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?
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
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()
其他回答
更新:在Python 2.6及以上版本中,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(原答案内容)为:
class Struct:
def __init__(self, **entries):
self.__dict__.update(entries)
然后,你可以使用:
>>> args = {'a': 1, 'b': 2}
>>> s = Struct(**args)
>>> s
<__main__.Struct instance at 0x01D6A738>
>>> s.a
1
>>> s.b
2
有一个 名为namedtuple的集合助手,可以为你做这些:
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
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]遗漏了在列表中处理字典的部分,而不仅仅是其他字典。添加修复。
令人惊讶的是,没有人提到邦奇。这个库专门用于提供对dict对象的属性样式访问,并完全符合OP的要求。一个示范:
>>> 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'
Python 3的库可以在https://github.com/Infinidat/munch上获得-来源是codyzu
>>> from munch import DefaultMunch
>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
>>> obj = DefaultMunch.fromDict(d)
>>> obj.b.c
2
>>> obj.a
1
>>> obj.d[1].foo
'bar'
你可以使用一个自定义对象钩子来利用标准库的json模块:
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)
使用示例:
>>> 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'
而且它不像namedtuple那样是严格只读的,也就是说,你可以改变值-而不是结构:
>>> o.b.c = 3
>>> o.b.c
3