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

用于无限级别的字典、列表、字典的列表和列表的字典的嵌套。

它还支持酸洗

这是这个答案的延伸。

class DotDict(dict):
    # https://stackoverflow.com/a/70665030/913098
    """
    Example:
    m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])

    Iterable are assumed to have a constructor taking list as input.
    """

    def __init__(self, *args, **kwargs):
        super(DotDict, self).__init__(*args, **kwargs)

        args_with_kwargs = []
        for arg in args:
            args_with_kwargs.append(arg)
        args_with_kwargs.append(kwargs)
        args = args_with_kwargs

        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.items():
                    self[k] = v
                    if isinstance(v, dict):
                        self[k] = DotDict(v)
                    elif isinstance(v, str) or isinstance(v, bytes):
                        self[k] = v
                    elif isinstance(v, Iterable):
                        klass = type(v)
                        map_value: List[Any] = []
                        for e in v:
                            map_e = DotDict(e) if isinstance(e, dict) else e
                            map_value.append(map_e)
                        self[k] = klass(map_value)



    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(DotDict, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(DotDict, self).__delitem__(key)
        del self.__dict__[key]

    def __getstate__(self):
        return self.__dict__

    def __setstate__(self, d):
        self.__dict__.update(d)


if __name__ == "__main__":
    import pickle
    def test_map():
        d = {
            "a": 1,
            "b": {
                "c": "d",
                "e": 2,
                "f": None
            },
            "g": [],
            "h": [1, "i"],
            "j": [1, "k", {}],
            "l":
                [
                    1,
                    "m",
                    {
                        "n": [3],
                        "o": "p",
                        "q": {
                            "r": "s",
                            "t": ["u", 5, {"v": "w"}, ],
                            "x": ("z", 1)
                        }
                    }
                ],
        }
        map_d = DotDict(d)
        w = map_d.l[2].q.t[2].v
        assert w == "w"

        pickled = pickle.dumps(map_d)
        unpickled = pickle.loads(pickled)
        assert unpickled == map_d

        kwargs_check = DotDict(a=1, b=[dict(c=2, d="3"), 5])
        assert kwargs_check.b[0].d == "3"

        kwargs_and_args_check = DotDict(d, a=1, b=[dict(c=2, d="3"), 5])
        assert kwargs_and_args_check.l[2].q.t[2].v == "w"
        assert kwargs_and_args_check.b[0].d == "3"



    test_map()

其他回答

我一直把它保存在util文件中。您也可以在自己的类中使用它作为mixin。

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'

基于epool的答案,这个版本允许你通过点操作符访问任何字典:

foo = {
    "bar" : {
        "baz" : [ {"boo" : "hoo"} , {"baba" : "loo"} ]
    }
}

例如,foo.bar.baz[1]。爸爸回答“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.items():
                    if isinstance(v, dict):
                        v = Map(v)
                    if isinstance(v, list):
                        self.__convert(v)
                    self[k] = v

        if kwargs:
            for k, v in kwargs.items():
                if isinstance(v, dict):
                    v = Map(v)
                elif isinstance(v, list):
                    self.__convert(v)
                self[k] = v

    def __convert(self, v):
        for elem in range(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]

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

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__。

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