一些python标准类是插槽,例如datetime.datetime
。这不是我可以更改的东西,很多库都希望使用datetime
对象。
我想更改现有__format__
对象的默认datetime
方法,但是很遗憾,由于这是插槽类,因此被禁止:
In [10]: import datetime
In [11]: datetime.datetime.now().__format__ = lambda s, f: ''
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-11-c98141136d9d> in <module>()
----> 1 datetime.datetime.now().__format__ = lambda s, f: ''
AttributeError: 'datetime.datetime' object attribute '__format__' is read-only
是否有可能滥用python的动态特性来实现这一目标?我想是这样。
答案 0 :(得分:0)
这是我的解决方法:
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)
它基本上执行以下操作:
__dict__
。在这种情况下,无需执行任何操作。copy.copy
一样减少提供的对象,但为简单起见仅支持__reduce_ex__(4)
。 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'
以随机顺序:
isinstance(d, datetime.datetime)
将是True
。__format__
有点特殊,因为由于how format works,您需要更改类实例,而不是绑定方法。