我正在寻找一种优雅的方式来获得数据使用属性访问字典与一些嵌套的字典和列表(即javascript风格的对象语法)。

例如:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

应该以这样的方式访问:

>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
bar

我想,如果没有递归,这是不可能的,但是有什么更好的方法来获得字典的对象样式呢?


当前回答

如果你想访问dict键作为一个对象(或作为一个dict难键),做递归,也能够更新原来的dict,你可以这样做:

class Dictate(object):
    """Object view of a dict, updating the passed in dict when values are set
    or deleted. "Dictate" the contents of a dict...: """

    def __init__(self, d):
        # since __setattr__ is overridden, self.__dict = d doesn't work
        object.__setattr__(self, '_Dictate__dict', d)

    # Dictionary-like access / updates
    def __getitem__(self, name):
        value = self.__dict[name]
        if isinstance(value, dict):  # recursively view sub-dicts as objects
            value = Dictate(value)
        return value

    def __setitem__(self, name, value):
        self.__dict[name] = value
    def __delitem__(self, name):
        del self.__dict[name]

    # Object-like access / updates
    def __getattr__(self, name):
        return self[name]

    def __setattr__(self, name, value):
        self[name] = value
    def __delattr__(self, name):
        del self[name]

    def __repr__(self):
        return "%s(%r)" % (type(self).__name__, self.__dict)
    def __str__(self):
        return str(self.__dict)

使用示例:

d = {'a': 'b', 1: 2}
dd = Dictate(d)
assert dd.a == 'b'  # Access like an object
assert dd[1] == 2  # Access like a dict
# Updates affect d
dd.c = 'd'
assert d['c'] == 'd'
del dd.a
del dd[1]
# Inner dicts are mapped
dd.e = {}
dd.e.f = 'g'
assert dd['e'].f == 'g'
assert d == {'c': 'd', 'e': {'f': 'g'}}

其他回答

以下是我认为前面例子中最好的方面:

class Struct:
    """The recursive class for building and representing objects with."""

    def __init__(self, obj):
        for k, v in obj.items():
            if isinstance(v, dict):
                setattr(self, k, Struct(v))
            else:
                setattr(self, k, v)

    def __getitem__(self, val):
        return self.__dict__[val]

    def __repr__(self):
        return '{%s}' % str(', '.join('%s : %s' % (k, repr(v)) for (k, v) in self.__dict__.items()))
x = type('new_dict', (object,), d)

然后再加上递归,就完成了。

编辑这是我如何实现它:

>>> d
{'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> def obj_dic(d):
    top = type('new', (object,), d)
    seqs = tuple, list, set, frozenset
    for i, j in d.items():
        if isinstance(j, dict):
            setattr(top, i, obj_dic(j))
        elif isinstance(j, seqs):
            setattr(top, i, 
                type(j)(obj_dic(sj) if isinstance(sj, dict) else sj for sj in j))
        else:
            setattr(top, i, j)
    return top

>>> x = obj_dic(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'

这个怎么样:

from functools import partial
d2o=partial(type, "d2o", ())

然后可以这样使用:

>>> o=d2o({"a" : 5, "b" : 3})
>>> print o.a
5
>>> print o.b
3

X.__dict__.update (d)应该没问题。

将字典转换为对象

from types import SimpleNamespace

def dict2obj(data):
    """将字典对象转换为可访问的对象属性"""
    if not isinstance(data, dict):
        raise ValueError('data must be dict object.')

    def _d2o(d):
        _d = {}
        for key, item in d.items():
            if isinstance(item, dict):
                _d[key] = _d2o(item)
            else:
                _d[key] = item
        return SimpleNamespace(**_d)

    return _d2o(data)

参考答案