Reputation: 113
I would like to write a split function for monix.reactive.Observable
. It should split a source Observable[A]
into a new pair (Observable[A], Observable[A])
, based on the value of a predicate, evaluated against each element in the source. I would like the split to work independently of whether the source Observable is hot or cold. In the case where the source is cold, the new pair of Observables should also be cold and where the source is hot the new pair of Observables will be hot. I would like to know if such an implementation is possible and, if so, how (I have pasted a failing testcase below).
The signature, as a method on an implicit class, would look like, or similar to
/**
* Split an observable by a predicate, placing values for which the predicate returns true
* to the right (and values for which the predicate returns false to the left).
* This is consistent with the convention adopted by Either.cond.
*/
def split(p: T => Boolean)(implicit scheduler: Scheduler, taskLike: TaskLike[Future]): (Observable[T], Observable[T]) = {
splitEither[T, T](elem => Either.cond(p(elem), elem, elem))
}
Currently, I have a naive implementation that consumes the source elements and pushes them to PublishSubject
. The new pair of Observables is thus hot. My tests for a cold Observable are failing.
import monix.eval.TaskLike
import monix.execution.{Ack, Scheduler}
import monix.reactive.{Observable, Observer}
import monix.reactive.subjects.PublishSubject
import scala.concurrent.Future
object ObservableOps {
implicit class ObservableExtensions[T](o: Observable[T]) {
/**
* Split an observable by a predicate, placing values for which the predicate returns true
* to the right (and values for which the predicate returns false to the left).
* This is consistent with the convention adopted by Either.cond.
*/
def split(p: T => Boolean)(implicit scheduler: Scheduler, taskLike: TaskLike[Future]): (Observable[T], Observable[T]) = {
splitEither[T, T](elem => Either.cond(p(elem), elem, elem))
}
/**
* Split an observable into a pair of Observables, one left, one right, according
* to a determinant function.
*/
def splitEither[U, V](f: T => Either[U, V])(implicit scheduler: Scheduler, taskLike: TaskLike[Future]): (Observable[U], Observable[V]) = {
val l = PublishSubject[U]()
val r = PublishSubject[V]()
o.subscribe(new Observer[T] {
override def onNext(elem: T): Future[Ack] = {
f(elem) match {
case Left(u) => l.onNext(u)
case Right(v) => r.onNext(v)
}
}
override def onError(ex: Throwable): Unit = {
l.onError(ex)
r.onError(ex)
}
override def onComplete(): Unit = {
l.onComplete()
r.onComplete()
}
})
(l, r)
}
}
}
//////////
import ObservableOps._
import monix.execution.Scheduler.Implicits.global
import monix.reactive.Observable
import monix.reactive.subjects.PublishSubject
import org.scalatest.FlatSpec
import org.scalatest.Matchers._
import org.scalatest.concurrent.ScalaFutures._
class ObservableOpsSpec extends FlatSpec {
val isEven: Int => Boolean = _ % 2 == 0
"Observable Ops" should "split a cold observable" in {
val o = Observable(1, 2, 3, 4, 5)
val (l, r) = o.split(isEven)
l.toListL.runToFuture.futureValue shouldBe List(1, 3, 5)
r.toListL.runToFuture.futureValue shouldBe List(2, 4)
}
"Observable Ops" should "split a hot observable" in {
val o = PublishSubject[Int]()
val (l, r) = o.split(isEven)
val lbuf = l.toListL.runToFuture
val rbuf = r.toListL.runToFuture
Observable.fromIterable(1 to 5).mapEvalF(i => o.onNext(i)).subscribe()
o.onComplete()
lbuf.futureValue shouldBe List(1, 3, 5)
rbuf.futureValue shouldBe List(2, 4)
}
}
I expect both testcases above to pass but "Observable Ops" should "split a cold observable"
is failing.
An implementation that passes both test cases is as follows:
import monix.execution.Scheduler
import monix.reactive.Observable
object ObservableOps {
implicit class ObservableExtension[T](o: Observable[T]) {
/**
* Split an observable by a predicate, placing values for which the predicate returns true
* to the right (and values for which the predicate returns false to the left).
* This is consistent with the convention adopted by Either.cond.
*/
def split(
p: T => Boolean
)(implicit scheduler: Scheduler): (Observable[T], Observable[T]) = {
splitEither[T, T](elem => Either.cond(p(elem), elem, elem))
}
/**
* Split an observable into a pair of Observables, one left, one right, according
* to a determinant function.
*/
def splitEither[U, V](
f: T => Either[U, V]
)(implicit scheduler: Scheduler): (Observable[U], Observable[V]) = {
val oo = o.map(f)
val l = oo.collect {
case Left(u) => u
}
val r = oo.collect {
case Right(v) => v
}
(l, r)
}
}
}
Upvotes: 4
Views: 719
Reputation: 7353
Cold observable, by definition, is lazily evaluated for each subscriber. You can't split it without either evaluating everything twice or converting it into hot one.
If you don't mind evaluating everything twice, just use .filter
two times.
If you don't mind converting to hot, do it with .publish
(or .publish.refCount
so you don't need to connect
manually).
If you want to preserve cold/hot property and process two pieces in parallel, there's a publishSelector
method that lets you treat any observable like a hot one in a limited scope:
coldOrHot.publishSelector { totallyHot =>
val s1 = totallyHot.filter(...).flatMap(...) // any processing
val s2 = totallyHot.filter(...).mapEval(...) // any processing 2
Observable(s1, s2).merge
}
It's limitation, apart from scope, is that result of inner lambda has to be another Observable (which will be returned from publishSelector), so you can't have the helper with the signature you want. But the result will still be cold if the original was cold.
Upvotes: 2
Reputation: 21
class ObservableOpsSpec extends FlatSpec {
val isEven: Int => Boolean = _ % 2 == 0
"Observable Ops" should "split a cold observable" in {
val o = Observable(1, 2, 3, 4, 5)
val o2 = o.publish
val (l, r) = o2.split(isEven)
val x= l.toListL.runToFuture
val y = r.toListL.runToFuture
o2.connect()
x.futureValue shouldBe List(1, 3, 5)
y.futureValue shouldBe List(2, 4)
}
"Observable Ops" should "split a hot observable" in {
val o = PublishSubject[Int]()
val (l, r) = o.split(isEven)
val lbuf = l.toListL.runToFuture
val rbuf = r.toListL.runToFuture
Observable.fromIterable(1 to 5).mapEvalF(i => o.onNext(i)).subscribe()
o.onComplete()
lbuf.futureValue shouldBe List(1, 3, 5)
rbuf.futureValue shouldBe List(2, 4)
}
}
Upvotes: 2