asyncio 协程并发发送网络请求，如何优先继续处理返回结果的任务，而不是开始发送新的请求？

import asyncio import httpx request_time = 10 limits = httpx.Limits(max_cOnnections=5) client = httpx.AsyncClient(limits=limits) async def request_baidu(i): # async with httpx.AsyncClient(limits=limits) as client: print(f"===> %d" % i) r = await client.get("http://www.baidu.com") # print(r.status_code) print(f"<=== %d" % i) # 如何保证优先执行此处代码，而不发送新的请求？ async def main(): request_list = [asyncio.create_task(request_baidu(i)) for i in range(request_time)] await asyncio.gather(*request_list) if __name__ == '__main__': asyncio.run(main()) """ ===> 0 ===> 1 ===> 2 ===> 3 ===> 4 ===> 5 ===> 6 ===> 7 ===> 8 ===> 9 <=== 3 <=== 4 <=== 0 # 可以看到发送完所有的网络请求，才开始处理返回的结果 <=== 1 <=== 2 <=== 5 <=== 6 <=== 7 <=== 8 <=== 9 期望的结果： ===> 0 ===> 1 <=== 0 # 在前面的请求放回结果的时候优先返回该处继续向下执行，而不是发送新的请求 ===> 2 ===> 3 <=== 1 <=== 2 ===> 4 ... """