Given an ngrx selector:
store.select('count')
I want to create an observable that will emit values emitted by the selector, then emit another specific value after a delay.
Using concat doesn't work as (I assume) the selector doesn't complete, so the 0 is never emitted:
this.count$ = concat(store.select('count'), of(0).pipe(delay(2000)));
Stackblitz: https://stackblitz.com/edit/so-selector-delay?file=src/app/my-counter/my-counter.component.ts
- click 'Increment' button - Current Count should change to 1 then back to 0 after 2 seconds.
If you want to emit the store.select('count') value, then essentially reset it to 0 after not receiving an emission for 2 seconds, you can use a switchMap to create a source that emits two values:
The emitted count
The "default" value of 0 after 2000ms
The trick here is that the second value (0) will NOT be emitted if another store.select('count') value is received, because switchMap will create a new source and "switch" to that:
this.count$ = store.select('count').pipe(
switchMap(count => concat(
of(count),
of(0).pipe(delay(2000))
))
);
Here's a working StackBlitz demo.
It might even be worth creating a custom operator:
this.count$ = this.store.select('count').pipe(
resetAfterDelay(0, 2000)
);
export function resetAfterDelay<T>(defaultValue: T, delayMs: number) {
return switchMap((value: T) => concat(
of(value),
of(defaultValue).pipe(delay(delayMs))
));
}
StackBlitz
Below is an approach using combineLatest and BehaviorSubject
We are hold a value in a subject and create a timer that emits value 0 after 2s. So we have two Observables, one emits immediately and the other after 2s. We combine this two and the effect is a single observable as desired
valueHolderSubject$ = new BehaviorSubject(0);
...
this.count$ = combineLatest([
store.select("count").pipe(
tap(x => this.valueHolderSubject$.next(x)),
tap(() =>
timer(2000).subscribe({
next: () => {
this.valueHolderSubject$.next(0);
}
})
),
map(() => this.valueHolderSubject$.value),
distinctUntilChanged()
),
this.valueHolderSubject$
]).pipe(map(([, x]) => x));
Demo Here
A per my comments on the answer from BizzyBob, I was getting unreliable results. But I refined it to work using:
this.count$ = merge(
store.select('count'),
store.select('count').pipe(delay(2000), map(_ => 0))
);
See stackblitz: https://stackblitz.com/edit/so-selector-delay-merge-working?file=src/app/my-counter/my-counter.component.ts
I was wondering if anyone could help me theorize a solution. I create an Observable from a wheel event, prevent the default action, throttle it by 200ms, map deltaY (which I can use to determine direction), and then I share it.
My problem is that it emits more values than I need creating a situation where my subscribers continue to fire even after the desired action has occurred. I'm new to RxJS so bear with me but... Is there a way for me to take the "first" value emitted in a series of values within say X amount of time passed and not have the observable complete?
Below is the code.
import { fromEvent } from 'rxjs';
const wheel$ = fromEvent(document, 'wheel')
.pipe(
tap((event) => event.preventDefault()),
// throttleTime(200), /* I have tried throttling and debouncing but that doesn't work - values will continue to be emitted */
map((event) => event.deltaY),
share()
)
// handles scrolling down //
wheel$.pipe(filter((val) => val > 0))
.subscribe((event) => {
if (this.props.isScrolling) return
this.scrollDown();
})
One solution would be "bufferCount()"
of(1,2,3,4,5,6,7,8,9).pipe(
bufferCount(3)
).subscribe(data => console.log(data) )
would create packages of 3 signals. So the Events would be
[1,2,3]
[4,5,6]
[7,8,9]
Or "throttleTime(xy)", than it will put the first signal through, will ignore for "xy" milliseconds every other signal, and then give the next signal a chance.
interval(500).pipe(
throttleTime(2000)
).subscribe(data => console.log(data) )
will result in something like
1 // ignore everything the next 2 seconds
5 // ignore everything the next 2 seconds
9 // ignore everything the next 2 seconds
...
warm regards
Found this question while looking for a solution.
I went with a switchMap
fromEvent(el, 'wheel').pipe(
switchMap(e =>
concat(
//wrap the original event, which fires immediately
of({kind: 'ON_WHEEL', e}),
// fire a onWheelEnd after a 200ms delay
// this event is unsubscribed by the switchMap if a new event comes in
// before the delay
of({kind: 'ON_WHEEL_END}).pipe(delay(200))
)
)
)
I have 2 different Observables, let's call them racer_1 and racer2, which emit when 2 different events occur.
If racer_1 occurs I have to perform the logic of function doStuff_1(), if racer_2 occurs I have to perform doStuff_2().
racer_1 and racer_2 are not mutually exclusive and they can occur one after the other, but what I want to achieve is that if any of them occur, I process just that one with its associated function.
It seems clearly the case of using the race function of RxJs.
race(
racer_1.pipe(
tap(() => doStuff_1()),
finalize(() => console.log('racer 1 completed'))
),
racer_2.pipe(
tap(() => doStuff_2()),
finalize(() => console.log('racer 2 completed'))
)
)
Unfortunately though it may happen that racer_1 emits first at t0 and doStuff_1() starts a synchronous execution which ends at t0+1000.
In the meantime, at t0+200 racer_2 emits and the execution of doStuff_2() is much shorter so that it ends at t0+300.
In this case what I see is that racer_2 "wins the race" and its emissions continue to get processed, while racer_1 gets completed.
On the contrary what I would like to see is that racer_1 wins since it occurred first, even if its processing can get long to be completed.
Is there a way to obtain such behavior?
You can just reorder operators and wrap the initial emissions from both source Observables and perform doStuff_1 or doStuff_2 later:
race( // wrap emissions with tmp objects
racer_1.pipe(map(result => { type: 'racer1', result })),
racer_2.pipe(map(result => { type: 'racer2', result })),
)
.tap(({ type, result }) => type === 'racer1'
? doStuff_1()
: doStuff_2()
)
map(({ type, result }) => result) // unwrap the object
I'm making use of the withLatestFrom operator in RxJS in the normal way:
var combined = source1.withLatestFrom(source2, source3);
...to actively collect the most recent emission from source2 and source3 and to emit all three value only when source1 emits.
But I cannot guarantee that source2 or source3 will have produced values before source1 produces a value. Instead I need to wait until all three sources produce at least one value each before letting withLatestFrom do its thing.
The contract needs to be: if source1 emits then combined will always eventually emit when the other sources finally produce. If source1 emits multiple times while waiting for the other sources we can use the latest value and discard the previous values. Edit: as a marble diagram:
--1------------2---- (source)
----a-----b--------- (other1)
------x-----y------- (other2)
------1ax------2by--
--1------------2---- (source)
------a---b--------- (other1)
--x---------y------- (other2)
------1ax------2by--
------1--------2---- (source)
----a-----b--------- (other1)
--x---------y------- (other2)
------1ax------2by--
I can make a custom operator for this, but I want to make sure I'm not missing an obvious way to do this using the vanilla operators. It feels almost like I want combineLatest for the initial emit and then to switch to withLatestFrom from then on but I haven't been able to figure out how to do that.
Edit: Full code example from final solution:
var Dispatcher = new Rx.Subject();
var source1 = Dispatcher.filter(x => x === 'foo');
var source2 = Dispatcher.filter(x => x === 'bar');
var source3 = Dispatcher.filter(x => x === 'baz');
var combined = source1.publish(function(s1) {
return source2.publish(function(s2) {
return source3.publish(function(s3) {
var cL = s1.combineLatest(s2, s3).take(1).do(() => console.log('cL'));
var wLF = s1.skip(1).withLatestFrom(s2, s3).do(() => console.log('wLF'));
return Rx.Observable.merge(cL, wLF);
});
});
});
var sub1 = combined.subscribe(x => console.log('x', x));
// These can arrive in any order
// and we can get multiple values from any one.
Dispatcher.onNext('foo');
Dispatcher.onNext('bar');
Dispatcher.onNext('foo');
Dispatcher.onNext('baz');
// combineLatest triggers once we have all values.
// cL
// x ["foo", "bar", "baz"]
// withLatestFrom takes over from there.
Dispatcher.onNext('foo');
Dispatcher.onNext('bar');
Dispatcher.onNext('foo');
// wLF
// x ["foo", "bar", "baz"]
// wLF
// x ["foo", "bar", "baz"]
I think the answer is more or less as you described, let the first value be a combineLatest, then switch to withLatestFrom. My JS is hazy, but I think it would look something like this:
var selector = function(x,y,z) {};
var combined = Rx.Observable.concat(
source1.combineLatest(source2, source3, selector).take(1),
source1.withLatestFrom(source2, source3, selector)
);
You should probably use publish to avoid multiple subscriptions, so that would look like this:
var combined = source1.publish(function(s1)
{
return source2.publish(function(s2)
{
return source3.publish(function(s3)
{
return Rx.Observable.concat(
s1.combineLatest(s2, s3, selector).take(1),
s1.withLatestFrom(s2, s3, selector)
);
});
});
});
or using arrow functions...
var combined = source1.publish(s1 => source2.publish(s2 => source3.publish(s3 =>
Rx.Observable.concat(
s1.combineLatest(s2, s3, selector).take(1),
s1.withLatestFrom(s2, s3, selector)
)
)));
EDIT:
I see the problem with concat, the withLatestFrom isn't getting the values. I think the following would work:
var combined = source1.publish(s1 => source2.publish(s2 => source3.publish(s3 =>
Rx.Observable.merge(
s1.combineLatest(s2, s3, selector).take(1),
s1.skip(1).withLatestFrom(s2, s3, selector)
)
)));
...so take one value using combineLatest, then get the rest using withLatestFrom.
I wasn't quite satisfied with the accepted answer, so I ended up finding another solution. Many ways to skin a cat!
My use-case involves just two streams - a "requests" stream and a "tokens" stream. I want requests to fire as soon as they are received, using the whatever the latest token is. If there is no token yet, then it should wait until the first token appears, and then fire off all the pending requests.
I wasn't quite satisfied with the accepted answer, so I ended up finding another solution. Essentially I split the request stream into two parts - before and after first token arrives. I buffer the first part, and then re-release everything in one go once I know that the token stream is non-empty.
const first = token$.first()
Rx.Observable.merge(
request$.buffer(first).mergeAll(),
request$.skipUntil(first)
)
.withLatestFrom(token$)
See it live here: https://rxviz.com/v/VOK2GEoX
For RxJs 7:
const first = token$.first()
merge(
request$.pipe(
buffer(first),
mergeAll()
),
request$.pipe(
skipUntil(first)
)
).pipe(
withLatestFrom(token$)
)
I had similar requirements but for just 2 observables.
I ended up using switchMap+first:
observable1
.switchMap(() => observable2.first(), (a, b) => [a, b])
.subscribe(([a, b]) => {...}));
So it:
waits until both observables emit some value
pulls the value from second observable only if the first one has changed (unlike combineLatest)
doesn't hang subscribed on second observable (because of .first())
In my case, second observable is a ReplaySubject. I'm not sure if it will work with other observable types.
I think that:
flatMap would probably work too
it might be possible to extend this approach to handle more than 2 observables
I was surprised that withLatestFrom will not wait on second observable.
In my mind, the most elegant way to achieve the different behavior of an existing RxJS operator is to wrap it into a custom operator. So that from the outside it looks just like any regular operator and doesn't require you to restructure your code each time you need this behavior.
Here is how you can create your own operator which behaves just like withLatestFrom, except that at the very beginning it will emit as soon as the first value of the target observable is emitted (unlike standard withLatestFrom, which will ignore the first emission of the source if the target hasn't yet emitted once). Let's call it delayedWithLatestFrom.
Note that it's written in TypeScript, but you can easily transform it to plain JS. Also, it's a simple version that supports only one target observable and no selector function - you can extend it as needed from here.
export function delayedWithLatestFrom<T, N>(
target$: Observable<N>
): OperatorFunction<T, [T, N]> {
// special value to avoid accidental match with values that could originate from target$
const uniqueSymbol = Symbol('withLatestFromIgnore');
return pipe(
// emit as soon target observable emits the first value
combineLatestWith<T, [N]>(target$.pipe(first())),
// skip the first emission because it's handled above, and then continue like a normal `withLatestFrom` operator
withLatestFrom(target$.pipe(skip(1), startWith(uniqueSymbol))),
map(([[rest, combineLatestValue], withLatestValue]) => {
// take combineLatestValue for the first time, and then always take withLatestValue
const appendedValue =
withLatestValue === uniqueSymbol ? combineLatestValue : withLatestValue;
return [rest, appendedValue];
})
);
}
// SAMPLE USAGE
source$.pipe(
delayedWithLatestFrom(target$)
).subscribe(console.log);
So if you compare it with the original marble diagram for withLatestFrom, it will differ only in one fact: while withLatestFrom ignores the first emissions and produces b1 as the first value, the delayedWithlatestFrom operator will emit one more value a1 at the beginning, as soon as the second observable emits 1.
a) Standard withLatestFrom:
b) Custom delayedWithLatestFrom:
Use combineLatest and filter to remove tuples before first full set is found then set a variable to stop filtering. The variable can be within the scope of a wrapping defer to do things properly (support resubscription). Here it is in java (but the same operators exist in RxJs):
Observable.defer(
boolean emittedOne = false;
return Observable.combineLatest(s1, s2, s3, selector)
.filter(x -> {
if (emittedOne)
return true;
else {
if (hasAll(x)) {
emittedOne = true;
return true;
} else
return false;
}
});
)
I wanted a version where tokens are fetched regularly - and where I want to retry the main data post on (network) failure. I found shareReplay to be the key. The first mergeWith creates a "muted" stream, which causes the first token to be fetched immediately, not when the first action arrives. In the unlikely event that the first token will still not be available in time, the logic also has a startWith with an invalid value. This causes the retry logic to pause and try again. (Some/map is just a Maybe-monad):
Some(fetchToken$.pipe(shareReplay({refCount: false, bufferSize: 1})))
.map(fetchToken$ =>
actions$.pipe(
// This line is just for starting the loadToken loop immediately, not waiting until first write arrives.
mergeWith(fetchToken$.pipe(map(() => true), catchError(() => of(false)), tap(x => loggers.info(`New token received, success: ${x}`)), mergeMap(() => of()))),
concatMap(action =>
of(action).pipe(
withLatestFrom(fetchToken$.pipe(startWith(""))),
mergeMap(([x, token]) => (!token ? throwError(() => "Token not ready") : of([x, token] as const))),
mergeMap(([{sessionId, visitId, events, eventIds}, token]) => writer(sessionId, visitId, events, token).pipe(map(() => <ISessionEventIdPair>{sessionId, eventIds}))),
retryWhen(errors =>
errors.pipe(
tap(err => loggers.warn(`Error writing data to WG; ${err?.message || err}`)),
mergeMap((_error: any, attemptIdx) => (attemptIdx >= retryPolicy.retryCount ? throwError(() => Error("It's enough now, already")) : of(attemptIdx))), // error?.response?.status (int, response code) error.code === "ENOTFOUND" / isAxiosError: true / response === undefined
delayWhen(attempt => timer(attempt < 2 ? retryPolicy.shortRetry : retryPolicy.longRetry, scheduler))
)
)
)
),
)
)
Thanks to everyone on this question-page for good inputs.
Based on the answer from #cjol
Here's a RxJs 7 implementation of a waitFor operator that will buffer the source stream until all input observables have emitted values, then emit all buffered events on the source stream. Any subsequent events on the source stream are emitted immediately.
// Copied from the definition of withLatestFrom() operator.
export function waitFor<T, O extends unknown[]>(
inputs: [...ObservableInputTuple<O>]
): OperatorFunction<T, [T, ...O]>;
/**
* Buffers the source until every observable in "from" have emitted a value. Then
* emit all buffered source values with the latest values of the "from" array.
* Any source events are emitted immediately after that.
* #param from Array of observables to wait for.
* #returns Observable that emits an array that concatenates the source and the observables to wait.
*/
export function waitFor(
from: Observable<unknown>[]
): (source$: Observable<unknown>) => Observable<unknown> {
const combined$ = combineLatest(from);
// This served as a conditional that switched on and off the streams that
// wait for the the other observables, or emits the source right away because
// the other observables have emitted.
const firstCombined$ = combined$.pipe(first());
return function (source$: Observable<unknown>): Observable<unknown> {
return merge(
// This stream will buffer the source until the other observables have all emitted.
source$.pipe(
takeUntil(firstCombined$), // without this it continues to buffer new values forever
buffer(firstCombined$),
mergeAll()
),
// This stream emits the source straight away and will take over when the other
// observables have emitted.
source$.pipe(skipUntil(firstCombined$))
).pipe(
withLatestFrom(combined$),
// Flatten it to behave like withLatestFrom() operator.
map(([source, combined]) => [source, ...combined])
);
};
}
All of the above solutions are not really on the point, therefore I made my own. Hope it helps someone out.
import {
combineLatest,
take,
map,
ObservableInputTuple,
OperatorFunction,
pipe,
switchMap
} from 'rxjs';
/**
* ### Description
* Works similar to {#link withLatestFrom} with the main difference that it awaits the observables.
* When all observables can emit at least one value, then takes the latest state of all observables and proceeds execution of the pipe.
* Will execute this pipe only once and will only retrigger pipe execution if source observable emits a new value.
*
* ### Example
* ```ts
* import { BehaviorSubject } from 'rxjs';
* import { awaitLatestFrom } from './await-latest-from.ts';
*
* const myNumber$ = new BehaviorSubject<number>(1);
* const myString$ = new BehaviorSubject<string>("Some text.");
* const myBoolean$ = new BehaviorSubject<boolean>(true);
*
* myNumber$.pipe(
* awaitLatestFrom([myString$, myBoolean$])
* ).subscribe(([myNumber, myString, myBoolean]) => {});
* ```
* ### Additional
* #param observables - the observables of which the latest value will be taken when all of them have a value.
* #returns a tuple which contains the source value as well as the values of the observables which are passed as input.
*/
export function awaitLatestFrom<T, O extends unknown[]>(
observables: [...ObservableInputTuple<O>]
): OperatorFunction<T, [T, ...O]> {
return pipe(
switchMap((sourceValue) =>
combineLatest(observables).pipe(
take(1),
map((values) => [sourceValue, ...values] as unknown as [T, ...O])
)
)
);
}
Actually withLatestFrom already
waits for every source
emits only when source1 emits
remembers only the last source1-message while the other sources are yet to start
// when source 1 emits the others have emitted already
var source1 = Rx.Observable.interval(500).take(7)
var source2 = Rx.Observable.interval(100, 300).take(10)
var source3 = Rx.Observable.interval(200).take(10)
var selector = (a,b,c) => [a,b,c]
source1
.withLatestFrom(source2, source3, selector)
.subscribe()
vs
// source1 emits first, withLatestFrom discards 1 value from source1
var source1 = Rx.Observable.interval(500).take(7)
var source2 = Rx.Observable.interval(1000, 300).take(10)
var source3 = Rx.Observable.interval(2000).take(10)
var selector = (a,b,c) => [a,b,c]
source1
.withLatestFrom(source2, source3, selector)
.subscribe()