我想就扩大了例如自动激活从前面的回答给nosklo允许通过元组字典的访问。

nosklo的解决方案是这样的：

class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

测试：

a = AutoVivification()

a[1][2][3] = 4
a[1][3][3] = 5
a[1][2]['test'] = 6

print a

输出：

{1: {2: {'test': 6, 3: 4}, 3: {3: 5}}}

我有我想要的设置节点定标中的一些任意元组的情况下。如果我不知道元组有多少层深的是，我怎么能设计一种方法来设置适当的节点？

我想，也许我可以使用类似下面的语法：

mytuple = (1,2,3)
a[mytuple] = 4

但我有拿出一个工作实现麻烦。

更新

我有一个基于@ JCash的回答完全工作示例：

class NestedDict(dict):
    """                                                                       
    Nested dictionary of arbitrary depth with autovivification.               

    Allows data access via extended slice notation.                           
    """
    def __getitem__(self, keys):
        # Let's assume *keys* is a list or tuple.                             
        if not isinstance(keys, basestring):
            try:
                node = self
                for key in keys:
                    node = dict.__getitem__(node, key)
                return node
            except TypeError:
            # *keys* is not a list or tuple.                              
                pass
        try:
            return dict.__getitem__(self, keys)
        except KeyError:
            raise KeyError(keys)
    def __setitem__(self, keys, value):
        # Let's assume *keys* is a list or tuple.                             
        if not isinstance(keys, basestring):
            try:
                node = self
                for key in keys[:-1]:
                    try:
                        node = dict.__getitem__(node, key)
                    except KeyError:
                        node[key] = type(self)()
                        node = node[key]
                return dict.__setitem__(node, keys[-1], value)
            except TypeError:
                # *keys* is not a list or tuple.                              
                pass
        dict.__setitem__(self, keys, value)

从而可以达到相同的输出如上使用扩展切片表示法：

d = NestedDict()
d[1,2,3] = 4
d[1,3,3] = 5
d[1,2,'test'] = 6

这似乎工作

def __setitem__(self, key, value):
    if isinstance(key, tuple):
        node = self
        for i in key[:-1]:
            try:
                node = dict.__getitem__(node, i)
            except KeyError:
                node = node[i] = type(self)()
        return dict.__setitem__(node, i, value)
    return dict.__setitem__(self, key, value)