我最近遇到了“Box”库，它也做同样的事情。

安装命令:pip install python-box

例子:

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(如dotmap)更有效，当你有大量嵌套字典时，dotmap会产生python递归错误。

链接到图书馆和详细信息:https://pypi.org/project/python-box/

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()

我最近遇到了“Box”库，它也做同样的事情。

安装命令:pip install python-box

例子:

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(如dotmap)更有效，当你有大量嵌套字典时，dotmap会产生python递归错误。

链接到图书馆和详细信息:https://pypi.org/project/python-box/

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__。

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]

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