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
from mock import Mock
d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}
my_data = Mock(**d)
# We got
# my_data.a == 1
其他回答
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'
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'}}
如果你想让它递归的话,在之前接受的答案所做的基础上。
class FullStruct:
def __init__(self, **kwargs):
for key, value in kwargs.items():
if isinstance(value, dict):
f = FullStruct(**value)
self.__dict__.update({key: f})
else:
self.__dict__.update({key: value})
最简单的方法是使用collections.namedtuple。
我发现下面的4行代码是最漂亮的,它支持嵌套字典:
def dict_to_namedtuple(typename, data):
return namedtuple(typename, data.keys())(
*(dict_to_namedtuple(typename + '_' + k, v) if isinstance(v, dict) else v for k, v in data.items())
)
输出看起来也会很好:
>>> nt = dict_to_namedtuple('config', {
... 'path': '/app',
... 'debug': {'level': 'error', 'stream': 'stdout'}
... })
>>> print(nt)
config(path='/app', debug=config_debug(level='error', stream='stdout'))
>>> print(nt.debug.level)
'error'
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。