老式问答，但我有更多的话题要谈。似乎没有人谈论递归字典。这是我的代码:

#!/usr/bin/env python

class Object( dict ):
    def __init__( self, data = None ):
        super( Object, self ).__init__()
        if data:
            self.__update( data, {} )

    def __update( self, data, did ):
        dataid = id(data)
        did[ dataid ] = self

        for k in data:
            dkid = id(data[k])
            if did.has_key(dkid):
                self[k] = did[dkid]
            elif isinstance( data[k], Object ):
                self[k] = data[k]
            elif isinstance( data[k], dict ):
                obj = Object()
                obj.__update( data[k], did )
                self[k] = obj
                obj = None
            else:
                self[k] = data[k]

    def __getattr__( self, key ):
        return self.get( key, None )

    def __setattr__( self, key, value ):
        if isinstance(value,dict):
            self[key] = Object( value )
        else:
            self[key] = value

    def update( self, *args ):
        for obj in args:
            for k in obj:
                if isinstance(obj[k],dict):
                    self[k] = Object( obj[k] )
                else:
                    self[k] = obj[k]
        return self

    def merge( self, *args ):
        for obj in args:
            for k in obj:
                if self.has_key(k):
                    if isinstance(self[k],list) and isinstance(obj[k],list):
                        self[k] += obj[k]
                    elif isinstance(self[k],list):
                        self[k].append( obj[k] )
                    elif isinstance(obj[k],list):
                        self[k] = [self[k]] + obj[k]
                    elif isinstance(self[k],Object) and isinstance(obj[k],Object):
                        self[k].merge( obj[k] )
                    elif isinstance(self[k],Object) and isinstance(obj[k],dict):
                        self[k].merge( obj[k] )
                    else:
                        self[k] = [ self[k], obj[k] ]
                else:
                    if isinstance(obj[k],dict):
                        self[k] = Object( obj[k] )
                    else:
                        self[k] = obj[k]
        return self

def test01():
    class UObject( Object ):
        pass
    obj = Object({1:2})
    d = {}
    d.update({
        "a": 1,
        "b": {
            "c": 2,
            "d": [ 3, 4, 5 ],
            "e": [ [6,7], (8,9) ],
            "self": d,
        },
        1: 10,
        "1": 11,
        "obj": obj,
    })
    x = UObject(d)


    assert x.a == x["a"] == 1
    assert x.b.c == x["b"]["c"] == 2
    assert x.b.d[0] == 3
    assert x.b.d[1] == 4
    assert x.b.e[0][0] == 6
    assert x.b.e[1][0] == 8
    assert x[1] == 10
    assert x["1"] == 11
    assert x[1] != x["1"]
    assert id(x) == id(x.b.self.b.self) == id(x.b.self)
    assert x.b.self.a == x.b.self.b.self.a == 1

    x.x = 12
    assert x.x == x["x"] == 12
    x.y = {"a":13,"b":[14,15]}
    assert x.y.a == 13
    assert x.y.b[0] == 14

def test02():
    x = Object({
        "a": {
            "b": 1,
            "c": [ 2, 3 ]
        },
        1: 6,
        2: [ 8, 9 ],
        3: 11,
    })
    y = Object({
        "a": {
            "b": 4,
            "c": [ 5 ]
        },
        1: 7,
        2: 10,
        3: [ 12 , 13 ],
    })
    z = {
        3: 14,
        2: 15,
        "a": {
            "b": 16,
            "c": 17,
        }
    }
    x.merge( y, z )
    assert 2 in x.a.c
    assert 3 in x.a.c
    assert 5 in x.a.c
    assert 1 in x.a.b
    assert 4 in x.a.b
    assert 8 in x[2]
    assert 9 in x[2]
    assert 10 in x[2]
    assert 11 in x[3]
    assert 12 in x[3]
    assert 13 in x[3]
    assert 14 in x[3]
    assert 15 in x[2]
    assert 16 in x.a.b
    assert 17 in x.a.c

if __name__ == '__main__':
    test01()
    test02()

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'

class obj(object):
    def __init__(self, d):
        for k, v in d.items():
            if isinstance(k, (list, tuple)):
                setattr(self, k, [obj(x) if isinstance(x, dict) else x for x in v])
            else:
                setattr(self, k, obj(v) if isinstance(v, dict) else v)

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = obj(d)
>>> x.b.c
2
>>> x.d[1].foo
'bar'

我偶然发现的情况下，我需要递归转换字典列表到对象列表，所以基于罗伯托的片段在这里为我做了什么工作:

def dict2obj(d):
    if isinstance(d, dict):
        n = {}
        for item in d:
            if isinstance(d[item], dict):
                n[item] = dict2obj(d[item])
            elif isinstance(d[item], (list, tuple)):
                n[item] = [dict2obj(elem) for elem in d[item]]
            else:
                n[item] = d[item]
        return type('obj_from_dict', (object,), n)
    elif isinstance(d, (list, tuple,)):
        l = []
        for item in d:
            l.append(dict2obj(item))
        return l
    else:
        return d

注意，由于显而易见的原因，任何元组都将被转换为与其列表相当的元素。

希望这能像你们的答案对我一样帮助到别人。

为dict寻找一个简单的包装器类，支持属性样式的键访问/赋值(点表示法)，我对现有选项不满意，原因如下。

数据类、pydantic等都很棒，但需要对内容进行静态定义。此外，它们不能在依赖dict的代码中替换dict，因为它们不共享相同的方法，并且不支持__getitem__()语法。

因此，我开发了MetaDict。它的行为完全类似于dict，但支持点表示法和IDE自动补全(如果对象被加载到RAM中)，而没有其他解决方案的缺点和潜在的名称空间冲突。所有功能和使用示例都可以在GitHub上找到(见上面的链接)。

完全披露:我是MetaDict的作者。

我在尝试其他解决方案时遇到的缺点/限制:

Addict No key autocompletion in IDE Nested key assignment cannot be turned off Newly assigned dict objects are not converted to support attribute-style key access Shadows inbuilt type Dict Prodict No key autocompletion in IDE without defining a static schema (similar to dataclass) No recursive conversion of dict objects when embedded in list or other inbuilt iterables AttrDict No key autocompletion in IDE Converts list objects to tuple behind the scenes Munch Inbuilt methods like items(), update(), etc. can be overwritten with obj.items = [1, 2, 3] No recursive conversion of dict objects when embedded in list or other inbuilt iterables EasyDict Only strings are valid keys, but dict accepts all hashable objects as keys Inbuilt methods like items(), update(), etc. can be overwritten with obj.items = [1, 2, 3] Inbuilt methods don't behave as expected: obj.pop('unknown_key', None) raises an AttributeError

注意:我在这个stackoverflow中写了一个类似的答案，这是相关的。

我认为字典由数字、字符串和字典组成，大多数时候就足够了。 所以我忽略了元组、列表和其他类型没有出现在字典的最后一个维度的情况。

考虑了继承，结合递归，方便地解决了打印问题，并提供了两种数据查询方式，一种数据编辑方式。

请看下面的例子，这是一个描述学生信息的词典:

group=["class1","class2","class3","class4",]
rank=["rank1","rank2","rank3","rank4","rank5",]
data=["name","sex","height","weight","score"]

#build a dict based on the lists above
student_dic=dict([(g,dict([(r,dict([(d,'') for d in data])) for r in rank ]))for g in group])

#this is the solution
class dic2class(dict):
    def __init__(self, dic):
        for key,val in dic.items():
            self.__dict__[key]=self[key]=dic2class(val) if isinstance(val,dict) else val


student_class=dic2class(student_dic)

#one way to edit:
student_class.class1.rank1['sex']='male'
student_class.class1.rank1['name']='Nan Xiang'

#two ways to query:
print student_class.class1.rank1
print student_class.class1['rank1']
print '-'*50
for rank in student_class.class1:
    print getattr(student_class.class1,rank)

结果:

{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
--------------------------------------------------
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': 'male', 'name': 'Nan Xiang', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}
{'score': '', 'sex': '', 'name': '', 'weight': '', 'height': ''}