不是对OP问题的直接回答，但受到启发，也许对一些人有用。我已经创建了一个基于对象的解决方案使用内部__dict__(在任何方式优化代码)

payload = {
    "name": "John",
    "location": {
        "lat": 53.12312312,
        "long": 43.21345112
    },
    "numbers": [
        {
            "role": "home",
            "number": "070-12345678"
        },
        {
            "role": "office",
            "number": "070-12345679"
        }
    ]
}


class Map(object):
    """
    Dot style access to object members, access raw values
    with an underscore e.g.

    class Foo(Map):
        def foo(self):
            return self.get('foo') + 'bar'

    obj = Foo(**{'foo': 'foo'})

    obj.foo => 'foobar'
    obj._foo => 'foo'

    """

    def __init__(self, *args, **kwargs):
        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.iteritems():
                    self.__dict__[k] = v
                    self.__dict__['_' + k] = v

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

    def __getattribute__(self, attr):
        if hasattr(self, 'get_' + attr):
            return object.__getattribute__(self, 'get_' + attr)()
        else:
            return object.__getattribute__(self, attr)

    def get(self, key):
        try:
            return self.__dict__.get('get_' + key)()
        except (AttributeError, TypeError):
            return self.__dict__.get(key)

    def __repr__(self):
        return u"<{name} object>".format(
            name=self.__class__.__name__
        )


class Number(Map):
    def get_role(self):
        return self.get('role')

    def get_number(self):
        return self.get('number')


class Location(Map):
    def get_latitude(self):
        return self.get('lat') + 1

    def get_longitude(self):
        return self.get('long') + 1


class Item(Map):
    def get_name(self):
        return self.get('name') + " Doe"

    def get_location(self):
        return Location(**self.get('location'))

    def get_numbers(self):
        return [Number(**n) for n in self.get('numbers')]


# Tests

obj = Item({'foo': 'bar'}, **payload)

assert type(obj) == Item
assert obj._name == "John"
assert obj.name == "John Doe"
assert type(obj.location) == Location
assert obj.location._lat == 53.12312312
assert obj.location._long == 43.21345112
assert obj.location.latitude == 54.12312312
assert obj.location.longitude == 44.21345112

for n in obj.numbers:
    assert type(n) == Number
    if n.role == 'home':
        assert n.number == "070-12345678"
    if n.role == 'office':
        assert n.number == "070-12345679"

使用namedtuple允许点访问。

它就像一个轻量级对象，也具有元组的属性。

它允许定义属性并使用点操作符访问它们。

from collections import namedtuple
Data = namedtuple('Data', ['key1', 'key2'])

dataObj = Data(val1, key2=val2) # can instantiate using keyword arguments and positional arguments

使用点运算符访问

dataObj.key1 # Gives val1
datObj.key2 # Gives val2

使用元组索引进行访问

dataObj[0] # Gives val1
dataObj[1] # Gives val2

但记住这是一个元组;不是字典。因此下面的代码将给出错误

dataObj['key1'] # Gives TypeError: tuple indices must be integers or slices, not str

参考:namedtuple

一个很微妙的解

class DotDict(dict):

    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__

    def __getattr__(self, key):

        def typer(candidate):
            if isinstance(candidate, dict):
                return DotDict(candidate)

            if isinstance(candidate, str):  # iterable but no need to iter
                return candidate

            try:  # other iterable are processed as list
                return [typer(item) for item in candidate]
            except TypeError:
                return candidate

            return candidate

        return typer(dict.get(self, key))

如果你想pickle你修改后的字典，你需要添加几个状态方法到上面的答案:

class DotDict(dict):
    """dot.notation access to dictionary attributes"""
    def __getattr__(self, attr):
        return self.get(attr)
    __setattr__= dict.__setitem__
    __delattr__= dict.__delitem__

    def __getstate__(self):
        return self

    def __setstate__(self, state):
        self.update(state)
        self.__dict__ = self

kaggle_environments使用的实现是一个名为structify的函数。

class Struct(dict):
    def __init__(self, **entries):
        entries = {k: v for k, v in entries.items() if k != "items"}
        dict.__init__(self, entries)
        self.__dict__.update(entries)

    def __setattr__(self, attr, value):
        self.__dict__[attr] = value
        self[attr] = value

# Added benefit of cloning lists and dicts.
def structify(o):
    if isinstance(o, list):
        return [structify(o[i]) for i in range(len(o))]
    elif isinstance(o, dict):
        return Struct(**{k: structify(v) for k, v in o.items()})
    return o

https://github.com/Kaggle/kaggle-environments/blob/master/kaggle_environments/utils.py

这可能有助于在《ConnectX》等游戏中测试AI模拟代理

from kaggle_environments import structify

obs  = structify({ 'remainingOverageTime': 60, 'step': 0, 'mark': 1, 'board': [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]})
conf = structify({ 'timeout': 2, 'actTimeout': 2, 'agentTimeout': 60, 'episodeSteps': 1000, 'runTimeout': 1200, 'columns': 7, 'rows': 6, 'inarow': 4, '__raw_path__': '/kaggle_simulations/agent/main.py' })

def agent(obs, conf):
  action = obs.step % conf.columns
  return action

通过pip安装dotmap

pip install dotmap

它能做你想让它做的所有事情，并继承dict的子类，所以它的操作就像一个普通的字典:

from dotmap import DotMap

m = DotMap()
m.hello = 'world'
m.hello
m.hello += '!'
# m.hello and m['hello'] now both return 'world!'
m.val = 5
m.val2 = 'Sam'

最重要的是，你可以将它转换为dict对象:

d = m.toDict()
m = DotMap(d) # automatic conversion in constructor

这意味着如果你想访问的东西已经是字典形式的，你可以把它转换成DotMap来方便访问:

import json
jsonDict = json.loads(text)
data = DotMap(jsonDict)
print data.location.city

最后，它会自动创建新的子DotMap实例，你可以这样做:

m = DotMap()
m.people.steve.age = 31

与Bunch的比较

完全公开:我是DotMap的创造者。我创建它是因为Bunch缺少这些功能

记住添加的顺序项并按此顺序迭代 自动创建子DotMap，当你有很多层次结构时，这节省了时间，并使代码更干净 从字典构造并递归地将所有子字典实例转换为DotMap