Python-如何从属性类获取类实例引用？

Question

class A()    
    att = B()    

class B()    
    ...

a = A()

b = B()

a.att = b

b如何获得a的引用？ 我需要在这里获取a的属性。

谢谢！

Answer 1

最简单的方法是向需要A的B中的方法添加一个额外的函数参数，并在调用时将其传递给它。 或者，仅使B的init以A作为参数，然后将A的init的位更改为att = B（self）

Answer 2

class A(object):
    def __init__(self):
        self.att = B(self)

class B(object):
    def __init__(self, a):
        self.a = a

a = A()

a.att.a is a

或者另一种方式

class A(object):
    def __init__(self, b):
        b.a = self
        self.att = b

class B(object):
    pass

a = A(B())

a.att.a is a

Answer 3

这段代码没有什么意义...但是如果我正确理解了您的问题...

class A(object):
    pass     #or whatever you like

class B(object):
    def __init__(self, ref):  #accept one argument
        self.ref = ref

a = A()

b = B(a) #pass `a` as that argument

a.att = b

可能是一个答案。

Answer 4

class A(object):

    def __init__(self):
        self._att=None

    @property
    def att(self):
        return self._att

    @att.setter
    def att(self, value):
        self._att = value
        value.parent = self


class B(object):

    pass

a = A()

b = B()

a.att = b
print b.parent

Answer 5

您可以创建一个通用的“ Reference（）”类，该类将自身的任何引用保留在属性字典中。

class Reference(object):
    def __init__(self):
        self.references = {}

    def __setattr__(self, key, value):
        if hasattr(self, 'references'):
            if isinstance(value, Reference):
                if not key in value.references:
                    value.references[key] = []
                value.references[key].append(self)

            elif value is None and hasattr(self, key):
                old = getattr(self, key).references
                if key in old and self in old[key]:
                    old[key].remove(self)

        super(Reference, self).__setattr__(key, value)

然后，创建您的类：

class A(Reference):
    def __init__(self):
        super(A, self).__init__()
        self.att = None

class B(Reference):
    def __init__(self):
        super(B, self).__init__()
        self.att = None

并使用它：

a = A()
b = B()

print 'A references', a.references
print 'B references', b.references
# A references {}
# B references {}

a.att = b

print 'A references', a.references
print 'B references', b.references
# A references {}
# B references {'att': [<__main__.A object at 0x7f731c8fc910>]}

最后，您将从任何属性中反向引用所有Reference类