Is there a performant way to transform `List<Fl...

# coroutines

[Marc Knaup]

Is there a performant way to transform List<Flow<T>> into Flow<List<T>> where List<T> contains the latest value of each Flow<T> at the same index. combine() is not an option.

[Marc Knaup]

I’ve written a simple one, in case someone needs it:
https://gist.github.com/fluidsonic/4330ebda9942c57a6660fadba1145362

[MiSikora]

I’m not sure why comine is not an option. I don’t know about performance but I recently had to do this and I used something like this.

fun <T> Iterable<Flow<T>>.unwrap(): Flow<List<T>> {
  val emptyFlow = emptyFlow<List<T>>()
  return fold(emptyFlow) { flows, flow ->
    if (flows === emptyFlow) {
      flow.map(::listOf)
    } else {
      flows.combine(flow) { group, item -> group + item }
    }
  }
}

[Marc Knaup]

Because the parallel combine is slow as heck: https://github.com/Kotlin/kotlinx.coroutines/issues/2296

[MiSikora]

Ok, thanks for the link 🙂

[bezrukov]

it's slow because under the hood it uses "fair channels" that calls yield each time when element offered with fast path. As far as I understand it's necessary for sequential flows.

[bezrukov]

when you write
listOf(flow1,flow2,flow3).combine() you can expect that it will use order of these flows. With your extension you can't expect this. It may lead to the following behavior:

time:   | 1 | 2 | 3 | 4 | 5
flow1:  | A |   | B |   | 
flow2:  | w | x | y | z | 
combine:| Aw| Ax| Ay| Az| Bz| <-your extension
stdlib: | Aw| Ax| Bx| By| Bz|

[Marc Knaup]

What would such a time unit be in Kotlin? There is no time frame defined in which values from multiple flows would be merged 🤔
Or is it elements that happen to be collectible in the same “tick”?

[Marc Knaup]

Or expressed differently: How is it defined that “B” and “y” arrive at the same time?

[bezrukov]

Or is it elements that happen to be collectible in the same “tick”?

yes, it's just shows that at some time called "3" B and y are ready for collect

[Marc Knaup]

Okay. So if that’s really necessary then it may be better to have two operators, one which is fast and doesn’t care about that, and one that is slow and maintains order if elements arrive “simultaneously”.
And clearly document each 🙂