这是我的解决方案:
def make_extendable(o):
"""
Return an object that can be extended via its __dict__
If it is a slot, the object type is copied and the object is pickled through
this new type, before returning it.
If there is already a __dict__, then the object is returned.
"""
if getattr(o, "__dict__", None) is not None:
return o
# Now for fun
# Don't take care of immutable types or constant for now
import copy
import copyreg
cls = o.__class__
new_cls = type(cls.__name__, (cls,), {"__module__": cls.__module__})
# Support only Python >= 3.4
pick = o.__reduce_ex__(4)
if pick[0] == cls:
# This is the case for datetime objects
pick = (new_cls, *pick[1:])
elif pick[0] in (copyreg.__newobj__, copyreg.__newobj_ex__):
# Now the second item in pick is (cls, )
# It should be rare though, it's only for slots
pick = (pick[0], (new_cls,), *pick[2:])
else:
return ValueError(f"Unable to extend {o} of type {type(o)}")
# Build new type
return copy._reconstruct(o, None, *pick)
它基本上做了以下事情:
- 测试对象是否已经有
__dict__。在这种情况下,没有什么可做的。
- 根据提供的对象类型创建一个新类型。这个新类型不是槽类,尽量模仿基类。
- 如
copy.copy 中所做的那样减少提供的对象,但为简单起见仅支持__reduce_ex__(4)。
- 修改缩减版以使用新创建的类型。
- 使用修改后的简化版本取消腌制新对象。
datetime 的结果:
In [13]: d = make_extendable(datetime.datetime.now())
In [14]: d
Out[14]: datetime(2019, 3, 29, 11, 24, 23, 285875)
In [15]: d.__class__.__mro__
Out[15]: (datetime.datetime, datetime.datetime, datetime.date, object)
In [16]: d.__str__ = lambda: 'Hello, world'
In [17]: d.__str__()
Out[17]: 'Hello, world'
注意事项
随机顺序:
- 某些类型可能无法减少。
- 返回的对象是一个副本,而不是初始对象。
- 类不一样,但
isinstance(d, datetime.datetime) 将是True。
- 类层次结构将背叛黑客。
- 它可能非常慢。
-
__format__ 有点特殊,因为how format works需要修改类实例,而不是绑定方法。
- 。