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

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。

其他回答

获得点访问(但不是数组访问)的一个简单方法是在Python中使用一个普通对象。是这样的:

class YourObject:
    def __init__(self, *args, **kwargs):
        for k, v in kwargs.items():
            setattr(self, k, v)

...像这样使用它:

>>> obj = YourObject(key="value")
>>> print(obj.key)
"value"

... 把它转换成字典:

>>> print(obj.__dict__)
{"key": "value"}

不喜欢。在Python中,属性访问和索引是分开的事情,您不应该希望它们执行相同的操作。创建一个类(可能是由namedtuple创建的),如果你有一些应该具有可访问属性的东西,并使用[]符号从字典中获取一个项。

使用__getattr__,非常简单,适用于 Python 3.4.3

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

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()

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'
>>