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
获得点访问(但不是数组访问)的一个简单方法是在Python中使用一个普通对象。是这样的:
class YourObject:
def __init__(self, *args, **kwargs):
for k, v in kwargs.items():
setattr(self, k, v)
...像这样使用它:
>>> obj = YourObject(key="value")
>>> print(obj.key)
"value"
... 把它转换成字典:
>>> print(obj.__dict__)
{"key": "value"}
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__。
基于Kugel的回答,并考虑到Mike Graham的警告,如果我们制作一个包装器呢?
class DictWrap(object):
""" Wrap an existing dict, or create a new one, and access with either dot
notation or key lookup.
The attribute _data is reserved and stores the underlying dictionary.
When using the += operator with create=True, the empty nested dict is
replaced with the operand, effectively creating a default dictionary
of mixed types.
args:
d({}): Existing dict to wrap, an empty dict is created by default
create(True): Create an empty, nested dict instead of raising a KeyError
example:
>>>dw = DictWrap({'pp':3})
>>>dw.a.b += 2
>>>dw.a.b += 2
>>>dw.a['c'] += 'Hello'
>>>dw.a['c'] += ' World'
>>>dw.a.d
>>>print dw._data
{'a': {'c': 'Hello World', 'b': 4, 'd': {}}, 'pp': 3}
"""
def __init__(self, d=None, create=True):
if d is None:
d = {}
supr = super(DictWrap, self)
supr.__setattr__('_data', d)
supr.__setattr__('__create', create)
def __getattr__(self, name):
try:
value = self._data[name]
except KeyError:
if not super(DictWrap, self).__getattribute__('__create'):
raise
value = {}
self._data[name] = value
if hasattr(value, 'items'):
create = super(DictWrap, self).__getattribute__('__create')
return DictWrap(value, create)
return value
def __setattr__(self, name, value):
self._data[name] = value
def __getitem__(self, key):
try:
value = self._data[key]
except KeyError:
if not super(DictWrap, self).__getattribute__('__create'):
raise
value = {}
self._data[key] = value
if hasattr(value, 'items'):
create = super(DictWrap, self).__getattribute__('__create')
return DictWrap(value, create)
return value
def __setitem__(self, key, value):
self._data[key] = value
def __iadd__(self, other):
if self._data:
raise TypeError("A Nested dict will only be replaced if it's empty")
else:
return other
我最近遇到了“Box”库,它也做同样的事情。
安装命令:pip install python-box
例子:
from box import Box
mydict = {"key1":{"v1":0.375,
"v2":0.625},
"key2":0.125,
}
mydict = Box(mydict)
print(mydict.key1.v1)
我发现它比其他现有的库(如dotmap)更有效,当你有大量嵌套字典时,dotmap会产生python递归错误。
链接到图书馆和详细信息:https://pypi.org/project/python-box/
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
派生自dict和并实现__getattr__和__setattr__。
或者你也可以用Bunch,非常相似。
我不认为这是可能的monkeypatch内置字典类。