Python 中的 itertools.count 使用什么类型？

Question

我试图itertool.count在 Python 中指定对象的类型，如下所示：

from itertools import count

c: count = count()

然而，运行mypy会出现以下错误：

test.py:3: error: Function "itertools.count" is not valid as a type
test.py:3: note: Perhaps you need "Callable[...]" or a callback protocol?
Found 1 error in 1 file (checked 1 source file)

这似乎是由于其itertools.count行为类似于函数而引起的。但是，它返回一个itertools.count对象，如下所示

test.py:3: error: Function "itertools.count" is not valid as a type
test.py:3: note: Perhaps you need "Callable[...]" or a callback protocol?
Found 1 error in 1 file (checked 1 source file)

那么，我应该如何指定结果的类型呢count()？

Answer 1

中有如下注释itertools.pyi：

_N = TypeVar('_N', int, float)

def count(start: _N = ...,
          step: _N = ...) -> Iterator[_N]: ...  # more general types?

因此，在您的代码中，您可以这样做：

_N = TypeVar('_N', int, float)

def count(start: _N = ...,
          step: _N = ...) -> Iterator[_N]: ...  # more general types?