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Immediate debounce in Rx

I am looking an operator to debounce a series of event, let us say user's click. The input and output should be like this:

interval :      ->     <-            ->     <-
in       : 1--2--3-------4--5--5--6-7-8--------
out      : 1-------------4---------------------

The idea is like underscore's debounce with immediate option on http://underscorejs.org/#debounce. The operator can be presented/implemented in any language that supports Reactive Extensions

Edit: Clarify the interval, let say 5 seconds (5 spaces between two arrows) : -> <-

Edit2: A more understandable version: I have a user, he clicks repeatedly a button (1, 2, 3); I want to catch the first click (1) and ignore the rest. After a while, he is tired and rest for 7 seconds (which is longer than the 5 seconds interval between two arrows) and continue clicking the button again (4, 5, 6, 7, 8) I want to catch the first click (4) and ignore the rest.

If he clicks after the forth arrow, I want to catch that click, too.

Edit3: Here is an image image which can be found at the original article

like image 672
Quang Linh Le Avatar asked Jan 31 '17 18:01

Quang Linh Le


4 Answers

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()))
            );
    }
}
like image 143
akarnokd Avatar answered Oct 09 '22 06:10

akarnokd


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.

like image 40
martin Avatar answered Oct 09 '22 07:10

martin


Because debounce() is inherently asynchronous, you need to bring the result back to the current thread explicitly.

seriesOfUnfortunateEvents
  .debounce( 14, TimeUnit.MILLISECONDS )
  .observeOn( Schedulers.immediate() )
  .subscribe( v -> yourStuff() );
like image 45
Bob Dalgleish Avatar answered Oct 09 '22 08:10

Bob Dalgleish


According to documentation there are two debounce operators in RxJS. You might be interested in debounceTime in particular.

debounceTime

From documentation

Emits a value from the source Observable only after a particular time span has passed without another source emission.

Example:

Rx.Observable
    .fromEvent(document.querySelector('button'), 'click')
    .debounceTime(200)
    .mapTo(() => 'clicked!')
    .subscribe(v => console.log(v));

It will emit one clicked! if button was clicked in given timespan (200ms in this example).

debounce

From documentation

Emits a value from the source Observable only after a particular time span determined by another Observable has passed without another source emission.

like image 34
Michał Pietraszko Avatar answered Oct 09 '22 06:10

Michał Pietraszko