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
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'
其他回答
kaggle_environments使用的实现是一个名为structify的函数。
class Struct(dict):
def __init__(self, **entries):
entries = {k: v for k, v in entries.items() if k != "items"}
dict.__init__(self, entries)
self.__dict__.update(entries)
def __setattr__(self, attr, value):
self.__dict__[attr] = value
self[attr] = value
# Added benefit of cloning lists and dicts.
def structify(o):
if isinstance(o, list):
return [structify(o[i]) for i in range(len(o))]
elif isinstance(o, dict):
return Struct(**{k: structify(v) for k, v in o.items()})
return o
https://github.com/Kaggle/kaggle-environments/blob/master/kaggle_environments/utils.py
这可能有助于在《ConnectX》等游戏中测试AI模拟代理
from kaggle_environments import structify
obs = structify({ 'remainingOverageTime': 60, 'step': 0, 'mark': 1, 'board': [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]})
conf = structify({ 'timeout': 2, 'actTimeout': 2, 'agentTimeout': 60, 'episodeSteps': 1000, 'runTimeout': 1200, 'columns': 7, 'rows': 6, 'inarow': 4, '__raw_path__': '/kaggle_simulations/agent/main.py' })
def agent(obs, conf):
action = obs.step % conf.columns
return action
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!
不是对OP问题的直接回答,但受到启发,也许对一些人有用。我已经创建了一个基于对象的解决方案使用内部__dict__(在任何方式优化代码)
payload = {
"name": "John",
"location": {
"lat": 53.12312312,
"long": 43.21345112
},
"numbers": [
{
"role": "home",
"number": "070-12345678"
},
{
"role": "office",
"number": "070-12345679"
}
]
}
class Map(object):
"""
Dot style access to object members, access raw values
with an underscore e.g.
class Foo(Map):
def foo(self):
return self.get('foo') + 'bar'
obj = Foo(**{'foo': 'foo'})
obj.foo => 'foobar'
obj._foo => 'foo'
"""
def __init__(self, *args, **kwargs):
for arg in args:
if isinstance(arg, dict):
for k, v in arg.iteritems():
self.__dict__[k] = v
self.__dict__['_' + k] = v
if kwargs:
for k, v in kwargs.iteritems():
self.__dict__[k] = v
self.__dict__['_' + k] = v
def __getattribute__(self, attr):
if hasattr(self, 'get_' + attr):
return object.__getattribute__(self, 'get_' + attr)()
else:
return object.__getattribute__(self, attr)
def get(self, key):
try:
return self.__dict__.get('get_' + key)()
except (AttributeError, TypeError):
return self.__dict__.get(key)
def __repr__(self):
return u"<{name} object>".format(
name=self.__class__.__name__
)
class Number(Map):
def get_role(self):
return self.get('role')
def get_number(self):
return self.get('number')
class Location(Map):
def get_latitude(self):
return self.get('lat') + 1
def get_longitude(self):
return self.get('long') + 1
class Item(Map):
def get_name(self):
return self.get('name') + " Doe"
def get_location(self):
return Location(**self.get('location'))
def get_numbers(self):
return [Number(**n) for n in self.get('numbers')]
# Tests
obj = Item({'foo': 'bar'}, **payload)
assert type(obj) == Item
assert obj._name == "John"
assert obj.name == "John Doe"
assert type(obj.location) == Location
assert obj.location._lat == 53.12312312
assert obj.location._long == 43.21345112
assert obj.location.latitude == 54.12312312
assert obj.location.longitude == 44.21345112
for n in obj.numbers:
assert type(n) == Number
if n.role == 'home':
assert n.number == "070-12345678"
if n.role == 'office':
assert n.number == "070-12345679"
@derek73的答案非常简洁,但它不能被pickle或(深度)复制,并且它在缺少键时返回None。下面的代码修复了这个问题。
编辑:我没有看到上面的答案解决了完全相同的问题(点赞)。我把答案留在这里供参考。
class dotdict(dict):
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__
def __getattr__(self, name):
try:
return self[name]
except KeyError:
raise AttributeError(name)
我只需要使用虚线路径字符串访问字典,所以我想到了:
def get_value_from_path(dictionary, parts):
""" extracts a value from a dictionary using a dotted path string """
if type(parts) is str:
parts = parts.split('.')
if len(parts) > 1:
return get_value_from_path(dictionary[parts[0]], parts[1:])
return dictionary[parts[0]]
a = {'a':{'b':'c'}}
print(get_value_from_path(a, 'a.b')) # c