Is there a performant way to transform `List lt Flow lt T gt

Question

Is there a performant way to transform `List lt Flow lt T gt gt ` into `Flow lt List lt T gt gt ` where `List lt T gt ` contains the latest value of each `Flow lt T gt ` at the same index `combine ` is not an option

Marc Knaup · Answer

I ve written a simple one in case someone needs it <https gist github com fluidsonic 4330ebda9942c57a6660fadba1145362>

MiSikora · Answer

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 lt T gt Iterable lt Flow lt T gt gt unwrap Flow lt List lt T gt gt val emptyFlow = emptyFlow lt List lt T gt gt return fold emptyFlow flows flow gt if flows === emptyFlow flow map listOf else flows combine flow group item gt group + item ```

Marc Knaup · Answer

Because the parallel combine is slow as heck <https github com Kotlin kotlinx coroutines issues 2296>

MiSikora · Answer

Ok thanks for the link slightly smiling face

bezrukov · Answer

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 · Answer

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| lt your extension stdlib | Aw| Ax| Bx| By| Bz|```

Marc Knaup · Answer

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

Marc Knaup · Answer

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

bezrukov · Answer

gt 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 · Answer

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 slightly smiling face