python - 使类常量的类型是它所在的类

Question

我有一个具有特殊值“EMPTY”和“UNIVERSE”的 Python 类:

class RealSet:
    """Continuous open, half-open, and closed regions and discreet values of the Reals"""

    # implementation placeholder
    def __init__(self, intervals, *, canonicalize):
        pass

# Outside the class

RealSet.EMPTY = RealSet(tuple(), canonicalize=False)  # type: ignore
RealSet.UNIVERSE = RealSet(((None, None),), canonicalize=False)  # type: ignore

但是，linting、代码完成等不喜欢这样，因为它们不被视为类的静态属性。即使设置它们也会报告为 mypy 错误，因此 # type:ignore。

下面的代码不起作用，因为我无法在类范围内构造 RealSet，因为它还不存在:

class RealSet:
    """Continuous open, half-open, and closed regions and discreet values of the Reals"""
    ...
    ...

    EMPTY = RealSet(tuple(), canonicalize=False)  # error
    UNIVERSE = RealSet(((None, None),), canonicalize=False)  # error

这不起作用，因为它定义了实例属性，而不是类属性:

class RealSet:
    """Continuous open, half-open, and closed regions and discreet values of the Reals"""
    ...
    ...

    EMPTY: "RealSet"
    UNIVERSE: "RealSet"

# Outside the class

RealSet.EMPTY = RealSet(tuple(), canonicalize=False)
RealSet.UNIVERSE = RealSet(((None, None),), canonicalize=False)

这似乎是 Python 类设计中的一个极端情况。如何创建类属性，其中属性的类型是它所在的类？奖励:使它们保持不变。

Answer 1

您可以使用typing.ClassVar注释类变量:

class RealSet:
    def __init__(self, intervals, *, canonicalize):
        pass

    EMPTY: ClassVar['RealSet']
    UNIVERSE: ClassVar['RealSet']


RealSet.EMPTY = RealSet(tuple(), canonicalize=False)
RealSet.UNIVERSE = RealSet(((None, None),), canonicalize=False)