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
不喜欢。在Python中,属性访问和索引是分开的事情,您不应该希望它们执行相同的操作。创建一个类(可能是由namedtuple创建的),如果你有一些应该具有可访问属性的东西,并使用[]符号从字典中获取一个项。
其他回答
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'
一个很微妙的解
class DotDict(dict):
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__
def __getattr__(self, key):
def typer(candidate):
if isinstance(candidate, dict):
return DotDict(candidate)
if isinstance(candidate, str): # iterable but no need to iter
return candidate
try: # other iterable are processed as list
return [typer(item) for item in candidate]
except TypeError:
return candidate
return candidate
return typer(dict.get(self, key))
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
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
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