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
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()
其他回答
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__。
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'
>>
可以使用dotsi来支持完整列表、dict和递归,并使用一些扩展方法
pip install dotsi
and
>>> import dotsi
>>>
>>> d = dotsi.Dict({"foo": {"bar": "baz"}}) # Basic
>>> d.foo.bar
'baz'
>>> d.users = [{"id": 0, "name": "Alice"}] # List
>>> d.users[0].name
'Alice'
>>> d.users.append({"id": 1, "name": "Becca"}); # Append
>>> d.users[1].name
'Becca'
>>> d.users += [{"id": 2, "name": "Cathy"}]; # `+=`
>>> d.users[2].name
'Cathy'
>>> d.update({"tasks": [{"id": "a", "text": "Task A"}]});
>>> d.tasks[0].text
'Task A'
>>> d.tasks[0].tags = ["red", "white", "blue"];
>>> d.tasks[0].tags[2];
'blue'
>>> d.tasks[0].pop("tags") # `.pop()`
['red', 'white', 'blue']
>>>
>>> import pprint
>>> pprint.pprint(d)
{'foo': {'bar': 'baz'},
'tasks': [{'id': 'a', 'text': 'Task A'}],
'users': [{'id': 0, 'name': 'Alice'},
{'id': 1, 'name': 'Becca'},
{'id': 2, 'name': 'Cathy'}]}
>>>
>>> type(d.users) # dotsi.Dict (AKA dotsi.DotsiDict)
<class 'dotsi.DotsiList'>
>>> type(d.users[0]) # dotsi.List (AKA dotsi.DotsiList)
<class 'dotsi.DotsiDict'>
>>>
我一直把它保存在util文件中。您也可以在自己的类中使用它作为mixin。
class dotdict(dict):
"""dot.notation access to dictionary attributes"""
__getattr__ = dict.get
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__
mydict = {'val':'it works'}
nested_dict = {'val':'nested works too'}
mydict = dotdict(mydict)
mydict.val
# 'it works'
mydict.nested = dotdict(nested_dict)
mydict.nested.val
# 'nested works too'
这是我从很久以前的一个项目里挖出来的。它可能还可以再优化一点,但就是这样了。
class DotNotation(dict):
__setattr__= dict.__setitem__
__delattr__= dict.__delitem__
def __init__(self, data):
if isinstance(data, str):
data = json.loads(data)
for name, value in data.items():
setattr(self, name, self._wrap(value))
def __getattr__(self, attr):
def _traverse(obj, attr):
if self._is_indexable(obj):
try:
return obj[int(attr)]
except:
return None
elif isinstance(obj, dict):
return obj.get(attr, None)
else:
return attr
if '.' in attr:
return reduce(_traverse, attr.split('.'), self)
return self.get(attr, None)
def _wrap(self, value):
if self._is_indexable(value):
# (!) recursive (!)
return type(value)([self._wrap(v) for v in value])
elif isinstance(value, dict):
return DotNotation(value)
else:
return value
@staticmethod
def _is_indexable(obj):
return isinstance(obj, (tuple, list, set, frozenset))
if __name__ == "__main__":
test_dict = {
"dimensions": {
"length": "112",
"width": "103",
"height": "42"
},
"meta_data": [
{
"id": 11089769,
"key": "imported_gallery_files",
"value": [
"https://example.com/wp-content/uploads/2019/09/unnamed-3.jpg",
"https://example.com/wp-content/uploads/2019/09/unnamed-2.jpg",
"https://example.com/wp-content/uploads/2019/09/unnamed-4.jpg"
]
}
]
}
dotted_dict = DotNotation(test_dict)
print(dotted_dict.dimensions.length) # => '112'
print(getattr(dotted_dict, 'dimensions.length')) # => '112'
print(dotted_dict.meta_data[0].key) # => 'imported_gallery_files'
print(getattr(dotted_dict, 'meta_data.0.key')) # => 'imported_gallery_files'
print(dotted_dict.meta_data[0].value) # => ['link1','link2','link2']
print(getattr(dotted_dict, 'meta_data.0.value')) # => ['link1','link2','link3']
print(dotted_dict.meta_data[0].value[2]) # => 'link3'
print(getattr(dotted_dict, 'meta_data.0.value.2')) # => 'link3'