Develop Reference

How to have a delay occur POST-emission?

I would like for every emission from my Observable to have a delay occur AFTER its emission. i.e. For every item emitted from Observable.fromIterable(listOf(1,2,3)) I would like a variable/dynamic delay AFTERWARDS.
i.e. given a list of timers or delays [3, 5, 10], I would like for the following timeline to occur with my observable:
1 -> Wait 3 seconds -> 2 -> Wait 5 seconds -> 3 -> Wait 10 seconds
Please note that this is NOT the same as:
Wait 3 seconds -> 1 -> Wait 5 seconds -> 2 -> Wait 10 seconds -> 3
which can be easily achieved with .zip + .delay or Observable.Timer

Using zip() is the right way to go. When you want the delay after the first value is emitted, then you can use startWith() to add a zero delay for the first combination. The code might look like this:
Observable<Integer> values = Observable.fromArray(1,2,3,4,5);
Observable<Long> delays = Observable.fromArray(300L, 400L, 500L, 600L);
Observable<Integer> delayedValues = values.zipWith(
delays.startWith(0L),
new BiFunction<Integer, Long, Integer>() {
#Override
public Integer apply(Integer v, Long d) throws Exception {
Thread.sleep(d);
return v;
}
}
);
System.out.println("Begin subscribing");
long startTime = System.currentTimeMillis();
delayedValues.subscribe(v -> {
long currentTime = System.currentTimeMillis();
long diff = currentTime - startTime;
System.out.println("["+diff+"] "+v);
});
System.out.println("After subscribing");
This will generate the following output:
Begin subscribing
[35] 1
[336] 2
[736] 3
[1236] 4
[1836] 5
After subscribing
As you see, the first value is printed immediately, following by a 300ms delay for the next value until you reached the end.

Flux.range waits to emit more element once 256 elements are reached

I wrote this code:
Flux.range(0, 300)
.doOnNext(i -> System.out.println("i = " + i))
.flatMap(i -> Mono.just(i)
.subscribeOn(Schedulers.elastic())
.delayElement(Duration.ofMillis(1000))
)
.doOnNext(i -> System.out.println("end " + i))
.blockLast();
When running it, the first System.out.println shows that the Flux stop emitting numbers at the 256th element, then it waits for the older to be completed before emitting new ones.
Why is this happening?
Why 256?

Why this happening?
The flatMap operator can be characterized as operator that (rephrased from javadoc):
subscribes to its inners eagerly
does not preserve ordering of elements.
lets values from different inners interleave.
For this question the first point is important. Project Reactor restricts the
number of in-flight inner sequences via concurrency parameter.
While flatMap(mapper) uses the default parameter the flatMap(mapper, concurrency) overload accepts this parameter explicitly.
The flatMaps javadoc describes the parameter as:
The concurrency argument allows to control how many Publisher can be subscribed to and merged in parallel
Consider the following code using concurrency = 500
Flux.range(0, 300)
.doOnNext(i -> System.out.println("i = " + i))
.flatMap(i -> Mono.just(i)
.subscribeOn(Schedulers.elastic())
.delayElement(Duration.ofMillis(1000)),
500
// ^^^^^^^^^^
)
.doOnNext(i -> System.out.println("end " + i))
.blockLast();
In this case there is no waiting:
i = 297
i = 298
i = 299
end 0
end 1
end 2
In contrast if you pass 1 as concurrency the output will be similar to:
i = 0
end 0
i = 1
end 1
Awaiting one second before emitting the next element.
Why 256?
256 is the default value for concurrency of flatMap.
Take a look at Queues.SMALL_BUFFER_SIZE:
public static final int SMALL_BUFFER_SIZE = Math.max(16,
Integer.parseInt(System.getProperty("reactor.bufferSize.small", "256")));

Why is CompletableFuture join/get faster in separate streams than using one stream

For the following program I am trying to figure out why using 2 different streams parallelizes the task and using the same stream and calling join/get on the Completable future makes them take longer time equivalent to as if they were sequentially processed).
public class HelloConcurrency {
private static Integer sleepTask(int number) {
System.out.println(String.format("Task with sleep time %d", number));
try {
TimeUnit.SECONDS.sleep(number);
} catch (InterruptedException e) {
e.printStackTrace();
return -1;
}
return number;
}
public static void main(String[] args) {
List<Integer> sleepTimes = Arrays.asList(1,2,3,4,5,6);
System.out.println("WITH SEPARATE STREAMS FOR FUTURE AND JOIN");
ExecutorService executorService = Executors.newFixedThreadPool(6);
long start = System.currentTimeMillis();
List<CompletableFuture<Integer>> futures = sleepTimes.stream()
.map(sleepTime -> CompletableFuture.supplyAsync(() -> sleepTask(sleepTime), executorService)
.exceptionally(ex -> { ex.printStackTrace(); return -1; }))
.collect(Collectors.toList());
executorService.shutdown();
List<Integer> result = futures.stream()
.map(CompletableFuture::join)
.collect(Collectors.toList());
long finish = System.currentTimeMillis();
long timeElapsed = (finish - start)/1000;
System.out.println(String.format("done in %d seconds.", timeElapsed));
System.out.println(result);
System.out.println("WITH SAME STREAM FOR FUTURE AND JOIN");
ExecutorService executorService2 = Executors.newFixedThreadPool(6);
start = System.currentTimeMillis();
List<Integer> results = sleepTimes.stream()
.map(sleepTime -> CompletableFuture.supplyAsync(() -> sleepTask(sleepTime), executorService2)
.exceptionally(ex -> { ex.printStackTrace(); return -1; }))
.map(CompletableFuture::join)
.collect(Collectors.toList());
executorService2.shutdown();
finish = System.currentTimeMillis();
timeElapsed = (finish - start)/1000;
System.out.println(String.format("done in %d seconds.", timeElapsed));
System.out.println(results);
}
}
Output
WITH SEPARATE STREAMS FOR FUTURE AND JOIN
Task with sleep time 6
Task with sleep time 5
Task with sleep time 1
Task with sleep time 3
Task with sleep time 2
Task with sleep time 4
done in 6 seconds.
[1, 2, 3, 4, 5, 6]
WITH SAME STREAM FOR FUTURE AND JOIN
Task with sleep time 1
Task with sleep time 2
Task with sleep time 3
Task with sleep time 4
Task with sleep time 5
Task with sleep time 6
done in 21 seconds.
[1, 2, 3, 4, 5, 6]

The two approaches are quite different, let me try to explain it clearly
1st approach : In the first approach you are spinning up all Async requests for all 6 tasks and then calling join function on each one of them to get the result
2st approach : But in the second approach you are calling the join immediately after spinning the Async request for each task. For example after spinning Async thread for task 1 calling join, make sure that thread to complete task and then only spin up the second task with Async thread
Note : Another side if you observe the output clearly, In the 1st approach output appears in random order since the all six tasks were executed asynchronously. But during second approach all tasks were executed sequentially one after the another.
I believe you have an idea how stream map operation is performed, or you can get more information from here or here
To perform a computation, stream operations are composed into a stream pipeline. A stream pipeline consists of a source (which might be an array, a collection, a generator function, an I/O channel, etc), zero or more intermediate operations (which transform a stream into another stream, such as filter(Predicate)), and a terminal operation (which produces a result or side-effect, such as count() or forEach(Consumer)). Streams are lazy; computation on the source data is only performed when the terminal operation is initiated, and source elements are consumed only as needed.

The stream framework does not define the order in which map operations are executed on stream elements, because it is not intended for use cases in which that might be a relevant issue. As a result, the particular way your second version is executing is equivalent, essentially, to
List<Integer> results = new ArrayList<>();
for (Integer sleepTime : sleepTimes) {
results.add(CompletableFuture
.supplyAsync(() -> sleepTask(sleepTime), executorService2)
.exceptionally(ex -> { ex.printStackTrace(); return -1; }))
.join());
}
...which is itself essentially equivalent to
List<Integer> results = new ArrayList<>()
for (Integer sleepTime : sleepTimes) {
results.add(sleepTask(sleepTime));
}

#Deadpool answered it pretty well, just adding my answer which can help someone understand it better.
I was able to get an answer by adding more printing to both methods.
TLDR
2 stream approach: We are starting up all 6 tasks asynchronously and then calling join function on each one of them to get the result in a separate stream.
1 stream approach: We are calling the join immediately after starting up each task. For example after spinning a thread for task 1, calling join makes sure the thread waits for completion of task 1 and then only spin up the second task with async thread.
Note: Also, if we observe the output clearly, in the 1 stream approach, output appears sequential order since the all six tasks were executed in order. But during second approach all tasks were executed in parallel, hence the random order.
Note 2: If we replace stream() with parallelStream() in the 1 stream approach, it will work identically to 2 stream approach.
More proof
I added more printing to the streams which gave the following outputs and confirmed the note above :
1 stream:
List<Integer> results = sleepTimes.stream()
.map(sleepTime -> CompletableFuture.supplyAsync(() -> sleepTask(sleepTime), executorService2)
.exceptionally(ex -> { ex.printStackTrace(); return -1; }))
.map(f -> {
int num = f.join();
System.out.println(String.format("doing join on task %d", num));
return num;
})
.collect(Collectors.toList());
WITH SAME STREAM FOR FUTURE AND JOIN
Task with sleep time 1
doing join on task 1
Task with sleep time 2
doing join on task 2
Task with sleep time 3
doing join on task 3
Task with sleep time 4
doing join on task 4
Task with sleep time 5
doing join on task 5
Task with sleep time 6
doing join on task 6
done in 21 seconds.
[1, 2, 3, 4, 5, 6]
2 streams:
List<CompletableFuture<Integer>> futures = sleepTimes.stream()
.map(sleepTime -> CompletableFuture.supplyAsync(() -> sleepTask(sleepTime), executorService)
.exceptionally(ex -> { ex.printStackTrace(); return -1; }))
.collect(Collectors.toList());
List<Integer> result = futures.stream()
.map(f -> {
int num = f.join();
System.out.println(String.format("doing join on task %d", num));
return num;
})
.collect(Collectors.toList());
WITH SEPARATE STREAMS FOR FUTURE AND JOIN
Task with sleep time 2
Task with sleep time 5
Task with sleep time 3
Task with sleep time 1
Task with sleep time 4
Task with sleep time 6
doing join on task 1
doing join on task 2
doing join on task 3
doing join on task 4
doing join on task 5
doing join on task 6
done in 6 seconds.
[1, 2, 3, 4, 5, 6]

Flux parallel-serial execution with groupBy

Say I have this:
Flux<GroupedFlux<Integer, Integer>> intsGrouped = Flux.range(0, 12)
.groupBy(i -> i % 3);
and say I have a method:
Mono<Integer> getFromService(Integer i);
I want to call getFromService in parallel for each of the groups, but make sure the calls are serial within each group.
For the above example that would be three parallel streams with these input values:
stream 1: 0 -> 3 -> 6 -> 9
stream 2: 1 -> 4 -> 7 -> 10
stream 3: 2 -> 5 -> 8 -> 11
I tried this, but it's not doing what I want:
Flux.range(0, 12)
.groupBy(i -> i % 3)
.flatMap(g -> g.flatMap(i -> getFromService(g.key(), i)))
This is calling the service in parallel for all the ints at once. How do I proceed?

Use either concatMap or flatMapSequential instead of the inner .flatMap
If you want sequential execution within each group (i.e. only one subscription to getFromService at a single time within each group), then use .concatMap, like this:
Flux.range(0, 12)
.groupBy(i -> i % 3)
.flatMap(g -> g.concatMap(i -> getFromService(g.key(), i)))
If parallel execution within a group is ok, but you just care about the order in which the sequence is emitted, then use flatMapSequential, like this:
Flux.range(0, 12)
.groupBy(i -> i % 3)
.flatMap(g -> g.flatMapSequential(i -> getFromService(g.key(), i)))
Another option is to use .flatMap with the concurrency argument set to 1, but I'd recommend one of the above instead.

Observable caches emitted items or does not?

Does Observable caches emitted items? I have two tests that lead me to different conclusions:
From the test #1 I make an conclusion that it does:
Test #1:
Observable<Long> clock = Observable
.interval(1000, TimeUnit.MILLISECONDS)
.take(10)
.map(i -> i++);
//subscribefor the first time
clock.subscribe(i -> System.out.println("a: " + i));
//subscribe with 2.5 seconds delay
Executors.newScheduledThreadPool(1).schedule(
() -> clock.subscribe(i -> System.out.println(" b: " + i)),
2500,
TimeUnit.MILLISECONDS
);
Output #1:
a: 0
a: 1
a: 2
b: 0
a: 3
b: 1
But the second test shows that we get different values for two observers:
Test #2:
Observable<Integer> observable = Observable
.range(1, 1000000)
.sample(7, TimeUnit.MILLISECONDS);
observable.subscribe(i -> System.out.println("Subscriber #1:" + i));
observable.subscribe(i -> System.out.println("Subscriber #2:" + i));
Output #2:
Subscriber #1:72745
Subscriber #1:196390
Subscriber #1:678171
Subscriber #2:336533
Subscriber #2:735521

There exist two kinds of Observables: hot and cold. Cold observables tend to generate the same sequence to its Observers unless you have external effects, such as a timer based action, associated with it.
In the first example, you get the same sequence twice because there are no external effects other than timer ticks you get one by one. In the second example, you sample a fast source and sampling with time has a non-deterministic effect: each nanosecond counts so even the slightest imprecision leads to different value reported.