来自Python 3异步for循环的TypeError

问题描述

我正在学习Python的相对较新的异步功能.我在 PEP 492 中找到了它:

I'm learning about Python's relatively new async features. I found this in PEP 492:

class AsyncIteratorWrapper:
    def __init__(self, obj):
        self._it = iter(obj)

    def __aiter__(self):
        return self

    async def __anext__(self):
        try:
            value = next(self._it)
        except StopIteration:
            raise StopAsyncIteration
        return value

async for letter in AsyncIteratorWrapper("abc"):
    print(letter)

我试图运行此代码，方法是将给定的 async for 循环添加到函数中，然后使用事件循环进行调用.

I attempted to run this code, by adding the given async for loop to a function, and then calling that using an event loop.

完整的示例代码(在解释器中运行):

Full example code (run in the interpreter):

class AsyncIteratorWrapper:
    def __init__(self, obj):
        self._it = iter(obj)
    def __aiter__(self):
        return self
    async def __anext__(self):
        try:
            value = next(self._it)
        except StopIteration:
            raise StopAsyncIteration
        return value

async def aprint(str):
  async for letter in AsyncIteratorWrapper(str):
    print(letter)

import asyncio
loop = asyncio.get_event_loop()
co = aprint("abcde")
loop.run_until_complete(co)

但是，我遇到一个错误:

However, I'm getting an error:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/rh/rh-python35/root/usr/lib64/python3.5/asyncio/base_events.py", line 337, in run_until_complete
    return future.result()
  File "/opt/rh/rh-python35/root/usr/lib64/python3.5/asyncio/futures.py", line 274, in result
    raise self._exception
  File "/opt/rh/rh-python35/root/usr/lib64/python3.5/asyncio/tasks.py", line 239, in _step
    result = coro.send(None)
  File "<stdin>", line 2, in aprint
TypeError: 'async for' received an invalid object from __aiter__: AsyncIteratorWrapper

我做错了什么?这个例子怎么解决?PEP之外的代码失败了，我感到有些惊讶.

What am I doing wrong? How can this example be fixed? I'm a little surprised that code right out of the PEP is failing.

我正在使用python 3.5.1版本.

I'm using python version 3.5.1.

推荐答案

您正在使用的代码适用于python 3.5.2 +.

The code you are using works with python 3.5.2+.

从Python 3.5.2开始， __ aiter __ 可以直接返回异步迭代器.在此处更多

From Python 3.5.2 __aiter__ can directly return asynchronous iterators. More here

您收到的错误是由于使用较旧的python(3.5.1)，因此返回错误的类型.

The error you were receiving was because of the older python(3.5.1) and it was therefore returning the wrong type.

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