Python 异步

1 看起来是有点异步的意思，实际上还是一个多线程

from threading import Thread
import time

def async(func):
    def wrapper(*args,**kwargs):
        t = Thread(target=func,args=args,kwargs=kwargs)
        t.start()
    return wrapper

@async
def tests():
    time.sleep(5)
    print('ssss')

if __name__ == '__main__':
    tests()

2 回调

import asyncio

async def do_some_work(x):
  print('Waiting '+str(x))
  await asyncio.sleep(x)

def done_callback(futu):
  print('Done')

def huidiao():
  """ 回调 """
  loop = asyncio.get_event_loop()
  futu = asyncio.ensure_future(do_some_work(3))
  futu.add_done_callback(done_callback)
  loop.run_until_complete(futu)

huidiao()

3 单个协程

import asyncio

async def do_some_work(x):
  print('Waiting '+str(x))
  await asyncio.sleep(x)

if __name__ == '__main__':
  loop = asyncio.get_event_loop()
  loop.run_until_complete(do_some_work(1))
  loop.close()

4 async / await 多个协程

import asyncio


async def do_some_work(x):
  print('Waiting '+str(x))
  await asyncio.sleep(x)

def many(size):
  """ 多个协程 """
  coros = [do_some_work(3) for i in range(size)]
  # loop.run_until_complete(asyncio.wait(coros))
  loop.run_until_complete(asyncio.gather(*coros))

if __name__ == '__main__':
  many(10000)  # 执行10000次
  # 如果使用循环执行 do_some_work 函数,将耗费8个小时
  # 如果使用多线程，由于线程的限制，一台32位的电脑也只能开启2000个以内的线程
  # 而此程序仅仅执行了3秒钟