基于Kugel的回答，并考虑到Mike Graham的警告，如果我们制作一个包装器呢?

class DictWrap(object):
  """ Wrap an existing dict, or create a new one, and access with either dot 
    notation or key lookup.

    The attribute _data is reserved and stores the underlying dictionary.
    When using the += operator with create=True, the empty nested dict is 
    replaced with the operand, effectively creating a default dictionary
    of mixed types.

    args:
      d({}): Existing dict to wrap, an empty dict is created by default
      create(True): Create an empty, nested dict instead of raising a KeyError

    example:
      >>>dw = DictWrap({'pp':3})
      >>>dw.a.b += 2
      >>>dw.a.b += 2
      >>>dw.a['c'] += 'Hello'
      >>>dw.a['c'] += ' World'
      >>>dw.a.d
      >>>print dw._data
      {'a': {'c': 'Hello World', 'b': 4, 'd': {}}, 'pp': 3}

  """

  def __init__(self, d=None, create=True):
    if d is None:
      d = {}
    supr = super(DictWrap, self)  
    supr.__setattr__('_data', d)
    supr.__setattr__('__create', create)

  def __getattr__(self, name):
    try:
      value = self._data[name]
    except KeyError:
      if not super(DictWrap, self).__getattribute__('__create'):
        raise
      value = {}
      self._data[name] = value

    if hasattr(value, 'items'):
      create = super(DictWrap, self).__getattribute__('__create')
      return DictWrap(value, create)
    return value

  def __setattr__(self, name, value):
    self._data[name] = value  

  def __getitem__(self, key):
    try:
      value = self._data[key]
    except KeyError:
      if not super(DictWrap, self).__getattribute__('__create'):
        raise
      value = {}
      self._data[key] = value

    if hasattr(value, 'items'):
      create = super(DictWrap, self).__getattribute__('__create')
      return DictWrap(value, create)
    return value

  def __setitem__(self, key, value):
    self._data[key] = value

  def __iadd__(self, other):
    if self._data:
      raise TypeError("A Nested dict will only be replaced if it's empty")
    else:
      return other

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/

语言本身不支持这一点，但有时这仍然是一个有用的需求。除了Bunch recipe，你还可以写一个小方法，可以使用虚线字符串访问字典:

def get_var(input_dict, accessor_string):
    """Gets data from a dictionary using a dotted accessor-string"""
    current_data = input_dict
    for chunk in accessor_string.split('.'):
        current_data = current_data.get(chunk, {})
    return current_data

这将支持如下内容:

>> test_dict = {'thing': {'spam': 12, 'foo': {'cheeze': 'bar'}}}
>> output = get_var(test_dict, 'thing.spam.foo.cheeze')
>> print output
'bar'
>>

最简单的解决方案。

定义一个只有pass语句的类。为该类创建对象并使用点表示法。

class my_dict:
    pass

person = my_dict()
person.id = 1 # create using dot notation
person.phone = 9999
del person.phone # Remove a property using dot notation

name_data = my_dict()
name_data.first_name = 'Arnold'
name_data.last_name = 'Schwarzenegger'

person.name = name_data
person.name.first_name # dot notation access for nested properties - gives Arnold

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

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)