched_a = Subject()
def noop():
pass
pending_as = [[]]
def handle_unmatched(a):
if a in pending_as[0]:
pending_as[0].remove(a)
print("unmatched a!")
unmatched_a.on_next((a, None))
def handle_a(a):
pending_as[0].append(a)
t = threading.Timer(some_timeout, handle_unmatched)
t.start()
return a
def handle_b(b):
if len(pending_as[0]):
a = pending_as[0].pop(0)
return (a, b)
else:
print("unmatched b!")
return (None, b)
stream_a.map(handle_a).subscribe(noop)
stream_b.map(handle_b).merge(unmatched_a).subscribe(print)
Not only is this rather hacky, but although I've not observed it I'm pretty sure there's a race condition when I check the pending queue using threading.Timer. Given the plethora of rx operators, I'm pretty sure some combination of them will let you do this without using Subject, but I can't figure it out. How does one accomplish this?
Edit
Although for organizational and operational reasons I'd prefer to stick to Python, I'll take a JavaScript rxjs answer and either port it or even possibly rewrite the entire script in node.
回答1:
You should be able to solve the problem using auditTime and buffer. Like this:
function matchWithinTime(a$, b$, N) {
const merged$ = Rx.Observable.merge(a$, b$);
// Use auditTime to compose a closing notifier for the buffer.
const audited$ = merged$.auditTime(N);
// Buffer emissions within an audit and filter out empty buffers.
return merged$
.buffer(audited$)
.filter(x => x.length > 0);
}
const a$ = new Rx.Subject();
const b$ = new Rx.Subject();
matchWithinTime(a$, b$, 50).subscribe(x => console.log(JSON.stringify(x)));
setTimeout(() => a$.next("a"), 0);
setTimeout(() => b$.next("b"), 0);
setTimeout(() => a$.next("a"), 100);
setTimeout(() => b$.next("b"), 125);
setTimeout(() => a$.next("a"), 200);
setTimeout(() => b$.next("b"), 275);
setTimeout(() => a$.next("a"), 400);
setTimeout(() => b$.next("b"), 425);
setTimeout(() => a$.next("a"), 500);
setTimeout(() => b$.next("b"), 575);
setTimeout(() => b$.next("b"), 700);
setTimeout(() => b$.next("a"), 800);.as-console-wrapper { max-height: 100% !important; top: 0; }<script src="https://unpkg.com/rxjs@5/bundles/Rx.min.js"></script>If it's possible for b values to be closely followed by a values and you do not want them to be matched, you could use a more specific audit, like this:
const audited$ = merged$.audit(x => x === "a" ?
// If an `a` was received, audit upcoming values for `N` milliseconds.
Rx.Observable.timer(N) :
// If a `b` was received, don't audit the upcoming values.
Rx.Observable.of(0, Rx.Scheduler.asap)
);
回答2:
I have developed a different strategy than Cartant, and clearly much less elegant, which may give you somehow a different result. I apologize if I have not understood the question and if my answer turns out to be useless.
My strategy is based on using switchMap on a$ and then bufferTime on b$.
This code emits at every timeInterval and it emits an object which contains the last a received and an array of bs representing the bs received during the time interval.
a$.pipe(
switchMap(a => {
return b$.pipe(
bufferTime(timeInterval),
mergeMap(arrayOfB => of({a, arrayOfB})),
)
})
)
If arrayOfB is empty, than it means that the last a in unmatched.
If arrayOfB has just one element, than it means that the last a has been matched by the b of the array.
If arrayOfB has more than one element, than it means that the last a has been matched by the first b of the array while all other bs are unmatched.
Now it is a matter of avoiding the emission of the same a more than once and this is where the code gets a bit messy.
In summary, the code could look like the following
const a$ = new Subject();
const b$ = new Subject();
setTimeout(() => a$.next("a1"), 0);
setTimeout(() => b$.next("b1"), 0);
setTimeout(() => a$.next("a2"), 100);
setTimeout(() => b$.next("b2"), 125);
setTimeout(() => a$.next("a3"), 200);
setTimeout(() => b$.next("b3"), 275);
setTimeout(() => a$.next("a4"), 400);
setTimeout(() => b$.next("b4"), 425);
setTimeout(() => b$.next("b4.1"), 435);
setTimeout(() => a$.next("a5"), 500);
setTimeout(() => b$.next("b5"), 575);
setTimeout(() => b$.next("b6"), 700);
setTimeout(() => b$.next("b6.1"), 701);
setTimeout(() => b$.next("b6.2"), 702);
setTimeout(() => a$.next("a6"), 800);
setTimeout(() => a$.complete(), 1000);
setTimeout(() => b$.complete(), 1000);
let currentA;
a$.pipe(
switchMap(a => {
currentA = a;
return b$.pipe(
bufferTime(50),
mergeMap(arrayOfB => {
let aVal = currentA ? currentA : null;
if (arrayOfB.length === 0) {
const ret = of({a: aVal, b: null})
currentA = null;
return ret;
}
if (arrayOfB.length === 1) {
const ret = of({a: aVal, b: arrayOfB[0]})
currentA = null;
return ret;
}
const ret = from(arrayOfB)
.pipe(
map((b, _indexB) => {
aVal = _indexB > 0 ? null : aVal;
return {a: aVal, b}
})
)
currentA = null;
return ret;
}),
filter(data => data.a !== null || data.b !== null)
)
})
)
.subscribe(console.log);
来源：https://stackoverflow.com/questions/50395441/how-to-delay-event-emission-with-rxpy-rxjs