你可以用我刚做的这个类来做。对于这个类，您可以像使用另一个字典(包括json序列化)一样使用Map对象，或者使用点表示法。希望对大家有所帮助:

class Map(dict):
    """
    Example:
    m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])
    """
    def __init__(self, *args, **kwargs):
        super(Map, self).__init__(*args, **kwargs)
        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.iteritems():
                    self[k] = v

        if kwargs:
            for k, v in kwargs.iteritems():
                self[k] = v

    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(Map, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(Map, self).__delitem__(key)
        del self.__dict__[key]

使用例子:

m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])
# Add new key
m.new_key = 'Hello world!'
# Or
m['new_key'] = 'Hello world!'
print m.new_key
print m['new_key']
# Update values
m.new_key = 'Yay!'
# Or
m['new_key'] = 'Yay!'
# Delete key
del m.new_key
# Or
del m['new_key']

我喜欢Munch，它提供了很多方便的选项在点访问。

进口蒙克 Temp_1 = {'person': {' fname': 'senthil'， 'lname': 'ramalingam'}} Dict_munch = munch.munchify(temp_1) dict_munch.person.fname

基于epool的答案，这个版本允许你通过点操作符访问任何字典:

foo = {
    "bar" : {
        "baz" : [ {"boo" : "hoo"} , {"baba" : "loo"} ]
    }
}

例如，foo.bar.baz[1]。爸爸回答“loo”。

class Map(dict):
    def __init__(self, *args, **kwargs):
        super(Map, self).__init__(*args, **kwargs)
        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.items():
                    if isinstance(v, dict):
                        v = Map(v)
                    if isinstance(v, list):
                        self.__convert(v)
                    self[k] = v

        if kwargs:
            for k, v in kwargs.items():
                if isinstance(v, dict):
                    v = Map(v)
                elif isinstance(v, list):
                    self.__convert(v)
                self[k] = v

    def __convert(self, v):
        for elem in range(0, len(v)):
            if isinstance(v[elem], dict):
                v[elem] = Map(v[elem])
            elif isinstance(v[elem], list):
                self.__convert(v[elem])

    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(Map, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(Map, self).__delitem__(key)
        del self.__dict__[key]

我最近遇到了“Box”库，它也做同样的事情。

安装命令:pip install python-box

例子:

from box import Box

mydict = {"key1":{"v1":0.375,
                    "v2":0.625},
          "key2":0.125,
          }
mydict = Box(mydict)

print(mydict.key1.v1)

我发现它比其他现有的库(如dotmap)更有效，当你有大量嵌套字典时，dotmap会产生python递归错误。

链接到图书馆和详细信息:https://pypi.org/project/python-box/

您可以使用SimpleNamespace来实现这一点

from types import SimpleNamespace
# Assign values
args = SimpleNamespace()
args.username = 'admin'

# Retrive values
print(args.username)  # output: admin

def dict_to_object(dick):
    # http://stackoverflow.com/a/1305663/968442

    class Struct:
        def __init__(self, **entries):
            self.__dict__.update(entries)

    return Struct(**dick)

如果一个人决定永久地将字典转换为对象，这应该做到。您可以在访问之前创建一个丢弃对象。

d = dict_to_object(d)