问题:如何使用点“。” 访问字典成员?
如何通过点“。”访问Python词典成员?
例如,mydict['val']我不想写,而是想写mydict.val。
我也想以这种方式访问嵌套的字典。例如
mydict.mydict2.val 
将指
mydict = { 'mydict2': { 'val': ... } }
回答 0
您可以使用我刚刚制作的此类进行操作。通过此类,您可以Map像其他字典(包括json序列化)一样使用该对象,也可以使用点符号。希望对您有所帮助:
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]
用法示例:
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']
回答 1
我一直将其保存在util文件中。您也可以在自己的类中将其用作混合。
class dotdict(dict):
    """dot.notation access to dictionary attributes"""
    __getattr__ = dict.get
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__
mydict = {'val':'it works'}
nested_dict = {'val':'nested works too'}
mydict = dotdict(mydict)
mydict.val
# 'it works'
mydict.nested = dotdict(nested_dict)
mydict.nested.val
# 'nested works too'
回答 2
dotmap通过安装pip
pip install dotmap
它可以完成您想要做的所有事情并成为其子类dict,因此它的作用类似于普通字典:
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,您可以将其与dict对象之间进行转换:
d = m.toDict()
m = DotMap(d) # automatic conversion in constructor
这意味着,如果您要访问的内容已经dict存在,可以将其转换DotMap为易于访问的内容:
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,因此您可以执行以下操作:
m = DotMap()
m.people.steve.age = 31
与束比较
全面披露:我是DotMap的创建者。我创建它是因为Bunch缺少这些功能
- 记住添加了订单项并按照该顺序进行迭代
 - 自动子 
DotMap创建,这样可以节省时间,并在您具有多个层次结构时使代码更简洁 - 从构造
dict并将所有子dict实例递归转换为DotMap 
回答 3
回答 4
Fabric具有非常好的,最小的实现。扩展它以允许嵌套访问,我们可以使用defaultdict,结果看起来像这样:
from collections import defaultdict
class AttributeDict(defaultdict):
    def __init__(self):
        super(AttributeDict, self).__init__(AttributeDict)
    def __getattr__(self, key):
        try:
            return self[key]
        except KeyError:
            raise AttributeError(key)
    def __setattr__(self, key, value):
        self[key] = value
如下使用它:
keys = AttributeDict()
keys.abc.xyz.x = 123
keys.abc.xyz.a.b.c = 234
这就详细说明了库格尔的回答:“从命令和实现中得出__getattr__和__setattr__”。现在您知道了!
回答 5
我尝试了这个:
class dotdict(dict):
    def __getattr__(self, name):
        return self[name]
你可以试试 __getattribute__。
让每个字典都使用点分隔符就足够了,如果您想从多层字典中初始化它,也可以尝试实现__init__。
回答 6
别。属性访问和索引编制在Python中是分开的,您不希望它们执行相同的操作。namedtuple如果您有一些应该具有可访问属性的东西,并使用[]表示法从字典中获取一项,则创建一个类(可能是由制成的类)。
回答 7
如果要腌制修改后的字典,则需要在上述答案中添加一些状态方法:
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
回答 8
以库格尔(Kugel)的答案为基础,并考虑迈克·格雷厄姆(Mike Graham)的警告语,如果我们做包装纸怎么办?
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
回答 9
用途SimpleNamespace:
>>> from types import SimpleNamespace   
>>> d = dict(x=[1, 2], y=['a', 'b'])
>>> ns = SimpleNamespace(**d)
>>> ns.x
[1, 2]
>>> ns
namespace(x=[1, 2], y=['a', 'b'])
回答 10
我喜欢Munch,它在点访问之外还提供了许多方便的选项。
进口午餐
temp_1 = {‘person’:{‘fname’:’senthil’,’lname’:’ramalingam’}}
dict_munch = munch.munchify(temp_1)
dict_munch.person.fname
回答 11
最近,我遇到了“ Box ”库,它做同样的事情。
安装命令: pip install python-box
例:
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(例如dotmap)更有效,如点阵图,当嵌套的字典较大时,它们会生成python递归错误。
链接到库和详细信息:https : //pypi.org/project/python-box/
回答 12
使用__getattr__非常简单,可在Python 3.4.3中使用
class myDict(dict):
    def __getattr__(self,val):
        return self[val]
blockBody=myDict()
blockBody['item1']=10000
blockBody['item2']="StackOverflow"
print(blockBody.item1)
print(blockBody.item2)
输出:
10000
StackOverflow
回答 13
语言本身不支持此功能,但有时这仍然是有用的要求。除了Bunch食谱之外,您还可以编写一些小方法,该方法可以使用点分字符串来访问字典:
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
这将支持以下内容:
>> test_dict = {'thing': {'spam': 12, 'foo': {'cheeze': 'bar'}}}
>> output = get_var(test_dict, 'thing.spam.foo.cheeze')
>> print output
'bar'
>>
回答 14
为了建立epool的答案,此版本允许您通过点运算符访问内部的所有dict:
foo = {
    "bar" : {
        "baz" : [ {"boo" : "hoo"} , {"baba" : "loo"} ]
    }
}
例如,foo.bar.baz[1].babareturn "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.iteritems():
                    if isinstance(v, dict):
                        v = Map(v)
                    if isinstance(v, list):
                        self.__convert(v)
                    self[k] = v
        if kwargs:
            for k, v in kwargs.iteritems():
                if isinstance(v, dict):
                    v = Map(v)
                elif isinstance(v, list):
                    self.__convert(v)
                self[k] = v
    def __convert(self, v):
        for elem in xrange(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]
回答 15
def dict_to_object(dick):
    # http://stackoverflow.com/a/1305663/968442
    class Struct:
        def __init__(self, **entries):
            self.__dict__.update(entries)
    return Struct(**dick)
如果决定永久将其转换dict为对象,则应该这样做。您可以在访问之前创建一个一次性对象。
d = dict_to_object(d)
回答 16
我最终都尝试了AttrDict和 Bunch库,发现它们是我使用速度减慢的方法。经过一个朋友和我的研究,我们发现编写这些库的主要方法导致该库通过嵌套对象积极地递归并在整个字典对象中进行复制。考虑到这一点,我们进行了两个关键更改。1)我们使属性延迟加载; 2)我们创建轻量级代理对象的副本,而不是创建字典对象的副本。这是最终的实现。使用此代码的性能提升令人难以置信。使用AttrDict或Bunch时,仅这两个库分别消耗了我的请求时间的1/2和1/3(什么!?)。这段代码将时间减少到几乎没有(在0.5ms范围内)。当然,这取决于您的需求,但是如果您在代码中大量使用此功能,
class DictProxy(object):
    def __init__(self, obj):
        self.obj = obj
    def __getitem__(self, key):
        return wrap(self.obj[key])
    def __getattr__(self, key):
        try:
            return wrap(getattr(self.obj, key))
        except AttributeError:
            try:
                return self[key]
            except KeyError:
                raise AttributeError(key)
    # you probably also want to proxy important list properties along like
    # items(), iteritems() and __len__
class ListProxy(object):
    def __init__(self, obj):
        self.obj = obj
    def __getitem__(self, key):
        return wrap(self.obj[key])
    # you probably also want to proxy important list properties along like
    # __iter__ and __len__
def wrap(value):
    if isinstance(value, dict):
        return DictProxy(value)
    if isinstance(value, (tuple, list)):
        return ListProxy(value)
    return value
通过https://stackoverflow.com/users/704327/michael-merickel查看此处的原始实现。
还要注意的另一件事是,此实现非常简单,并没有实现您可能需要的所有方法。您需要根据需要在DictProxy或ListProxy对象上编写这些内容。
回答 17
我想将自己的解决方案付诸实践:
https://github.com/skorokithakis/jsane
它使您可以将JSON解析为可以访问的内容with.attribute.lookups.like.this.r(),主要是因为在开始使用JSON 之前我还没有看到这个答案。
回答 18
不是直接回答OP的问题,而是受到某些人的启发,也许对某些人有用。.我已经使用内部__dict__方法创建了基于对象的解决方案(绝不优化代码)
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"
回答 19
获得点访问(而不是数组访问)的一种简单方法是在Python中使用普通对象。像这样:
class YourObject:
    def __init__(self, *args, **kwargs):
        for k, v in kwargs.items():
            setattr(self, k, v)
…并像这样使用它:
>>> obj = YourObject(key="value")
>>> print(obj.key)
"value"
…将其转换为字典:
>>> print(obj.__dict__)
{"key": "value"}
回答 20
此解决方案是对epool提供的解决方案的改进,以解决OP以一致方式访问嵌套dict的要求。epool的解决方案不允许访问嵌套字典。
class YAMLobj(dict):
    def __init__(self, args):
        super(YAMLobj, self).__init__(args)
        if isinstance(args, dict):
            for k, v in args.iteritems():
                if not isinstance(v, dict):
                    self[k] = v
                else:
                    self.__setattr__(k, YAMLobj(v))
    def __getattr__(self, attr):
        return self.get(attr)
    def __setattr__(self, key, value):
        self.__setitem__(key, value)
    def __setitem__(self, key, value):
        super(YAMLobj, self).__setitem__(key, value)
        self.__dict__.update({key: value})
    def __delattr__(self, item):
        self.__delitem__(item)
    def __delitem__(self, key):
        super(YAMLobj, self).__delitem__(key)
        del self.__dict__[key]
有了这一堂课,您现在可以执行以下操作:A.B.C.D。
回答 21
这也适用于嵌套字典,并确保稍后附加的字典具有相同的行为:
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__
回答 22
@ derek73的答案非常简洁,但是不能被腌制或(深度)复制,并且返回None因缺少键。下面的代码解决了这个问题。
编辑:我没有看到上面的答案完全相同的点(已投票)。我将答案留在这里以供参考。
class dotdict(dict):
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__
    def __getattr__(self, name):
        try:
            return self[name]
        except KeyError:
            raise AttributeError(name)
回答 23
一种精致的解决方案
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))
