我发现它更方便访问字典键作为obj。foo而不是obj['foo']，所以我写了这个片段:
class AttributeDict(dict):
def __getattr__(self, attr):
return self[attr]
def __setattr__(self, attr, value):
self[attr] = value
然而，我认为一定有一些原因，Python没有提供开箱即用的功能。以这种方式访问字典键的注意事项和缺陷是什么?
当前回答
这个答案摘自Luciano Ramalho的《流利的Python》一书。这要归功于那个家伙。
class AttrDict:
"""A read-only façade for navigating a JSON-like object
using attribute notation
"""
def __init__(self, mapping):
self._data = dict(mapping)
def __getattr__(self, name):
if hasattr(self._data, name):
return getattr(self._data, name)
else:
return AttrDict.build(self._data[name])
@classmethod
def build(cls, obj):
if isinstance(obj, Mapping):
return cls(obj)
elif isinstance(obj, MutableSequence):
return [cls.build(item) for item in obj]
else:
return obj
in the init we are taking the dict and making it a dictionary. when getattr is used we try to get the attribute from the dict if the dict already has that attribute. or else we are passing the argument to a class method called build. now build does the intresting thing. if the object is dict or a mapping like that, the that object is made an attr dict itself. if it's a sequence like list, it's passed to the build function we r on right now. if it's anythin else, like str or int. return the object itself.
其他回答
product怎么样，我写了一个小小的Python类来统治它们:)
此外，您还可以获得自动代码完成、递归对象实例化和自动类型转换!
你完全可以做到你所要求的:
p = Prodict()
p.foo = 1
p.bar = "baz"
例1:类型提示
class Country(Prodict):
name: str
population: int
turkey = Country()
turkey.name = 'Turkey'
turkey.population = 79814871
例2:自动类型转换
germany = Country(name='Germany', population='82175700', flag_colors=['black', 'red', 'yellow'])
print(germany.population) # 82175700
print(type(germany.population)) # <class 'int'>
print(germany.flag_colors) # ['black', 'red', 'yellow']
print(type(germany.flag_colors)) # <class 'list'>
如果使用数组表示法，则可以将所有合法字符串字符作为键的一部分。 例如，obj['!#$%^&*()_']
如果你想要一个方法的键，比如__eq__或__getattr__呢?
而且不能有一个不以字母开头的条目，因此使用0343853作为键是无效的。
如果你不想使用字符串呢?
最简单的方法是定义一个类，我们称之为Namespace。在字典上使用对象dict.update()。然后，字典将被视为一个对象。
class Namespace(object):
'''
helps referencing object in a dictionary as dict.key instead of dict['key']
'''
def __init__(self, adict):
self.__dict__.update(adict)
Person = Namespace({'name': 'ahmed',
'age': 30}) #--> added for edge_cls
print(Person.name)
从另一个SO问题中，有一个很好的实现示例，可以简化现有的代码。如何:
class AttributeDict(dict):
__slots__ = ()
__getattr__ = dict.__getitem__
__setattr__ = dict.__setitem__
更加简洁，并且不会为将来的__getattr__和__setattr__函数留下任何额外的麻烦空间。
