Question

我的程序执行以下操作：

获取txt文件的文件夹
：
- 阅读文件
- 使用文件内容对localhost中的API执行POST请求
- 解析XML响应（不在下面的示例中）

我关注程序同步版本的性能，所以尝试使用aiohttp使其异步（这是我在Python中除了Scrapy之外的第一次异步编程尝试）。原来，异步代码花了2倍的时间，我不明白为什么。

同步代码（152秒）

url = "http://localhost:6090/api/analyzexml"
package = #name of the package I send in each requests
with open("template.txt", "r", encoding="utf-8") as f:
    template = f.read()

articles_path = #location of my text files

def fetch(session, url, article_text):
    data = {"package": package, "data": template.format(article_text)}
    response = session.post(url, data=json.dumps(data))
    print(response.text)

files = glob(os.path.join(articles_path, "*.txt"))

with requests.Session() as s:
    for file in files:
        with open(file, "r", encoding="utf-8") as f:
                article_text = f.read()
        fetch(s, url, article_text)

分析结果：

+--------+---------+----------+---------+----------+-------------------------------------------------------+
| ncalls | tottime | percall  | cumtime | percall  |               filename:lineno(function)               |
+--------+---------+----------+---------+----------+-------------------------------------------------------+
|    849 |   145.6 |   0.1715 |   145.6 |   0.1715 | ~:0(<method 'recv_into' of '_socket.socket' objects>) |
|      2 |   1.001 |   0.5007 |   1.001 |   0.5007 | ~:0(<method 'connect' of '_socket.socket' objects>)   |
|    365 |   0.772 | 0.002115 |   1.001 | 0.002742 | ~:0(<built-in method builtins.print>)                 |
+--------+---------+----------+---------+----------+-------------------------------------------------------+

（WANNABE）异步代码（327秒）

async def fetch(session, url, article_text):
    data = {"package": package, "data": template.format(article_text)}
    async with session.post(url, data=json.dumps(data)) as response:
        return await response.text()

async def process_files(articles_path):
    tasks = []

    async with ClientSession() as session:
        files = glob(os.path.join(articles_path, "*.txt"))
        for file in files:
            with open(file, "r", encoding="utf-8") as f:
                article_text = f.read()
            task = asyncio.ensure_future(fetch(session=session, 
                                        url=url, 
                                        article_text=article_text
                                        ))
            tasks.append(task)
            responses = await asyncio.gather(*tasks)
            print(responses)


loop = asyncio.get_event_loop()
future = asyncio.ensure_future(process_files(articles_path))
loop.run_until_complete(future)

分析结果：

 +--------+---------+---------+---------+---------+-----------------------------------------------+
    | ncalls | tottime | percall | cumtime | percall |           filename:lineno(function)           |
    +--------+---------+---------+---------+---------+-----------------------------------------------+
    |   2278 |     156 | 0.06849 |     156 | 0.06849 | ~:0(<built-in method select.select>)          |
    |    365 |   128.3 |  0.3516 |   168.9 |  0.4626 | ~:0(<built-in method builtins.print>)         |
    |    730 |   40.54 | 0.05553 |   40.54 | 0.05553 | ~:0(<built-in method _codecs.charmap_encode>) |
    +--------+---------+---------+---------+---------+-----------------------------------------------+

我显然在这个概念中遗漏了一些东西。有人也可以帮助我理解为什么在异步版本中打印需要花费这么多时间（参见分析）。

Answer 1

因为它不是异步的：）

查看您的代码：您为每个文件执行responses = await asyncio.gather(*tasks) ，因此您每次都支付协同处理的所有代价时，基本上都会同步运行。

我认为它只是一个缩进错误;如果您取消responses = await asyncio.gather(*tasks)以使其超过for file in files循环，您将真正并行启动tasks。

异步慢于同步

1 个答案: