What is the difference between merge and mergeAll? - rxjs

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();

Related

How to recompute observable once for a single event loop tick?

I have a use case where I have one observable shared across multiple observables where finally those 2 observable merged into a single one.
The issue is, whenever I click the document, I get subscription callback called twice, but I need it to be called once containing the last value:
import { fromEvent, of, combineLatest } from "rxjs";
import { map, switchMap, startWith } from "rxjs/operators";
const sharedState$ = fromEvent(document, "click").pipe(
map((event) => event.clientX),
startWith(0)
);
const stateA$ = of(5).pipe(
switchMap((a) => sharedState$.pipe(map((x) => x + a)))
);
const stateB$ = of(7).pipe(
switchMap((b) => sharedState$.pipe(map((x) => x + b)))
);
const result$ = combineLatest([stateA$, stateB$]).pipe(map(([a, b]) => a + b));
result$.subscribe(console.log);
Here it is the codesandbox link for the convenience:
Technically, I understand why it's happening. First sharedState$ update is propagated to stateA$ and therefore the whole result$ is recomputed. And next the same thing is happening with stateB$.
But both these updates are happening synchronously one immediately after another in a single event loop tick, which make sense to result in a single subscriber notification of result$ stream.
Finally, I came up applying debounceTime(0) operator on the result$ stream, which solved the problem. But I have a feeling that it's more a workaround than the actual solution:
// works, but I'm not sure whether it's the right way
const result$ = combineLatest([stateA$, stateB$]).pipe(
map(([a, b]) => a + b),
debounceTime(0)
);
What's the idiomatic way of handling such things?
PS. Obviously, for that specific use-case it can be solved differently, which will result in a single subscriber call, but my real use case is quite complicated and unfortunately the pieces being merged together into a final observable composed dynamically and can include the same observable inside.
This will solve the issue you mentioned:
import { zip } from "rxjs";
import { distinctUntilChanged } from "rxjs/operators";
const result$ = zip([stateA$, stateB$]).pipe(map(([a, b]) => a + b), distinctUntilChanged());
For the cases where we need to trigger when any observable emits:
Just adding an operator to your initial solution, so it will not emit multiple times on clicking the same point.
const result$ = combineLatest([stateA$, stateB$]).pipe(
map(([a, b]) => a + b),
debounceTime(0),
distinctUntilChanged()
);

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.

Replace Observable which already has subscribers with other Observable

I have an Observable which gets later gets "replace" with another Observable.
How can I swap the Observable without loosing my subscribers?
const source = NEVER
const source2 = interval(1000);
source.subscribe(x => console.log(x));
// source.switch(source2)
source.switch(source2) is obviously not a valid operation. But it demonstrates, what I'dlike to achieve.
Same example on StackBlitz:
https://stackblitz.com/edit/rxjs-76a7ew
What would I need to do after the subscribtion, so this code will start printing the numbers from interval?
so you want switch to source2
source.pipe(
switchMap(() => source2)
).subscribe(x => console.log(x)); // x here is source2
you could use mergeMap or concatMap as well but I would recommend to use switchMap in this case as it's going to cancel the previous emit

rxjs concatMap with array

There is Higher Order Observable observable
const obs1 = interval(1000).pipe(take(5));
const obs2 = interval(500).pipe(take(2));
const obs3 = interval(2000).pipe(take(1));
//emit three observables
const source = of(obs1, obs2, obs3);
How to use concatMap in order to as soon as the previous observable is completed, the next one immediately begins. After it will return result array of Observable.Need to use concatMap.
You'll need to use toArray() to collect all emissions from all source Observables.
source.pipe(
concatMap(observable => observable),
toArray(),
);

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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