How can I make an observable based on 2 others, but emit only when the 2 observables emit new values

[MiniSuperDev]

How can I make an observable based on 2 others observables, but emit only when the 2 observables emit new values (regardless of whether it is equal to the previous value emitted), similar to combineLatest, but I want it to emit only when the 2 observables emit new values

input:

import { of, delay, concatMap } from 'rxjs';

const observable1$ = of('a', 'a', 'b').pipe(
  concatMap((v) => of(v).pipe(delay(100)))
);
const observable2$ = of(1, 2, 3, 4).pipe(
  concatMap((v) => of(v).pipe(delay(100)))
);

That is:

observable1$:  'a'--'a'--'b'-----
observable2$:  '1'--'2'--'3'--'4'

Expexted output:

["a", "1"]
["a", "2"]
["b", "3"]

With combineLatestWith i got other output

import { combineLatestWith, of, delay, concatMap } from 'rxjs';

const observable1$ = of('a', 'a', 'b').pipe(
  concatMap((v) => of(v).pipe(delay(100)))
);
const observable2$ = of('1', '2', '3', '4').pipe(
  concatMap((v) => of(v).pipe(delay(100)))
);

observable1$.pipe(combineLatestWith(observable2$)).subscribe(console.log);

["a", "1"]
["a", "1"]
["a", "2"]
["b", "2"]
["b", "3"]
["b", "4"]

[XuankangLin]

Pipe each one using “distinctUntilChanged” then “combineLatest”?

[MiniSuperDev]

@XuankangLin Thanks, but it doesn't work for my case, I already updated the question, because initially I didn't make myself understood, thanks

[XuankangLin]

From your updated example, maybe you are looking for withLatestFrom?

[josepot]

Hi @MiniSuperDev !

If I understood the question correctly, then I think that the operator that you are looking for is zip:

import { zip, of, delay, concatMap } from 'rxjs';

const observable1$ = of('a', 'a', 'b').pipe(
  concatMap((v) => of(v).pipe(delay(100)))
);
const observable2$ = of('1', '2', '3', '4').pipe(
  concatMap((v) => of(v).pipe(delay(100)))
);

zip(observabl1, observable2).subscribe(console.log);

Just be cautious of backpressure.

[MiniSuperDev]

Thanks @josepot, it works as expected.

What do you mean by "backpressure"?

[josepot]

Thanks @josepot, it works as expected.

🎉 if that's the case, then could you please mark the answer as "accepted"?

What do you mean by "backpressure"?

Check this out