在 Python 中设置嵌套字典：为什么类方法的行为与独立函数不同？

Question

考虑这个简化的类：

class test(object):

    def __init__(self):

        self.inner_dict = {}

    def nested_set_method(self, keys,value=None):

        end = len(keys) - 1

        for index, component in enumerate(keys):

            if index < end or value is None:

                self.inner_dict = self.inner_dict.setdefault(component, {})

            else:

                self.inner_dict[component] = value

这个函数与nested_set_method上面的类相同：

def nested_set_standalone(input_dict, keys,value=None):

    end = len(keys) - 1

    for index, component in enumerate(keys):

        if index < end or value is None:

            input_dict = input_dict.setdefault(component, {})

        else:

            input_dict[component] = value

这是该类的示例用法：

>>> a = test()

>>> a.inner_dict

{}

>>> a.nested_set_method([1,2,3,4],'l')

>>> a.inner_dict

{4: 'l'}

这是该函数在类的实例上的示例用法：

>>> b = test()

>>> b.inner_dict

{}

>>> nested_set_standalone(b.inner_dict,[1,2,3,4],'l')

>>> b.inner_dict

{1: {2: {3: {4: 'l'}}}}

我期望类的nested_set_method这种输出{4: 'l'}有输出功能相同的nested_set_standalone是{1: {2: {3: {4: 'l'}}}}。

但它们为什么不同呢？

Answer 1

inner_dict在函数中是一个局部变量，但在方法中它改变了属性。简单地说，也使用局部变量：class test(object):&nbsp; &nbsp; def __init__(self):&nbsp; &nbsp; &nbsp; &nbsp; self.inner_dict = {}&nbsp; &nbsp; def get_nested_dict(self, keys):&nbsp; &nbsp; &nbsp; &nbsp; inner_dict = self.inner_dict&nbsp; &nbsp; &nbsp; &nbsp; for component in keys:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; inner_dict = inner_dict.setdefault(component, {})&nbsp; &nbsp; &nbsp; &nbsp; return inner_dict&nbsp; &nbsp; def nested_set_method(self, keys,value=None):&nbsp; &nbsp; &nbsp; &nbsp; if value is None:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return self.get_nested_dict(keys)&nbsp; &nbsp; &nbsp; &nbsp; else:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; inner_dict = self.get_nested_dict(keys[:-1])&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; inner_dict[keys[-1]] = value