在 Python 中将信号量与 asyncio 结合使用
Using a semaphore with asyncio in Python
我正在尝试使用信号量限制同步异步函数的数量 运行ning,但我无法让它工作。我的代码归结为:
import asyncio
async def send(i):
print(f"starting {i}")
await asyncio.sleep(4)
print(f"ending {i}")
async def helper():
async with asyncio.Semaphore(value=5):
await asyncio.gather(*[
send(1),
send(2),
send(3),
send(4),
send(5),
send(6),
send(7),
send(8),
send(9),
send(10),
])
if __name__ == "__main__":
loop = asyncio.get_event_loop()
loop.run_until_complete(helper())
loop.close()
输出为:
starting 1
starting 2
starting 3
starting 4
starting 5
starting 6
starting 7
starting 8
starting 9
starting 10
ending 1
ending 2
ending 3
ending 4
ending 5
ending 6
ending 7
ending 8
ending 9
ending 10
我希望并期望一次只有 5 个 运行,但是所有 10 个都同时启动和停止。我做错了什么?
请在下面找到工作示例,随时提问:
import asyncio
async def send(i: int, semaphore: asyncio.Semaphore):
# to demonstrate that all tasks start nearly together
print(f"Hello: {i}")
# only two tasks can run code inside the block below simultaneously
async with semaphore:
print(f"starting {i}")
await asyncio.sleep(4)
print(f"ending {i}")
async def async_main():
s = asyncio.Semaphore(value=2)
await asyncio.gather(*[send(i, semaphore=s) for i in range(1, 11)])
if __name__ == "__main__":
loop = asyncio.get_event_loop()
loop.run_until_complete(async_main())
loop.close()
我正在尝试使用信号量限制同步异步函数的数量 运行ning,但我无法让它工作。我的代码归结为:
import asyncio
async def send(i):
print(f"starting {i}")
await asyncio.sleep(4)
print(f"ending {i}")
async def helper():
async with asyncio.Semaphore(value=5):
await asyncio.gather(*[
send(1),
send(2),
send(3),
send(4),
send(5),
send(6),
send(7),
send(8),
send(9),
send(10),
])
if __name__ == "__main__":
loop = asyncio.get_event_loop()
loop.run_until_complete(helper())
loop.close()
输出为:
starting 1
starting 2
starting 3
starting 4
starting 5
starting 6
starting 7
starting 8
starting 9
starting 10
ending 1
ending 2
ending 3
ending 4
ending 5
ending 6
ending 7
ending 8
ending 9
ending 10
我希望并期望一次只有 5 个 运行,但是所有 10 个都同时启动和停止。我做错了什么?
请在下面找到工作示例,随时提问:
import asyncio
async def send(i: int, semaphore: asyncio.Semaphore):
# to demonstrate that all tasks start nearly together
print(f"Hello: {i}")
# only two tasks can run code inside the block below simultaneously
async with semaphore:
print(f"starting {i}")
await asyncio.sleep(4)
print(f"ending {i}")
async def async_main():
s = asyncio.Semaphore(value=2)
await asyncio.gather(*[send(i, semaphore=s) for i in range(1, 11)])
if __name__ == "__main__":
loop = asyncio.get_event_loop()
loop.run_until_complete(async_main())
loop.close()