使用defaultdict调试python异步竞争条件

Question

我正在尝试编写一些东西，以将在一定时间范围内发送的一组输入整理为单个请求。我的代码类似于：

True

但是，似乎在in_dict = defaultdict(list) out_dict = defaultdict(list) guard = asyncio.Lock() async def collate(k, v): async with guard: in_dict[k].append(v) await asyncio.sleep(wait_time) async with guard: if k in in_dict and in_dict[k]: out_dict[k] = out_dict[k] + in_dict.pop(k) elif k in out_dict: handle_collated(out_dict.pop(k)) 仍包含已经处理过的元素的地方存在竞争状况。

谢谢

Answer 1

...out_dict[k] still contains elements that have been processed.-您的示例意图不明确，因此无法找到解决方案。从表面上看，这似乎是一个逻辑问题。

撇开异步：

>>> in_dict = collections.defaultdict(list)
>>> out_dict = collections.defaultdict(list)

代码始终首先在k:v中使它们成为in_dict项

>>> k, v = 1, ''
>>> in_dict[k].append(v)

因此if k in in_dict and in_dict[k]将始终为True，elif套件将永远不会执行。

>>> k in in_dict
True
>>> bool(in_dict[k])
True
>>>
>>> if k in in_dict and in_dict[k]:
...     out_dict[k] = out_dict[k] + in_dict.pop(k)
... elif k in out_dict:
...     print(f'elif executed: {out_dict.pop(k)}')

>>> in_dict
defaultdict(<class 'list'>, {})
>>> out_dict
defaultdict(<class 'list'>, {1: ['']})

下一个通话

>>> k,v = 1,'q'
>>> in_dict[k].append(v)
>>>
>>> if k in in_dict and in_dict[k]:
...     out_dict[k] = out_dict[k] + in_dict.pop(k)
... elif k in out_dict:
...     print(f'elif executed: {out_dict.pop(k)}')
>>>
>>> in_dict
defaultdict(<class 'list'>, {})
>>> out_dict
defaultdict(<class 'list'>, {1: ['', 'q']})
>>>

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The if statement