[英]Redefine Method of an Object
I've got a class, where a method should only run once.我有一个类,其中一个方法应该只运行一次。 Of course, it could easily be done with artificial
has_executed = True/False
flag, but why use it, if you can just delete the method itself?当然,它可以很容易地用人工
has_executed = True/False
标志来完成,但为什么要使用它,如果你可以删除方法本身呢? python
'sa duck-typed language, everything is a reference, bla-bla-bla , what can go wrong? python
是一种鸭式语言,一切都是参考, bla-bla-bla ,会出什么问题?
At least it was the thought.至少是这样的想法。 I couldn't actually do it:
我实际上无法做到:
class A:
def b(self):
print("empty")
self.__delattr__('b')
a = A()
a.b()
raises AttributeError: b
.引发
AttributeError: b
。 However, executing self.__getattribute__('b')
returns <bound method Ab of <__main__.A object at 0x000001CDC6742FD0>>
, which sounds stupid to me: why is a method
any different from an attribute
, since everything in python
is just a reference to an object?但是,执行
self.__getattribute__('b')
返回<bound method Ab of <__main__.A object at 0x000001CDC6742FD0>>
,这对我来说听起来很愚蠢:为什么method
与attribute
有任何不同,因为python
中的所有内容都只是一个对对象的引用? And why can I __getattribute__
, but not __delattr__
?为什么我可以
__getattribute__
而不是__delattr__
?
The same goes to redefinition.重新定义也是如此。 I can easily set any attribute, but methods are a no-no?
我可以轻松设置任何属性,但方法是禁忌吗?
class A:
def b(self):
print("first")
self.__setattr__('b', lambda self: print(f"second"))
a = A()
a.b()
a.b()
results into TypeError: <lambda>() missing 1 required positional argument: 'self'
.结果为
TypeError: <lambda>() missing 1 required positional argument: 'self'
。 Which, of course, means, that now python
isn't using dot-notation as intended.这当然意味着,现在
python
没有按预期使用点符号。 Of course, we could ditch the self
attribute in the lambda altogether, considering we've got the reference to it already in b
.当然,我们可以完全放弃 lambda 中的
self
属性,考虑到我们已经在b
获得了对它的引用。 But isn't it incorrect by design?但这不是设计不正确吗?
The further I'm trying to take python
to the limit, the more frustrated I become.我越是试图将
python
发挥到极致,我就越沮丧。 Some imposed limitations ( or seemingly imposed? ) seem so unnatural, considering the way the language is marketed.考虑到语言的营销方式,一些强加的限制(或看似强加的? )似乎很不自然。 Shouldn't it allow this?
这不应该允许吗? Why doesn't it work?
为什么不起作用?
UPD UPD
Ok, consider this:好的,考虑一下:
class A:
def __init__(self):
self.variable = 1
def b(self):
print("old")
self.variable += 1
def new_b():
print("new")
self.variable += 15
self.__setattr__('b', new_b)
It will work and do what we want: none of other objects will have their Ab
method redefined once one object kind of overlays its b
definition.它会工作并做我们想要的:一旦一种对象覆盖了它的
b
定义,其他任何对象都不会重新定义它们的Ab
方法。 ( overlays , since everyone so far says that you cannot redefine a method for an object, but instead only kind of hide it from the caller behind another attribute with the same name, as far as I understand). ( overlays ,因为到目前为止每个人都说你不能为一个对象重新定义一个方法,而只是将它隐藏在另一个同名属性后面的调用者中,据我所知)。
Is this good?这个好吗?
It doesn't work because b
isn't an attribute belonging to the instance, it belongs to the class.它不起作用,因为
b
不是属于实例的属性,它属于类。 So you can't delete it on the instance because it isn't there to be deleted.所以你不能在实例上删除它,因为它不是要删除的。
>>> a = A()
>>> list(a.__dict__)
[]
>>> list(A.__dict__)
['__module__', 'b', '__dict__', '__weakref__', '__doc__']
When ab
is evaluated, Python will see that a
has no instance attribute named b
and fall back to the class.当
ab
被求值时,Python 将看到a
没有名为b
实例属性并回退到该类。 (It's a little more complicated because when falling back to the class, it will not simply return the method itself, but a version of the method which is bound to the instance a
.) (这有点复杂,因为当回退到类时,它不会简单地返回方法本身,而是绑定到实例
a
的方法版本。)
Since you don't want to delete the method on the class, the way to go is to replace the method on the instance.既然不想删除类上的方法,那么要走的路就是替换实例上的方法。 I don't know why you tried to do this with
__setattr__
- there is no need for that, simply assign self.b = ...
as normal.我不知道你为什么试图用
__setattr__
来做到这__setattr__
- 没有必要,只需像往常一样分配self.b = ...
。 The reason your attempt failed is because your lambda requires a positional parameter named self
, but this parameter will not be automatically bound to the instance when you look it up, because it is an instance attribute, not a class attribute.你的尝试失败的原因是你的 lambda 需要一个名为
self
的位置参数,但是当你查找它时,这个参数不会自动绑定到实例,因为它是一个实例属性,而不是一个类属性。
class A:
def b(self):
print('first')
self.b = lambda: print('second')
Usage:用法:
>>> a = A()
>>> a.b()
first
>>> a.b()
second
Well in python you have 2 types of attributes那么在python中你有两种类型的属性
instance/object attribute
is a variable that belongs to one (and only one) object. instance/object attribute
是属于一个(且仅一个)对象的变量。 Every instance of a class points to its own attributes variables. In case of a class attribute its part of the class descriptor, so you cannot delete it from the object attributes like self.__deleteattr__
or add new one with __setattr__
as it alters the class descriptor and reflects on all objects.如果类属性是类描述符的一部分,因此您不能将其从对象属性中删除,例如
self.__deleteattr__
或使用__setattr__
添加新属性,因为它会更改类描述符并反映在所有对象上。 Such an operation can have devastating effects.这样的操作可能会产生破坏性的影响。
Its very similar to a class variable as well.它也非常类似于类变量。 You can however change the behavior with overriding or reassigning like below
但是,您可以通过覆盖或重新分配来更改行为,如下所示
class A:
def b(self):
print("empty")
A.b = lambda self: print(f"second")
a = A()
a.b()
a.b()
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