RxJava delay for each item of list emitted

Question

I\'m struggling to implement something I assumed would be fairly simple in Rx.

I have a list of items, and I want to have each item emitted with a delay.

It

Answer 1

A not so clean way is to make the delay change with the iteration using the .delay(Func1) operator.

Observable.range(1, 5)
            .delay(n -> n*50)
            .groupBy(n -> n % 5)
            .flatMap(g -> g.toList())
            .doOnNext(item -> {
                System.out.println(System.currentTimeMillis() - timeNow);
                System.out.println(item);
                System.out.println(" ");
            }).toList().toBlocking().first();

Answer 2

There is other way to do it using concatMap as concatMap returns observable of source items. so we can add delay on that observable.

here what i have tried.

Observable.range(1, 5)
          .groupBy(n -> n % 5)
          .concatMap(integerIntegerGroupedObservable ->
          integerIntegerGroupedObservable.delay(2000, TimeUnit.MILLISECONDS))
          .doOnNext(item -> {
                    System.out.println(System.currentTimeMillis() - timeNow);
                    System.out.println(item);
                    System.out.println(" ");
                }).toList().toBlocking().first(); 

Answer 3

You can implement a custom rx operator such as MinRegularIntervalDelayOperator and then use this with the lift function

Observable.range(1, 5)
    .groupBy(n -> n % 5)
    .flatMap(g -> g.toList())
    .lift(new MinRegularIntervalDelayOperator<Integer>(50L))
    .doOnNext(item -> {
      System.out.println(System.currentTimeMillis() - timeNow);
      System.out.println(item);
      System.out.println(" ");
    }).toList().toBlocking().first();

Answer 4

Observable.just("A", "B", "C", "D", "E", "F")
    .flatMap { item -> Thread.sleep(2000)
        Observable.just( item ) }
    .subscribe { println( it ) }

Answer 5

To introduce delay between each item emitted is useful:

List<String> letters = new ArrayList<>(Arrays.asList("a", "b", "c", "d"));

Observable.fromIterable(letters)
                .concatMap(item -> Observable.interval(1, TimeUnit.SECONDS)
                        .take(1)
                        .map(second -> item))
                .subscribe(System.out::println);

More good options at https://github.com/ReactiveX/RxJava/issues/3505

Answer 6

For kotlin users, I wrote an extension function for the 'zip with interval' approach

import io.reactivex.Observable
import io.reactivex.functions.BiFunction
import java.util.concurrent.TimeUnit

fun <T> Observable<T>.delayEach(interval: Long, timeUnit: TimeUnit): Observable<T> =
    Observable.zip(
        this, 
        Observable.interval(interval, timeUnit), 
        BiFunction { item, _ -> item }
    )

It works the same way, but this makes it reusable. Example:

Observable.range(1, 5)
    .delayEach(1, TimeUnit.SECONDS)