Question

我尝试编写一个包装asyncio.coroutine的装饰器函数，并返回完成所需的时间。下面的配方包含正常工作的代码。我唯一的问题是，尽管使用了@functools.wraps，但我还是失去了装饰函数的名称。如何保留原始协程的名称？我检查了asyncio.

的来源

import asyncio
import functools
import random
import time

MULTIPLIER = 5

def time_resulted(coro):
    @functools.wraps(coro)
    @asyncio.coroutine
    def wrapper(*args, **kargs):
        time_before = time.time()
        result = yield from coro(*args, **kargs)
        if result is not None:
            raise TypeError('time resulted coroutine can '
                'only return None')
        return time_before, time.time()
    print('= wrapper.__name__: {!r} ='.format(wrapper.__name__))
    return wrapper

@time_resulted
@asyncio.coroutine
def random_sleep():
    sleep_time = random.random() * MULTIPLIER
    print('{} -> {}'.format(time.time(), sleep_time))
    yield from asyncio.sleep(sleep_time)

if __name__ == '__main__':
    loop = asyncio.get_event_loop()
    tasks = [asyncio.Task(random_sleep()) for i in range(5)]
    loop.run_until_complete(asyncio.wait(tasks))
    loop.close()
    for task in tasks:
        print(task, task.result()[1] - task.result()[0])
    print('= random_sleep.__name__: {!r} ='.format(
        random_sleep.__name__))
    print('= random_sleep().__name__: {!r} ='.format(
        random_sleep().__name__))

结果：

= wrapper.__name__: 'random_sleep' =
1397226479.00875 -> 4.261069174838891
1397226479.00875 -> 0.6596335046471768
1397226479.00875 -> 3.83421163259601
1397226479.00875 -> 2.5514027672929713
1397226479.00875 -> 4.497471439365472
Task(<wrapper>)<result=(1397226479.00875, 1397226483.274884)> 4.266134023666382
Task(<wrapper>)<result=(1397226479.00875, 1397226479.6697)> 0.6609499454498291
Task(<wrapper>)<result=(1397226479.00875, 1397226482.844265)> 3.835515022277832
Task(<wrapper>)<result=(1397226479.00875, 1397226481.562422)> 2.5536720752716064
Task(<wrapper>)<result=(1397226479.00875, 1397226483.51523)> 4.506479978561401
= random_sleep.__name__: 'random_sleep' =
= random_sleep().__name__: 'wrapper' =

如您所见random_sleep()返回具有不同名称的生成器对象。我想保留装饰的协程的名称。我不知道这是否是asyncio.coroutines特有的问题。我也尝试了不同装饰器命令的代码，但都有相同的结果。如果我对@functools.wraps(coro)发表评论，那么即使random_sleep.__name__也会像我预期的那样变成wrapper。

编辑：我已将此问题发布到Python问题跟踪器，并收到了R. David Murray的以下答案：“我认为这是一个特殊情况，需要更好地改进'包装'，这是在python上讨论的 - 不久前的开发。“

Answer 1

问题是functools.wraps仅更改wrapper.__name__和wrapper().__name__仅保留wrapper。 __name__是一个只读生成器属性。您可以使用exec设置适当的名称：

import asyncio
import functools
import uuid
from textwrap import dedent

def wrap_coroutine(coro, name_prefix='__' + uuid.uuid4().hex):
    """Like functools.wraps but preserves coroutine names."""
    # attribute __name__ is not writable for a generator, set it dynamically
    namespace = {
        # use name_prefix to avoid an accidental name conflict
        name_prefix + 'coro': coro,
        name_prefix + 'functools': functools,
        name_prefix + 'asyncio': asyncio,
    }
    exec(dedent('''
        def {0}decorator({0}wrapper_coro):
            @{0}functools.wraps({0}coro)
            @{0}asyncio.coroutine
            def {wrapper_name}(*{0}args, **{0}kwargs):
                {0}result = yield from {0}wrapper_coro(*{0}args, **{0}kwargs)
                return {0}result
            return {wrapper_name}
        ''').format(name_prefix, wrapper_name=coro.__name__), namespace)
    return namespace[name_prefix + 'decorator']

用法：

def time_resulted(coro):
    @wrap_coroutine(coro)
    def wrapper(*args, **kargs):
        # ...
    return wrapper

虽然有效，但可能比使用exec()更好。

Answer 2

自从提出这个问题以来，更改协程的名称成为可能。这是通过设置 SELECT（不是 __qualname__）来完成的：

__name__

async def my_coro(): pass c = my_coro() print(repr(c)) # <coroutine object my_coro at 0x7ff8a7d52bc0> c.__qualname__ = 'flimflam' print(repr(c)) # <coroutine object flimflam at 0x7ff8a7d52bc0> import asyncio print(repr(asyncio.ensure_future(c))) # <Task pending name='Task-737' coro=<flimflam() running at <ipython-input>:1>> 在协程对象的 __qualname__ 中的用法定义为 in the CPython source

如何装饰asyncio.coroutine以保留其name？

2 个答案:

如何装饰asyncio.coroutine以保留其__name__？

2 个答案:

如何装饰asyncio.coroutine以保留其name？