RxSwift combineLatest emits all replayed elements from last Observable - rx-swift

I'm seeing some unexpected results when using combineLatest on a collection of cold observables. It emits the latest from all but the last Observable and instead combines the latest from the first (n-1) Observables with each element from the nth Observable.
let observable = ReplaySubject<Int>.createUnbounded()
let observable2 = ReplaySubject<String>.createUnbounded()
observable.onNext(1)
observable.onNext(2)
observable.onNext(3)
observable.onNext(4)
observable2.onNext("bed")
observable2.onNext("book")
observable2.onNext("table")
let latestObserver = Observable.combineLatest(observable, observable2)
_ = latestObserver
.subscribe(onNext: {
print($0)
})
.disposed(by: disposeBag)
Produces the output:
(4, "bed")
(4, "book")
(4, "table")
I had expected to see output of just (4, "table").
If I change the order of the observables like so:
let latestObserver = Observable.combineLatest(observable2, observable)
The I get the output:
("table", 1)
("table", 2)
("table", 3)
("table", 4)
If I add a final arbitrary Observable then I see just the latest from each of the first ones:
let observable = ReplaySubject<Int>.createUnbounded()
let observable2 = ReplaySubject<String>.createUnbounded()
let observable3 = Observable<Int>.just(42)
observable.onNext(1)
observable.onNext(2)
observable.onNext(3)
observable.onNext(4)
observable2.onNext("bed")
observable2.onNext("book")
observable2.onNext("table")
let latestObserver = Observable.combineLatest(observable, observable2, observable3)
_ = latestObserver
.subscribe(onNext: {
print($0)
})
.disposed(by: disposeBag)
produces the output: (4, "table", 42)
Is this really the expected behaviour?

Let's break down what is happening in your first example...
You could use Observable.from instead of the subject and get the same results... Steps in the code are as follows,
create two ReplaySubjects and load them up with events.
The combineLatest operator subscribes to the first subject.
That first subject immediately replays all of its values.
Since the second subject hasn't been subscribed to yet, the combineLatest operator doesn't emit anything, instead it silently absorbs the values while always storing the "latest" one.
The combineLatest operator then subscribes to the second subject.
That subject the replays all of its values.
Since the combineLatest operator has now received a next event from each of its sources, it emits the values emitted from the second source, combined with the latest (i.e., last) from the first source.
Your replay subjects are inherently synchronous. They emit all the values to the subscriber immediately upon subscription.
In your last example code, since the last observable of the three emits only one value and only after the other two have emitted all of their values, you only see one output containing the latest from the previous two observables.

Related

zip 2 observable everytime one of them emit

I have the following
const timer1 = interval(1000).pipe(take(10));
const timer2 = interval(2000).pipe(take(6));
const merged = merge(timer1, timer2);
merged.subscribe(x => console.log(x));
Now this will only provide the value of the latest emitting observable.
I would like that, when one emit, I get the new value, and the latest from the other one.
I tried to use zip but it will not use the latest. Check the screenshot below for the missing value I wish to have. How can I achieve this ?
Join, would work, but join do not emit if the same observable emitted twice before the other one emitted a value.
You are looking for combineLatest.
combineLatest will emit whenever either of the source observables emit. Note: it will not emit for the first time until each observable has emitted at least once.
So with your example, it could look like this (StackBlitz):
const timer1 = interval(1000).pipe(take(10));
const timer2 = interval(2000).pipe(take(6));
const merged = combineLatest([timer1, timer2]);
merged.subscribe(([one, two]) => console.log(`${one}-${two}`));

What is the difference between tap and map in RxJS?

I read the difference from the article but the main points look like this.
so with tap I can change the variables such as that if I put x=3+4 then it changes the values of variable then I can say there is one side effect.
But with map I can change the value looping each value, isn't it?
Can you pinpoint what outstanding differences they have?
tap
RxJS tap performs side effects for every value emitted by source Observable and returns an Observable identical to the source Observable until there is no error.
map
map is a RxJS pipeable operator. map applies a given function to each element emitted by the source Observable and emits the resulting values as an Observable
A mapping function takes a thing and returns another thing. e.g. I can build a function that takes 10 and returns 11, that takes 11 and returns 12, etc.
const inc = n => n + 1;
Array#map applies such mapping function to all elements of an array but "map" doesn't mean "iteration".
In RxJS, when a data is sent to the stream it goes through a series of operators:
The map operator will simply apply a function to that data and return the result.
The tap operator however takes a data, apply a function to that data but returns the original data, if the function bothered to return a result, tap just ignores it.
Here's an example:
We push 10 to stream a$, tap just log the value. We know that console.log always return undefined but that's fine because tap simply returns its parameter.
We push 10 to stream b$, it goes through map(inc) which applies inc to 10 returning 11.
const a$ = of(10).pipe(tap(n => console.log(`tap: ${n}`)));
const b$ = of(10).pipe(map(inc));
a$.subscribe(n => console.log(`n from a$: ${n}`));
b$.subscribe(n => console.log(`n from b$: ${n}`));
<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.5.5/rxjs.umd.min.js"></script>
<script>
const {of} = rxjs;
const {map, tap} = rxjs.operators;
const inc = n => n + 1;
</script>
Tap should be Used for Notification, logging non-contextual/critical side effects.
It's like a "peek" into the "pipe". The data stays the same, You can do something with it. Some data goes in, you look, same data comes out.
Map is for transformation/mapping of the Data in the "pipe". Some data comes in, different/transformed data comes out.
The purpose of tap is to execute an action keeping the same value
of the observable
The purpose of map is to transform the emitted values of the
observable
const messagesCount$ = newMessages$
.pipe(tap(messages => notificationService.notify('You have ' + message.length + ' message(s)')))
.pipe(map(messages => messages.length))
The tap and map are both RxJS operators, RxJS operators are just function that performs some manipulation over the data.
Both of them are pipeable operators which takes input as Observable, perform some action and return an output observable.
Difference between map and tap:
The map is a pipeable operator that takes an input observable, performs some manipulation on it and returns a new manipulated observable. For example
const source$ = of(1,2,3) // observable which will emit 1,2,3
// It take an input observable and return a new observable which will emit square of input values.
// So, the output observable will emit 1,4,9
const mapSource$ = of(1,2,3)
.pipe(map(value => value * value))
The tap operator on another hand takes an input observable perform some action and returns the same input observable.
const source$ = of(1,2,3) // observable which will emit 1,2,3
// It take an input observable and return a same observable after console value.
// So, the output observable will emit 1,2,3
const tapSource$ = of(1,2,3)
.pipe(tap(value => console.log(value)))
you can think of tap operator as a void function that whatever it does to the input value it does not change the original value
const source = of(1, 2, 3, 4, 5);
// here we are manipulating the input value but the output value of the observable still the same
const example = source.pipe(
tap(val => val + 100),
);
// output: 1, 2, 3, 4, 5
const subscribe = example.subscribe(val => console.log(val));
in the other hand if we made any manipulation of the input values of the observable using the map operator it will change the output values
const example = source.pipe(
map(val => val + 100)
);
// output: 101, 102, 103, 104, 105
const subscribe = example.subscribe(val => console.log(val));
I addition to what the others are saying, in Rxjs 7.4 tap now has three more subscribe handlers, so you can use it to get notified on subscribe, unsubscribe and finalize:
https://github.com/ReactiveX/rxjs/commit/eb26cbc4488c9953cdde565b598b1dbdeeeee9ea#diff-93cd3ac7329d72ed4ded62c6cbae17b6bdceb643fa7c1faa6f389729773364cc
This is great for debugging purposes, so you can use tap to find out much more about what is happening with your stream.
Example:
const subscription = subject
.pipe(
tap({
subscribe: () => console.log('subscribe'),
next: (value) => console.log(`next ${value}`),
error: (err) => console.log(`error: ${err.message}`),
complete: () => console.log('complete'),
unsubscribe: () => console.log('unsubscribe'),
finalize: () => console.log('finalize'),
})
)
.subscribe();
TAP, can NOT transform:
interval(1000).pipe(tap(el=> el*2)).subscribe(console.log); // 0,1,2,3
MAP, CAN transform:
interval(1000).pipe(map(el=> el*2)).subscribe(console.log); // 0,2,4,6
If you do NOT need to transform the value, just console.log it or run external function to pass the Original value = TAP is good.
If you NEED TO TRANSFORM/CHANGE the value = MAP is the way to go.

What is the difference between merge and mergeAll?

What is the difference between merge and mergeAll? They both seem identical to me:
http://reactivex.io/rxjs/class/es6/Observable.js~Observable.html#instance-method-mergeAll
http://reactivex.io/rxjs/class/es6/Observable.js~Observable.html#static-method-merge
merge is a static creation method that flattens group of observable.
according to the docs
Flattens multiple Observables together by blending their values into one Observable.
simply it will take a group of observables, and flattens them within one, so whenever any observable emits a value, the output will emit a value.
mergeAll However is different, it is an instance method that works with higher order observables (an observable that emits observables), according to docs
Converts a higher-order Observable into a first-order Observable which concurrently delivers all values that are emitted on the inner Observables.
I think that sums it up, but mergeAll can be confusing, so let's look at this example provided by rxjs docs
import { fromEvent, interval } from 'rxjs';
import { take, map, mergeAll } from 'rxjs/operators';
const higherOrder = fromEvent(document, 'click').pipe(
map((ev) => interval(1000).pipe(take(10))),
);
const firstOrder = higherOrder.pipe(mergeAll(2));
firstOrder.subscribe(x => console.log(x));
you have a document click observable (higher order) which return an interval observable (inner observable) that emits a value every second, it will complete after 10 intervals emits, which means every time you click on the document, a new interval will be returned, here where merge all comes in, it will subscribe to these intervals returned by the higher order observable, and flattens them into one observable, the first order observable, the argument 2, is to limit to 2 concurrent intervals at a time, so if you clicked 3 times, only 2 will run, but since these 2 intervals will complete after 10 seconds, then you can click again and mergeAll will subscribe to the new intervals.
Both merge and mergeAll inherit from mergeMap !
mergeAll
mergeAll is the same as calling mergeMap with an identity function(const identity = x => x)
mergeAll() === mergeMap(obs$ => obs$)
Example:
of(a$, b$, c$)
.pipe(
mergeAll(),
)
.subscribe()
// Same as
of(a$, b$, c$)
.pipe(
mergeMap(obs$ => obs$)
)
.subscribe()
Both will subscribe to the incoming observables(a$, b$ and c$) and will pass along to their values to the data consumer. Thus, a$, b$ and c$ are considered inner observables.
merge
Armed with the knowledge from the previous section, understanding merge should not be difficult.
merge(a$, b$, c$).subscribe() is essentially the same as
const observables = [a$, b$, c$];
new Observable(subscriber => {
for (let i = 0; i < observables.length; i++) {
subscriber.next(observables[i]);
}
subscriber.complete();
}).pipe(
mergeAll()
).subscribe();

why concatMap is not subscribing to all groupBy Observables rxjs?

I have a question why is this not writing to console the numbers 2,4,6? what is the explanation?
Observable.range(1, 6)
.groupBy(n => n % 2 === 0)
.concatMap(obs => obs)
.subscribe((n) => console.log(n), null, () => console.log('complete concatMap'))
// this is the output
1 -
3 -
5 -
complete concatMap
The basic problem is that you're using concatMap that subscribes to the next Observable only when the previous one completed. groupBy emits two GroupedObservables so it subscribes to the first one and I think before it can subscribe to the second one the chain completes. This means the observer receives the complete notification from the first GroupedObservable and therefore you never see values from the second GroupedObservable (to be honest I'm not 100% sure it really happens like this but that makes sense to be without further investigating you example).
So if you want only the second group you could do:
import { Observable } from 'rxjs';
Observable.range(1, 6)
.groupBy(n => n % 2 === 0)
.filter(o => o.key === true)
.concatMap(obs => obs)
.subscribe((n) => console.log(n), null, () => console.log('complete concatMap'))
See live demo (open console): https://stackblitz.com/edit/rxjs5-sfused
I checked the source code and groupBy completes all groups after receiving the complete notification (which it does after receiving all values from range) and therefore there's never space for concatMap to subscribe to the second Observable.
See this: https://github.com/ReactiveX/rxjs/blob/master/src/internal/operators/groupBy.ts#L200-L210
Problem is that groupBy operator emits Subjects for each key.
Concat map subscribes to subsequent Subject only after the first one is completed, i.e. it misses a chance to catch items from subsequent Subject because all of the sub-streams emit values in the same time.
Kudos to: https://blog.angularindepth.com/those-hidden-gotchas-within-rxjs-7d5c57406041
TL;DR:
GroupBy receives subjectSelector as a 4th argument. You can use it to force using ReplaySubject instead of Subject (default).
Observable.range(1, 6)
.groupBy(
n => n % 2 === 0,
null,
null,
() => new ReplaySubject() // <-- Here we go
)
.concatMap(obs => obs)
.subscribe((n) => console.log(n))
Demo on RxViz

Have withLatestFrom wait until all sources have produced one value

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()

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