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

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

其他回答

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"

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))

我试了一下:

class dotdict(dict):
    def __getattr__(self, name):
        return self[name]

你也可以尝试__getattribute__。

使每个字典都是一种类型的dotdict就足够了,如果你想从多层字典初始化它,也可以尝试实现__init__。

如果你已经在使用pandas,你可以构造一个pandas Series或DataFrame,从中你可以通过点语法访问项目:

1级字典:

import pandas as pd

my_dictionary = pd.Series({
  'key1': 'value1',
  'key2': 'value2'
})

print(my_dictionary.key1)
# Output: value1

2级字典:

import pandas as pd

my_dictionary = pd.DataFrame({
  'key1': {
    'inner_key1': 'value1'
  },
  'key2': {
    'inner_key2': 'value2'
  }
})

print(my_dictionary.key1.inner_key1)
# Output: value1

请注意,这可能在规范化数据结构(其中每个字典条目都具有相同的结构)下工作得更好。在上面的第二个例子中,得到的DataFrame是:

              key1    key2
inner_key1  value1     NaN
inner_key2     NaN  value2

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