我如何使Python字典成员访问通过点“。”?
例如,我想写mydict.val而不是mydict['val']。
我还想以这种方式访问嵌套字典。例如
mydict.mydict2.val
会提到
mydict = { 'mydict2': { 'val': ... } }
我如何使Python字典成员访问通过点“。”?
例如,我想写mydict.val而不是mydict['val']。
我还想以这种方式访问嵌套字典。例如
mydict.mydict2.val
会提到
mydict = { 'mydict2': { 'val': ... } }
当前回答
语言本身不支持这一点,但有时这仍然是一个有用的需求。除了Bunch recipe,你还可以写一个小方法,可以使用虚线字符串访问字典:
def get_var(input_dict, accessor_string):
"""Gets data from a dictionary using a dotted accessor-string"""
current_data = input_dict
for chunk in accessor_string.split('.'):
current_data = current_data.get(chunk, {})
return current_data
这将支持如下内容:
>> test_dict = {'thing': {'spam': 12, 'foo': {'cheeze': 'bar'}}}
>> output = get_var(test_dict, 'thing.spam.foo.cheeze')
>> print output
'bar'
>>
其他回答
Fabric有一个非常好的、最小的实现。将其扩展为允许嵌套访问,我们可以使用defaultdict,结果看起来像这样:
from collections import defaultdict
class AttributeDict(defaultdict):
def __init__(self):
super(AttributeDict, self).__init__(AttributeDict)
def __getattr__(self, key):
try:
return self[key]
except KeyError:
raise AttributeError(key)
def __setattr__(self, key, value):
self[key] = value
可以这样使用它:
keys = AttributeDict()
keys.abc.xyz.x = 123
keys.abc.xyz.a.b.c = 234
这详细阐述了Kugel的回答“从dict和派生并实现__getattr__和__setattr__”。现在你知道怎么做了!
用于无限级别的字典、列表、字典的列表和列表的字典的嵌套。
它还支持酸洗
这是这个答案的延伸。
class DotDict(dict):
# https://stackoverflow.com/a/70665030/913098
"""
Example:
m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])
Iterable are assumed to have a constructor taking list as input.
"""
def __init__(self, *args, **kwargs):
super(DotDict, self).__init__(*args, **kwargs)
args_with_kwargs = []
for arg in args:
args_with_kwargs.append(arg)
args_with_kwargs.append(kwargs)
args = args_with_kwargs
for arg in args:
if isinstance(arg, dict):
for k, v in arg.items():
self[k] = v
if isinstance(v, dict):
self[k] = DotDict(v)
elif isinstance(v, str) or isinstance(v, bytes):
self[k] = v
elif isinstance(v, Iterable):
klass = type(v)
map_value: List[Any] = []
for e in v:
map_e = DotDict(e) if isinstance(e, dict) else e
map_value.append(map_e)
self[k] = klass(map_value)
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(DotDict, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(DotDict, self).__delitem__(key)
del self.__dict__[key]
def __getstate__(self):
return self.__dict__
def __setstate__(self, d):
self.__dict__.update(d)
if __name__ == "__main__":
import pickle
def test_map():
d = {
"a": 1,
"b": {
"c": "d",
"e": 2,
"f": None
},
"g": [],
"h": [1, "i"],
"j": [1, "k", {}],
"l":
[
1,
"m",
{
"n": [3],
"o": "p",
"q": {
"r": "s",
"t": ["u", 5, {"v": "w"}, ],
"x": ("z", 1)
}
}
],
}
map_d = DotDict(d)
w = map_d.l[2].q.t[2].v
assert w == "w"
pickled = pickle.dumps(map_d)
unpickled = pickle.loads(pickled)
assert unpickled == map_d
kwargs_check = DotDict(a=1, b=[dict(c=2, d="3"), 5])
assert kwargs_check.b[0].d == "3"
kwargs_and_args_check = DotDict(d, a=1, b=[dict(c=2, d="3"), 5])
assert kwargs_and_args_check.l[2].q.t[2].v == "w"
assert kwargs_and_args_check.b[0].d == "3"
test_map()
如果你想pickle你修改后的字典,你需要添加几个状态方法到上面的答案:
class DotDict(dict):
"""dot.notation access to dictionary attributes"""
def __getattr__(self, attr):
return self.get(attr)
__setattr__= dict.__setitem__
__delattr__= dict.__delitem__
def __getstate__(self):
return self
def __setstate__(self, state):
self.update(state)
self.__dict__ = self
我试了一下:
class dotdict(dict):
def __getattr__(self, name):
return self[name]
你也可以尝试__getattribute__。
使每个字典都是一种类型的dotdict就足够了,如果你想从多层字典初始化它,也可以尝试实现__init__。
使用__getattr__,非常简单,适用于 Python 3.4.3
class myDict(dict):
def __getattr__(self,val):
return self[val]
blockBody=myDict()
blockBody['item1']=10000
blockBody['item2']="StackOverflow"
print(blockBody.item1)
print(blockBody.item2)
输出:
10000
StackOverflow