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
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
其他回答
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'
>>
I ended up trying BOTH the AttrDict and the Bunch libraries and found them to be way to slow for my uses. After a friend and I looked into it, we found that the main method for writing these libraries results in the library aggressively recursing through a nested object and making copies of the dictionary object throughout. With this in mind, we made two key changes. 1) We made attributes lazy-loaded 2) instead of creating copies of a dictionary object, we create copies of a light-weight proxy object. This is the final implementation. The performance increase of using this code is incredible. When using AttrDict or Bunch, these two libraries alone consumed 1/2 and 1/3 respectively of my request time(what!?). This code reduced that time to almost nothing(somewhere in the range of 0.5ms). This of course depends on your needs, but if you are using this functionality quite a bit in your code, definitely go with something simple like this.
class DictProxy(object):
def __init__(self, obj):
self.obj = obj
def __getitem__(self, key):
return wrap(self.obj[key])
def __getattr__(self, key):
try:
return wrap(getattr(self.obj, key))
except AttributeError:
try:
return self[key]
except KeyError:
raise AttributeError(key)
# you probably also want to proxy important list properties along like
# items(), iteritems() and __len__
class ListProxy(object):
def __init__(self, obj):
self.obj = obj
def __getitem__(self, key):
return wrap(self.obj[key])
# you probably also want to proxy important list properties along like
# __iter__ and __len__
def wrap(value):
if isinstance(value, dict):
return DictProxy(value)
if isinstance(value, (tuple, list)):
return ListProxy(value)
return value
参见https://stackoverflow.com/users/704327/michael-merickel的原始实现。
另一件需要注意的事情是,这个实现非常简单,并且没有实现您可能需要的所有方法。您需要根据需要在DictProxy或ListProxy对象上写入这些内容。
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
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__。