有没有办法在Python中创建类属性？

发布于 2024-08-19 23:19:15 字数 452 浏览 3 评论 0原文

由于某种原因，以下内容不起作用：

>>> class foo(object):
...     @property
...     @classmethod
...     def bar(cls):
...             return "asdf"
... 
>>> foo.bar
<property object at 0x1da8d0>
>>> foo.bar + '\n'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'property' and 'str'

有没有办法做到这一点，或者是我诉诸某种元类技巧的唯一选择？

原文

The following doesn't work for some reason:

>>> class foo(object):
...     @property
...     @classmethod
...     def bar(cls):
...             return "asdf"
... 
>>> foo.bar
<property object at 0x1da8d0>
>>> foo.bar + '\n'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'property' and 'str'

Is there a way to do this, or is my only alternative to resort to some kind of metaclass trickery?

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何以心动 2024-08-26 23:19:15

如果您希望在从对象 X 获取属性时触发描述符 property，则必须将该描述符放入 type(X) 中。因此，如果 X 是一个类，则描述符必须位于该类的类型中，也称为该类的元类——不涉及“欺骗”，这只是一个完全通用的规则问题。

或者，您可以编写自己的专用描述符。请参阅此处，了解有关描述符的出色“操作方法”条约。编辑例如：

class classprop(object):
  def __init__(self, f):
    self.f = classmethod(f)
  def __get__(self, *a):
    return self.f.__get__(*a)()

class buh(object):
  @classprop
  def bah(cls): return 23

print buh.bah

根据需要发出23。

If you want the descriptor property to trigger when you get an attribute from object X, then you must put the descriptor in type(X). So if X is a class, the descriptor must go in the class's type, also known as the class's metaclass -- no "trickery" involved, it's just a matter of completely general rules.

Alternatively, you might write your own special-purpose descriptor. See here for an excellent "how-to" treaty on descriptors. Edit for example:

class classprop(object):
  def __init__(self, f):
    self.f = classmethod(f)
  def __get__(self, *a):
    return self.f.__get__(*a)()

class buh(object):
  @classprop
  def bah(cls): return 23

print buh.bah

emits 23, as desired.

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