一次安排所有 100 万个任务
这是您正在谈论的代码。它最多需要 3 GB RAM,因此如果您的可用内存不足,它很容易被操作系统终止。
import asyncio
from aiohttp import ClientSession
MAX_SIM_CONNS = 50
LAST_ID = 10**6
async def fetch(url, session):
async with session.get(url) as response:
return await response.read()
async def bound_fetch(sem, url, session):
async with sem:
await fetch(url, session)
async def fetch_all():
url = "http://localhost:8080/?id={}"
tasks = set()
async with ClientSession() as session:
sem = asyncio.Semaphore(MAX_SIM_CONNS)
for i in range(1, LAST_ID + 1):
task = asyncio.create_task(bound_fetch(sem, url.format(i), session))
tasks.add(task)
return await asyncio.gather(*tasks)
if __name__ == '__main__':
asyncio.run(fetch_all())
使用队列简化工作
这是我对如何使用asyncio.Queue将 URL 传递给工作任务的建议。队列按需要填充,没有预先制作的 URL 列表。
它只需要 30 MB RAM :)
import asyncio
from aiohttp import ClientSession
MAX_SIM_CONNS = 50
LAST_ID = 10**6
async def fetch(url, session):
async with session.get(url) as response:
return await response.read()
async def fetch_worker(url_queue):
async with ClientSession() as session:
while True:
url = await url_queue.get()
try:
if url is None:
# all work is done
return
response = await fetch(url, session)
# ...do something with the response
finally:
url_queue.task_done()
# calling task_done() is necessary for the url_queue.join() to work correctly
async def fetch_all():
url = "http://localhost:8080/?id={}"
url_queue = asyncio.Queue(maxsize=100)
worker_tasks = []
for i in range(MAX_SIM_CONNS):
wt = asyncio.create_task(fetch_worker(url_queue))
worker_tasks.append(wt)
for i in range(1, LAST_ID + 1):
await url_queue.put(url.format(i))
for i in range(MAX_SIM_CONNS):
# tell the workers that the work is done
await url_queue.put(None)
await url_queue.join()
await asyncio.gather(*worker_tasks)
if __name__ == '__main__':
asyncio.run(fetch_all())