Immediate debounce in Rx
The behavior you want is not what debounce
operator does in Rx.
This is called throttle
, throttleTime
or throttleWithTimeout
(however, it falls under debounce
category of operators). I don't know what language you use but in RxJS it looks like the following image:
See http://reactivex.io/documentation/operators/debounce.html.
Edit: Based on the clarifications, RxJava doesn't have an operator for this type of flow but it can be composed from a non-trivial set of other operators:
import java.util.concurrent.TimeUnit;
import rx.Observable;
public class DebounceFirst {
public static void main(String[] args) {
Observable.just(0, 100, 200, 1500, 1600, 1800, 2000, 10000)
.flatMap(v -> Observable.timer(v, TimeUnit.MILLISECONDS).map(w -> v))
.doOnNext(v -> System.out.println("T=" + v))
.compose(debounceFirst(500, TimeUnit.MILLISECONDS))
.toBlocking()
.subscribe(v -> System.out.println("Debounced: " + v));
}
static <T> Observable.Transformer<T, T> debounceFirst(long timeout, TimeUnit unit) {
return f ->
f.publish(g ->
g.take(1)
.concatWith(
g.switchMap(u -> Observable.timer(timeout, unit).map(w -> u))
.take(1)
.ignoreElements()
)
.repeatWhen(h -> h.takeUntil(g.ignoreElements()))
);
}
}