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:

enter image description here

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()))
            );
    }
}