Combine a fixed number of mergeMapped observables - RxJS

I'm using RxJS v6, but this question applies to v5 as well.

When using mergeMap, my original array disappears and while I can do a number of operations in parallel, I no longer have a way of monitoring when all those observables I sent into mergeMap are complete.

mergeMap

mergeMap

Example

of([1, 2, 3, 4])
.pipe(
mergeMap(values => values),
)
.subscribe(console.log)

// 1
// 2
// 3
// 4


I'd like to see:

// [1, 2, 3, 4]


The only way I've come up with so far requires getting the length of the array, but I'm sure there's gotta be some operator I'm missing:

of([1, 2, 3, 4])
.pipe(
switchMap(values => (
of(values)
.pipe(
mergeMap(value => value),
bufferCount(values.length),
)
))
)
.subscribe(console.log)


toArray

forkJoin

1 Answer
1

When using mergeMap, my original array disappears

The reason is that mergeMap accepts an ObservableInput as parameter of the function you pass in. A javascript Array is an ObservableInput and therefore works in mergeMap and mergeMap does its job, which is to flatten the ObservableInput (consider that mergeMap was previously called flatMap).

mergeMap

ObservableInput

Array

ObservableInput

mergeMap

mergeMap

ObservableInput

mergeMap

flatMap

So, as @cartant says, if you want to get back to an array, you have to use toArray operator.
In other words

toArray

of([1, 2, 3, 4])
.pipe(
mergeMap(value => // do stuff with a value of an array),
toArray()
)
.subscribe(console.log)


is equivalent to

of([1, 2, 3, 4])
.pipe(
map(values => values.map(value => // do stuff with a value of an array)),
)


If your array though contains Observables and you want to eventually get the values they notify when all of them emit, than you have to use forkJoin. This is a simple example

forkJoin

of([1, 2, 3, 4].map(n => of(n)))
.pipe(
switchMap(observablesOfValues => forkJoin(observablesOfValues))
)
.subscribe(console.log)


toArray

Thanks for contributing an answer to Stack Overflow!

But avoid …

To learn more, see our tips on writing great answers.

Required, but never shown

Required, but never shown

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.