如何用另一个对象替换各处的 python 对象?
我有两个班级,
SimpleObject
和FancyObject
。我创建了一个 SimpleObject
,并有多个参考文献。现在我想创建一个 FancyObject
,并使所有这些引用都指向新对象。
a = SimpleObject()
some_list.append(a)
b = FancyObject()
a = b
不是我想要的,它只是改变了a指向的内容。我读到以下内容会起作用,但不起作用。我收到错误“属性 __dict__ 不可写”:
a.__dict__ = b.__dict__
我想要的是(伪C)的等价物:
*a = *b
我知道这很hacky,但是有什么办法可以做到这一点吗?
没有办法。它会让你改变不可变的对象并导致各种肮脏的事情。
x = 1
y = (x,)
z = {x: 3}
magic_replace(x, [1])
# x is now a list!
# The contents of y have changed, and z now has an unhashable key.
x = 1 + 1
# Is x 2, or [1, 1], or something stranger?
您可以将该对象放在单独模块的全局命名空间中,然后在需要时对其进行猴子修补。
objstore.py
:
replaceable = object()
sample.py
:
import objstore
b = object()
def isB():
return objstore.replaceable is b
if __name__ == '__main__':
print isB()#False
objstore.replaceable = b
print isB()#True
附注依赖猴子补丁是糟糕设计的症状
PyJack 有一个函数
replace_all_refs
可以替换内存中对象的所有引用。
文档中的示例:
>>> item = (100, 'one hundred')
>>> data = {item: True, 'itemdata': item}
>>>
>>> class Foobar(object):
... the_item = item
...
>>> def outer(datum):
... def inner():
... return ("Here is the datum:", datum,)
...
... return inner
...
>>> inner = outer(item)
>>>
>>> print item
(100, 'one hundred')
>>> print data
{'itemdata': (100, 'one hundred'), (100, 'one hundred'): True}
>>> print Foobar.the_item
(100, 'one hundred')
>>> print inner()
('Here is the datum:', (100, 'one hundred'))
调用replace_all_refs
>>> new = (101, 'one hundred and one')
>>> org_item = pyjack.replace_all_refs(item, new)
>>>
>>> print item
(101, 'one hundred and one')
>>> print data
{'itemdata': (101, 'one hundred and one'), (101, 'one hundred and one'): True}
>>> print Foobar.the_item
(101, 'one hundred and one')
>>> print inner()
('Here is the datum:', (101, 'one hundred and one'))
您有多种选择:
使用调试、GC 和自省功能来寻找每个符合您标准的对象并在运行时更改变量。缺点是变量的值会在代码执行期间发生变化,而无法从受影响的代码中发现。即使更改是原子的(消除一类错误),因为这可能会在执行确定值属于不同类型的代码后更改变量的类型,因此会引入在该代码中无法合理预期的错误。例如
a = iter(b) # will blow up if not iterable
[x for x in b] # before change, was iterable, but between two lines, b was changed to an int.
更微妙的是,当区分字符串和非字符串序列时(因为字符串的定义特征是迭代它们也会产生字符串,而字符串本身是可迭代的),当展平结构时,代码可能会被破坏。
另一个答案提到了 pyjack ,它实现了选项 3。虽然它可能有效,但它具有提到的所有问题。这可能仅适用于调试和开发。
利用可变对象,例如列表。
a = [SimpleObject()]
some_list.append(a)
b = FancyObject()
a[0] = b
证明这有效:
class SimpleObject():
def Who(self):
print 'SimpleObject'
class FancyObject():
def Who(self):
print 'FancyObject'
>>> a = [SimpleObject()]
>>> a[0].Who()
SimpleObject
>>> some_list = []
>>> some_list.append(a)
>>> some_list[0][0].Who()
SimpleObject
>>> b = FancyObject()
>>> b.Who()
FancyObject
>>> a[0] = b
>>> some_list[0][0].Who()
FancyObject
我遇到了同样的需求,我使用这个技巧解决了它:
class Reference:
directions = dict()
@staticmethod
def direct(reference): # -> Reference:
return Reference.directions.get(reference, reference)
@staticmethod
def route(origin, destination) -> None:
Reference.directions.update({origin: destination})
return None
def __init__(self) -> None:
self._reference = self
Reference.route(self, (lambda: self._reference))
return None
@property
def reference(self): # -> Reference:
return Reference.direct(self._reference)()
@reference.setter
def reference(self, obj) -> None:
Reference.route(self._reference, (lambda: obj.reference))
return None
class Example(Reference):
def __init__(self) -> None:
super().__init__()
return None
满足以下条件:
obj1 = Example()
obj2 = Example()
obj3 = obj1
obj4 = Example()
print(obj1.reference == obj3.reference)
print(obj1.reference == obj2.reference)
print(obj3.reference == obj2.reference)
print()
obj1.reference = obj2.reference
print(obj1.reference == obj2.reference)
print(obj3.reference == obj2.reference)
print()
obj2.reference = obj4.reference
print(obj1.reference == obj4.reference)
print(obj2.reference == obj4.reference)
print(obj3.reference == obj4.reference)
输出:
True
False
False
True
True
True
True
True