将嵌套的Python dict转换为对象？-Java 学习之路

468

我正在寻找一种优雅的方法，使用一些嵌套的dicts和列表（即javascript样式的对象语法）在dict上使用属性访问来获取数据 .

例如：

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

应该可以通过这种方式访问：

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

我认为，如果没有递归，这是不可能的，但是什么是获得dicts的对象样式的好方法？

30 回答

有一个名为namedtuple的集合助手，可以为你做到这一点：

from collections import namedtuple

d_named = namedtuple('Struct', d.keys())(*d.values())

In [7]: d_named
Out[7]: Struct(a=1, b={'c': 2}, d=['hi', {'foo': 'bar'}])

In [8]: d_named.a
Out[8]: 1

回复于 2024-05-10T04:37:44+08:00

Update: 在Python 2.6及更高版本中，考虑namedtuple数据结构是否适合您的需求：

>>> from collections import namedtuple
>>> MyStruct = namedtuple('MyStruct', 'a b d')
>>> s = MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s
MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s.a
1
>>> s.b
{'c': 2}
>>> s.c
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'MyStruct' object has no attribute 'c'
>>> s.d
['hi']

替代方案（原始答案内容）是：

class Struct:
    def __init__(self, **entries):
        self.__dict__.update(entries)

然后，您可以使用：

>>> args = {'a': 1, 'b': 2}
>>> s = Struct(**args)
>>> s
<__main__.Struct instance at 0x01D6A738>
>>> s.a
1
>>> s.b
2

回复于 2024-05-10T04:37:44+08:00

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

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

回复于 2024-05-10T04:37:44+08:00

1
令人惊讶的是没有人提到Bunch . 这个库专门用于提供对dict对象的属性样式访问，并且完全符合OP的要求 . 示范：
```
>>> from bunch import bunchify
>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = bunchify(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'
```
一个Python 3库可以在https://github.com/Infinidat/munch获得 - Credit to codyzu
回复于 2024-05-10T04:37:44+08:00

x = type('new_dict', (object,), d)

然后添加递归到这个，你就完成了 .

edit 这就是我实现它的方式：

>>> 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'

回复于 2024-05-10T04:37:44+08:00

我认为这是前面例子中最好的方面，这就是我想出的：

class Struct:
  '''The recursive class for building and representing objects with.'''
  def __init__(self, obj):
    for k, v in obj.iteritems():
      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__.iteritems()))

回复于 2024-05-10T04:37:44+08:00

class Struct(object):
    """Comment removed"""
    def __init__(self, data):
        for name, value in data.iteritems():
            setattr(self, name, self._wrap(value))

    def _wrap(self, value):
        if isinstance(value, (tuple, list, set, frozenset)): 
            return type(value)([self._wrap(v) for v in value])
        else:
            return Struct(value) if isinstance(value, dict) else value

可以与任何深度的任何序列/字典/值结构一起使用 .

回复于 2024-05-10T04:37:44+08:00

2
如果您的dict来自 json.loads() ，您可以在一行中将其转换为对象（而不是dict）：
```
import json
from collections import namedtuple

json.loads(data, object_hook=lambda d: namedtuple('X', d.keys())(*d.values()))
```
另见How to convert JSON data into a Python object .
回复于 2024-05-10T04:37:44+08:00

如果你想要将dict键作为一个对象（或者作为一个困难键的dict）来访问，那么递归执行它，并且还能够更新原始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)

用法示例：

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'}}

回复于 2024-05-10T04:37:44+08:00

>>> def dict2obj(d):
        if isinstance(d, list):
            d = [dict2obj(x) for x in d]
        if not isinstance(d, dict):
            return d
        class C(object):
            pass
        o = C()
        for k in d:
            o.__dict__[k] = dict2obj(d[k])
        return o


>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
'bar'

回复于 2024-05-10T04:37:44+08:00

2
我最终尝试了AttrDict和Bunch库，发现它们对我的使用来说太慢了 . 在我和朋友调查之后，我们发现编写这些库的主要方法导致库通过嵌套对象进行主动递归，并在整个过程中复制字典对象 . 考虑到这一点，我们做了两个关键的改变 . 1）我们使属性延迟加载2）而不是创建字典对象的副本，我们创建轻量级代理对象的副本 . 这是最终的实施 . 使用此代码的性能提升令人难以置信 . 当使用AttrDict或Bunch时，这两个库分别占用了我的请求时间的1/2和1/3（什么！？） . 这段代码将时间缩短到几乎为零（在0.5ms的范围内） . 这当然取决于您的需求，但如果您在代码中使用此功能，请务必使用这样的简单方法 .
```
class DictProxy(object):
    def __init__(self, obj):
        self.obj = obj

    def __getitem__(self, key):
        return wrap(self.obj[key])

    def __getattr__(self, key):
        try:
            return wrap(getattr(self.obj, key))
        except AttributeError:
            try:
                return self[key]
            except KeyError:
                raise AttributeError(key)

    # you probably also want to proxy important list properties along like
    # items(), iteritems() and __len__

class ListProxy(object):
    def __init__(self, obj):
        self.obj = obj

    def __getitem__(self, key):
        return wrap(self.obj[key])

    # you probably also want to proxy important list properties along like
    # __iter__ and __len__

def wrap(value):
    if isinstance(value, dict):
        return DictProxy(value)
    if isinstance(value, (tuple, list)):
        return ListProxy(value)
    return value
```
请参阅https://stackoverflow.com/users/704327/michael-merickel的原始实现here .

另外需要注意的是，这个实现非常简单，并没有实现您可能需要的所有方法 . 您需要在DictProxy或ListProxy对象上根据需要编写它们 .
回复于 2024-05-10T04:37:44+08:00
3

x.__dict__.update(d) 应该没问题 .

回复于 2024-05-10T04:37:44+08:00

这应该让你开始：

class dict2obj(object):
    def __init__(self, d):
        self.__dict__['d'] = d

    def __getattr__(self, key):
        value = self.__dict__['d'][key]
        if type(value) == type({}):
            return dict2obj(value)

        return value

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

x = dict2obj(d)
print x.a
print x.b.c
print x.d[1].foo

它不需要将列表包装在UserList中并重载 __getitem__ 以包装dicts .

回复于 2024-05-10T04:37:44+08:00

旧的问答，但我得到更多的话题 . 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()

回复于 2024-05-10T04:37:44+08:00

您可以使用 custom object hook 来利用标准库的 json module ：

import json

class obj(object):
    def __init__(self, dict_):
        self.__dict__.update(dict_)

def dict2obj(d):
    return json.loads(json.dumps(d), object_hook=obj)

用法示例：

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ['hi', {'foo': 'bar'}]}
>>> o = dict2obj(d)
>>> o.a
1
>>> o.b.c
2
>>> o.d[0]
u'hi'
>>> o.d[1].foo
u'bar'

并且 not strictly read-only 与 namedtuple 一样，即你可以改变值 - 而不是结构：

>>> o.b.c = 3
>>> o.b.c
3

回复于 2024-05-10T04:37:44+08:00

我偶然发现了我需要以递归方式将一个dicts列表转换为对象列表的情况，所以根据Roberto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

请注意，出于显而易见的原因，任何元组都将转换为其等效列表 .

希望这可以帮助别人，就像你的所有答案一样，对我来说，伙计们 .

回复于 2024-05-10T04:37:44+08:00

想上传我的这个小范例的版本 .

class Struct(dict):
  def __init__(self,data):
    for key, value in data.items():
      if isinstance(value, dict):
        setattr(self, key, Struct(value))
      else:   
        setattr(self, key, type(value).__init__(value))

      dict.__init__(self,data)

它保留了导入到类中的类型的属性 . 我唯一关心的是从字典中覆盖你解析的方法 . 但除此之外看似坚实！

回复于 2024-05-10T04:37:44+08:00

from mock import Mock
d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
my_data = Mock(**d)

# We got
# my_data.a == 1

回复于 2024-05-10T04:37:44+08:00

将 dict 分配给空对象的 __dict__ 怎么样？

class Object:
    """If your dict is "flat", this is a simple way to create an object from a dict

    >>> obj = Object()
    >>> obj.__dict__ = d
    >>> d.a
    1
    """
    pass

当然，除非你递归地遍历dict，否则你的嵌套dict示例会失败：

# For a nested dict, you need to recursively update __dict__
def dict2obj(d):
    """Convert a dict to an object

    >>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
    >>> obj = dict2obj(d)
    >>> obj.b.c
    2
    >>> obj.d
    ["hi", {'foo': "bar"}]
    """
    try:
        d = dict(d)
    except (TypeError, ValueError):
        return d
    obj = Object()
    for k, v in d.iteritems():
        obj.__dict__[k] = dict2obj(v)
    return obj

你的示例列表元素可能是 Mapping ，这是（键，值）对的列表，如下所示：

>>> d = {'a': 1, 'b': {'c': 2}, 'd': [("hi", {'foo': "bar"})]}
>>> obj = dict2obj(d)
>>> obj.d.hi.foo
"bar"

回复于 2024-05-10T04:37:44+08:00

601

我知道这里已经有很多答案了，我迟到了，但是这个方法会递归并且“就地”将字典转换为类似对象的结构...在3.x.x中工作

def dictToObject(d):
    for k,v in d.items():
        if isinstance(v, dict):
            d[k] = dictToObject(v)
    return namedtuple('object', d.keys())(*d.values())

# Dictionary created from JSON file
d = {
    'primaryKey': 'id', 
    'metadata': 
        {
            'rows': 0, 
            'lastID': 0
        }, 
    'columns': 
        {
            'col2': {
                'dataType': 'string', 
                'name': 'addressLine1'
            }, 
            'col1': {
                'datatype': 'string', 
                'name': 'postcode'
            }, 
            'col3': {
                'dataType': 'string', 
                'name': 'addressLine2'
            }, 
            'col0': {
                'datatype': 'integer', 
                'name': 'id'
            }, 
            'col4': {
                'dataType': 'string', 
                'name': 'contactNumber'
            }
        }, 
        'secondaryKeys': {}
}

d1 = dictToObject(d)
d1.columns.col1 # == object(datatype='string', name='postcode')
d1.metadata.rows # == 0

回复于 2024-05-10T04:37:44+08:00

这是实现SilentGhost原始建议的另一种方法：

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)
  else:
    return d

回复于 2024-05-10T04:37:44+08:00

让我解释一下我前一段时间使用过的解决方案 . 但首先，我没有的原因通过以下代码说明：

d = {'from': 1}
x = dict2obj(d)

print x.from

给出了这个错误：

File "test.py", line 20
    print x.from == 1
                ^
SyntaxError: invalid syntax

因为“from”是一个Python关键字，所以你不能使用某些字典键允许 .

现在我的解决方案允许直接使用他们的名字访问字典项 . 但它也允许你使用“字典语义” . 以下是带有示例用法的代码：

class dict2obj(dict):
    def __init__(self, dict_):
        super(dict2obj, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj(it)
            elif isinstance(item, dict):
                self[key] = dict2obj(item)

    def __getattr__(self, key):
        return self[key]

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

x = dict2obj(d)

assert x.a == x['a'] == 1
assert x.b.c == x['b']['c'] == 2
assert x.d[1].foo == x['d'][1]['foo'] == "bar"

回复于 2024-05-10T04:37:44+08:00

这是另一个实现：

class DictObj(object):
    def __init__(self, d):
        self.__dict__ = d

def dict_to_obj(d):
    if isinstance(d, (list, tuple)): return map(dict_to_obj, d)
    elif not isinstance(d, dict): return d
    return DictObj(dict((k, dict_to_obj(v)) for (k,v) in d.iteritems()))

[编辑]错过了关于在列表中处理dicts的一点，而不仅仅是其他的dicts . 添加了修复 .

回复于 2024-05-10T04:37:44+08:00

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

回复于 2024-05-10T04:37:44+08:00

我认为dict由数字，字符串和dict组成，大部分时间都足够了 . 所以我忽略了元组，列表和其他类型没有出现在dict的最终维度中的情况 .

考虑到继承，结合递归，它方便地解决了打印问题，并提供了两种查询数据的方法，一种是编辑数据的方法 .

请参阅下面的示例，该描述描述了有关学生的一些信息：

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': ''}

回复于 2024-05-10T04:37:44+08:00

这也很有效

class DObj(object):
    pass

dobj = Dobj()
dobj.__dict__ = {'a': 'aaa', 'b': 'bbb'}

print dobj.a
>>> aaa
print dobj.b
>>> bbb

回复于 2024-05-10T04:37:44+08:00

class Struct(dict):
    def __getattr__(self, name):
        try:
            return self[name]
        except KeyError:
            raise AttributeError(name)

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

    def copy(self):
        return Struct(dict.copy(self))

用法：

points = Struct(x=1, y=2)
# Changing
points['x'] = 2
points.y = 1
# Accessing
points['x'], points.x, points.get('x') # 2 2 2
points['y'], points.y, points.get('y') # 1 1 1
# Accessing inexistent keys/attrs 
points['z'] # KeyError: z
points.z # AttributeError: z
# Copying
points_copy = points.copy()
points.x = 2
points_copy.x # 1

回复于 2024-05-10T04:37:44+08:00

29
Build 我对“python: How to add property to a class dynamically?”的回答：
```
class data(object):
    def __init__(self,*args,**argd):
        self.__dict__.update(dict(*args,**argd))

def makedata(d):
    d2 = {}
    for n in d:
        d2[n] = trydata(d[n])
    return data(d2)

def trydata(o):
    if isinstance(o,dict):
        return makedata(o)
    elif isinstance(o,list):
        return [trydata(i) for i in o]
    else:
        return o
```
您在要转换的字典上调用 makedata ，或者根据您期望的输入调用 trydata ，它会吐出一个数据对象 .

笔记：
- 如果需要更多功能，可以将elif添加到 trydata .
- 显然，如果你想要 x.a = {} 或类似的话，这将不起作用 .
- 如果需要只读版本，请使用the original answer中的类数据 .
回复于 2024-05-10T04:37:44+08:00

我有一些问题 __getattr__ 没有被调用所以我构建了一个新的样式类版本：

class Struct(object):
    '''The recursive class for building and representing objects with.'''
    class NoneStruct(object):
        def __getattribute__(*args):
            return Struct.NoneStruct()

        def __eq__(self, obj):
            return obj == None

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

    def __getattribute__(*args):
        try:
            return object.__getattribute__(*args)
        except:            
            return Struct.NoneStruct()

    def __repr__(self):
        return '{%s}' % str(', '.join('%s : %s' % (k, repr(v)) for 
(k, v) in self.__dict__.iteritems()))

此版本还添加了 NoneStruct ，在调用未设置的属性时返回 . 这允许None测试以查看属性是否存在 . 当确切的dict输入未知时（设置等）非常有用 .

bla = Struct({'a':{'b':1}})
print(bla.a.b)
>> 1
print(bla.a.c == None)
>> True

回复于 2024-05-10T04:37:44+08:00

我的字典是这种格式：

addr_bk = {
    'person': [
        {'name': 'Andrew', 'id': 123, 'email': 'andrew@mailserver.com',
         'phone': [{'type': 2, 'number': '633311122'},
                   {'type': 0, 'number': '97788665'}]
        },
        {'name': 'Tom', 'id': 456,
         'phone': [{'type': 0, 'number': '91122334'}]}, 
        {'name': 'Jack', 'id': 7788, 'email': 'jack@gmail.com'}
    ]
}

可以看出，我有 nested dictionaries 和 list of dicts . 这是因为addr_bk是从使用lwpb.codec转换为python dict的协议缓冲区数据解码的 . 存在可选字段（例如，电子邮件=>其中密钥可能不可用）和重复字段（例如，电话=>转换为字典列表） .

我尝试了以上提出的所有解决方案 . 有些不能很好地处理嵌套字典 . 其他人无法轻松打印对象细节 .

只有Dawie Strauss的解决方案dict2obj（dict）效果最好 .

当无法找到密钥时，我已经增强了一点处理：

# Work the best, with nested dictionaries & lists! :)
# Able to print out all items.
class dict2obj_new(dict):
    def __init__(self, dict_):
        super(dict2obj_new, self).__init__(dict_)
        for key in self:
            item = self[key]
            if isinstance(item, list):
                for idx, it in enumerate(item):
                    if isinstance(it, dict):
                        item[idx] = dict2obj_new(it)
            elif isinstance(item, dict):
                self[key] = dict2obj_new(item)

    def __getattr__(self, key):
        # Enhanced to handle key not found.
        if self.has_key(key):
            return self[key]
        else:
            return None

然后，我测试了它：

# Testing...
ab = dict2obj_new(addr_bk)

for person in ab.person:
  print "Person ID:", person.id
  print "  Name:", person.name
  # Check if optional field is available before printing.
  if person.email:
    print "  E-mail address:", person.email

  # Check if optional field is available before printing.
  if person.phone:
    for phone_number in person.phone:
      if phone_number.type == codec.enums.PhoneType.MOBILE:
        print "  Mobile phone #:",
      elif phone_number.type == codec.enums.PhoneType.HOME:
        print "  Home phone #:",
      else:
        print "  Work phone #:",
      print phone_number.number

回复于 2024-05-10T04:37:44+08:00

将嵌套的Python dict转换为对象？

30 回答

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