我如何使Python字典成员访问通过点“。”?

例如,我想写mydict.val而不是mydict['val']。

我还想以这种方式访问嵌套字典。例如

mydict.mydict2.val 

会提到

mydict = { 'mydict2': { 'val': ... } }

当前回答

这也适用于嵌套字典,并确保后面追加的字典行为相同:

class DotDict(dict):

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        # Recursively turn nested dicts into DotDicts
        for key, value in self.items():
            if type(value) is dict:
                self[key] = DotDict(value)

    def __setitem__(self, key, item):
        if type(item) is dict:
            item = DotDict(item)
        super().__setitem__(key, item)

    __setattr__ = __setitem__
    __getattr__ = dict.__getitem__

其他回答

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

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"

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

派生自dict和并实现__getattr__和__setattr__。

或者你也可以用Bunch,非常相似。

我不认为这是可能的monkeypatch内置字典类。

我只需要使用虚线路径字符串访问字典,所以我想到了:

def get_value_from_path(dictionary, parts):
    """ extracts a value from a dictionary using a dotted path string """

    if type(parts) is str:
        parts = parts.split('.')

    if len(parts) > 1:
        return get_value_from_path(dictionary[parts[0]], parts[1:])

    return dictionary[parts[0]]

a = {'a':{'b':'c'}}
print(get_value_from_path(a, 'a.b')) # c