如何通过点“。”访问Python字典成员?
例如,我不想写mydict['val']
,而是想写mydict.val
。
此外,我想以这种方式访问嵌套的dicts。例如
mydict.mydict2.val
会参考
mydict = { 'mydict2': { 'val': ... } }
你可以使用我刚刚创建的这个类来完成它。使用此类,您可以像使用另一个字典(包括json序列化)或点符号一样使用Map
对象。我希望能帮助你:
class Map(dict):
"""
Example:
m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])
"""
def __init__(self, *args, **kwargs):
super(Map, self).__init__(*args, **kwargs)
for arg in args:
if isinstance(arg, dict):
for k, v in arg.iteritems():
self[k] = v
if kwargs:
for k, v in kwargs.iteritems():
self[k] = v
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(Map, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(Map, self).__delitem__(key)
del self.__dict__[key]
用法示例:
m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])
# Add new key
m.new_key = 'Hello world!'
# Or
m['new_key'] = 'Hello world!'
print m.new_key
print m['new_key']
# Update values
m.new_key = 'Yay!'
# Or
m['new_key'] = 'Yay!'
# Delete key
del m.new_key
# Or
del m['new_key']
我喜欢Munch,它在点访问之上提供了许多方便的选项。
进口蒙克
temp_1 = {'person':{'fname':'senthil','lname':'ramalingam'}}
dict_munch = munch.munchify(temp_1)
dict_munch.person.fname
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
语言本身不支持这一点,但有时这仍然是一个有用的要求。除了束配方,您还可以编写一个可以使用虚线字符串访问字典的方法:
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
这将支持这样的事情:
>> test_dict = {'thing': {'spam': 12, 'foo': {'cheeze': 'bar'}}}
>> output = get_var(test_dict, 'thing.spam.foo.cheeze')
>> print output
'bar'
>>
def dict_to_object(dick):
# http://stackoverflow.com/a/1305663/968442
class Struct:
def __init__(self, **entries):
self.__dict__.update(entries)
return Struct(**dick)
如果一个人决定将dict
永久转换为对象,那么应该这样做。您可以在访问之前创建一次性对象。
d = dict_to_object(d)
为了建立epool的答案,这个版本允许你通过点运算符访问里面的任何dict:
foo = {
"bar" : {
"baz" : [ {"boo" : "hoo"} , {"baba" : "loo"} ]
}
}
例如,foo.bar.baz[1].baba
返回"loo"
。
class Map(dict):
def __init__(self, *args, **kwargs):
super(Map, self).__init__(*args, **kwargs)
for arg in args:
if isinstance(arg, dict):
for k, v in arg.iteritems():
if isinstance(v, dict):
v = Map(v)
if isinstance(v, list):
self.__convert(v)
self[k] = v
if kwargs:
for k, v in kwargs.iteritems():
if isinstance(v, dict):
v = Map(v)
elif isinstance(v, list):
self.__convert(v)
self[k] = v
def __convert(self, v):
for elem in xrange(0, len(v)):
if isinstance(v[elem], dict):
v[elem] = Map(v[elem])
elif isinstance(v[elem], list):
self.__convert(v[elem])
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(Map, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(Map, self).__delitem__(key)
del self.__dict__[key]
我最终尝试了AttrDict和Bunch库,并发现它们可以减缓我的用途。在我和朋友调查之后,我们发现编写这些库的主要方法导致库通过嵌套对象进行主动递归,并在整个过程中复制字典对象。考虑到这一点,我们做了两个关键的改变。 1)我们使属性延迟加载2)而不是创建字典对象的副本,我们创建轻量级代理对象的副本。这是最终的实施。使用此代码的性能提升令人难以置信。当使用AttrDict或Bunch时,这两个库分别占用了我的请求时间的1/2和1/3(什么!?)。这段代码将时间缩短到几乎没有(在0.5ms的范围内)。这当然取决于您的需求,但如果您在代码中使用此功能,请务必使用这样的简单方法。
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
请参阅here的原始实现https://stackoverflow.com/users/704327/michael-merickel。
另外需要注意的是,这个实现非常简单,并没有实现您可能需要的所有方法。您需要在DictProxy或ListProxy对象上根据需要编写它们。
我想把自己的解决方案扔进戒指:
https://github.com/skorokithakis/jsane
它允许您将JSON解析为可以访问with.attribute.lookups.like.this.r()
的东西,主要是因为在开始处理它之前我没有看到这个答案。
不是对OP问题的直接回答,而是受到某些人的启发,也许对某些人有用..我使用内部__dict__
创建了一个基于对象的解决方案(绝不是优化代码)
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"
获得点访问(但不是数组访问)的一种简单方法是在Python中使用普通对象。像这样:
class YourObject:
def __init__(self, *args, **kwargs):
for k, v in kwargs.items():
setattr(self, k, v)
......并像这样使用它:
>>> obj = YourObject(key="value")
>>> print(obj.key)
"value"
...将其转换为字典:
>>> print(obj.__dict__)
{"key": "value"}
该解决方案是对epool提供的解决方案的改进,以解决OP以一致方式访问嵌套dicts的要求。 epool的解决方案不允许访问嵌套的dicts。
class YAMLobj(dict):
def __init__(self, args):
super(YAMLobj, self).__init__(args)
if isinstance(args, dict):
for k, v in args.iteritems():
if not isinstance(v, dict):
self[k] = v
else:
self.__setattr__(k, YAMLobj(v))
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(YAMLobj, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(YAMLobj, self).__delitem__(key)
del self.__dict__[key]
通过这个课程,人们现在可以做类似的事情:A.B.C.D
。
我一直把它保存在一个util文件中。您也可以在自己的课程中使用它作为mixin。
class dotdict(dict):
"""dot.notation access to dictionary attributes"""
__getattr__ = dict.get
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__
mydict = {'val':'it works'}
nested_dict = {'val':'nested works too'}
mydict = dotdict(mydict)
mydict.val
# 'it works'
mydict.nested = dotdict(nested_dict)
mydict.nested.val
# 'nested works too'
一种解决方案的细腻
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))
这也适用于嵌套的dicts,并确保后面附加的dicts行为相同:
class DotDict(dict):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
# Recursively turn nested dicts into DotDicts
for key, value in self.items():
if type(value) is dict:
self[key] = DotDict(value)
def __setitem__(self, key, item):
if type(item) is dict:
item = DotDict(item)
super().__setitem__(key, item)
__setattr__ = __setitem__
__getattr__ = dict.__getitem__
通过dotmap
安装pip
pip install dotmap
它完成了你想要它做的所有事情和子类dict
,所以它像普通的字典一样运行:
from dotmap import DotMap
m = DotMap()
m.hello = 'world'
m.hello
m.hello += '!'
# m.hello and m['hello'] now both return 'world!'
m.val = 5
m.val2 = 'Sam'
最重要的是,您可以将它转换为dict
对象:
d = m.toDict()
m = DotMap(d) # automatic conversion in constructor
这意味着如果您要访问的内容已经是dict
形式,您可以将其转换为DotMap
以便于访问:
import json
jsonDict = json.loads(text)
data = DotMap(jsonDict)
print data.location.city
最后,它会自动创建新的子DotMap
实例,以便您可以执行以下操作:
m = DotMap()
m.people.steve.age = 31
完全披露:我是DotMap的创造者。我创建它是因为Bunch
缺少这些功能
DotMap
创建,当你有很多层次结构时,它可以节省时间并使代码更清晰dict
构建并递归转换所有孩子dict
实例到DotMap
我试过这个:
class dotdict(dict):
def __getattr__(self, name):
return self[name]
你也可以试试__getattribute__
。
使每个dict成为一种dotdict就足够了,如果你想从多层dict中初始化,请尝试实现__init__
。
Fabric有一个非常好的,最小的implementation。扩展它以允许嵌套访问,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
按如下方式使用它:
keys = AttributeDict()
keys.abc.xyz.x = 123
keys.abc.xyz.a.b.c = 234
这详细阐述了库格尔对“从字典中得出并实施__getattr__
和__setattr__
”的回答。现在你知道了!
如果你想修改你修改过的字典,你需要在上面的答案中添加一些状态方法:
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
别。属性访问和索引是Python中的独立事物,您不应该希望它们执行相同的操作。如果你有一些应该具有可访问属性并使用namedtuple
表示法从dict中获取项目,那么创建一个类(可能由[]
创建)。
在Kugel的回答的基础上,考虑到Mike Graham的谨慎言辞,如果我们制作一个包装器怎么办?
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