asynchronous python itertools chain multiple generators

Question

UPDATED QUESTION FOR CLARITY:

suppose I have 2 processing generator functions:

def gen1(): # just for examples,
  yield 1   # yields actually carry 
  yield 2   # different computation weight 
  yield 3   # in my case

def gen2():
  yield 4
  yield 5
  yield 6

I can chain them with itertools

from itertools import chain

mix = chain(gen1(), gen2())

and then I can create another generator function object with it,

def mix_yield():
   for item in mix:
      yield item

or simply if I just want to next(mix), it's there.

My question is, how can I do the equivalent in asynchronous code?

Because I need it to:

• return in yield (one by one), or with next iterator
• the fastest resolved yield first (async)

PREV. UPDATE:

After experimenting and researching, I found aiostream library which states as async version of itertools, so what I did:

import asyncio
from aiostream import stream

async def gen1(): 
     await asyncio.sleep(0) 
     yield 1 
     await asyncio.sleep(0) 
     yield 2 
     await asyncio.sleep(0) 
     yield 3 

async def gen2(): 
     await asyncio.sleep(0) 
     yield 4 
     await asyncio.sleep(0) 
     yield 5 
     await asyncio.sleep(0) 
     yield 6 

a_mix = stream.combine.merge(gen1(),gen2())

async def a_mix_yield():
   for item in a_mix:
      yield item

but I still can't do next(a_mix)

TypeError: 'merge' object is not an iterator

or next(await a_mix)

raise StreamEmpty()

Although I still can make it into a list:

print(await stream.list(a_mix))
# [1, 2, 4, 3, 5, 6]

so one goal is completed, one more to go:

• return in yield (one by one), or with next iterator

  - the fastest resolved yield first (async)

Answer 1

Python's next built-in function is just a convenient way of invoking the underlying __next__ method on the object. The async equivalent of __next__ is the __anext__ method on the async iterator. There is no anext global function in the standard library (the aiostream library provides one), but one could easily write it:

async def anext(aiterator):
    return await aiterator.__anext__()

But the savings is so small that, in rare situations when this is needed, one may as well invoke __anext__ directly. The async iterator is in turn obtained from an async iterable by calling the __aiter__ (in analogy to __iter__ provided by regular iterables). Async iteration driven manually looks like this:

a_iterator = obj.__aiter__()          # regular method
elem1 = await a_iterator.__anext__()  # async method
elem2 = await a_iterator.__anext__()  # async method
...

__anext__ will raise StopAsyncIteration when no more elements are available. To loop over async iterators one should use async for.

Here is a runnable example, based on your code, using both __anext__ and async for to exhaust the stream set up with aiostream.stream.combine.merge:

async def main():
    a_mix = stream.combine.merge(gen1(), gen2())
    async with a_mix.stream() as streamer:
        mix_iter = streamer.__aiter__()    
        print(await mix_iter.__anext__())
        print(await mix_iter.__anext__())
        print('remaining:')
        async for x in mix_iter:
            print(x)

asyncio.get_event_loop().run_until_complete(main())

Answer 2

I came across this answer and I looked at the aiostream library. Here is the code I came up with to merge multiple async generators. It does not use any library.

async def merge_generators(gens:Set[AsyncGenerator[Any, None]]) -> AsyncGenerator[Any, None]:
    pending = gens.copy()
    pending_tasks = { asyncio.ensure_future(g.__anext__()): g for g in pending }
    while len(pending_tasks) > 0:
        done, _ = await asyncio.wait(pending_tasks.keys(), return_when="FIRST_COMPLETED")
        for d in done:
            try:
                result = d.result()
                yield result
                dg = pending_tasks[d]
                pending_tasks[asyncio.ensure_future(dg.__anext__())] = dg
            except StopAsyncIteration as sai:
                print("Exception in getting result", sai)
            finally:
                del pending_tasks[d]

Hope this helps you and let me know if there are any bugs in this.