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
我一直把它保存在util文件中。您也可以在自己的类中使用它作为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'
其他回答
不是对OP问题的直接回答,但受到启发,也许对一些人有用。我已经创建了一个基于对象的解决方案使用内部__dict__(在任何方式优化代码)
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"
我一直把它保存在util文件中。您也可以在自己的类中使用它作为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'
Fabric有一个非常好的、最小的实现。将其扩展为允许嵌套访问,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
可以这样使用它:
keys = AttributeDict()
keys.abc.xyz.x = 123
keys.abc.xyz.a.b.c = 234
这详细阐述了Kugel的回答“从dict和派生并实现__getattr__和__setattr__”。现在你知道怎么做了!
这是一个老问题,但我最近发现sklearn有一个可通过键访问的实现版本字典,即Bunch https://scikit-learn.org/stable/modules/generated/sklearn.utils.Bunch.html#sklearn.utils.Bunch
如果你已经在使用pandas,你可以构造一个pandas Series或DataFrame,从中你可以通过点语法访问项目:
1级字典:
import pandas as pd
my_dictionary = pd.Series({
'key1': 'value1',
'key2': 'value2'
})
print(my_dictionary.key1)
# Output: value1
2级字典:
import pandas as pd
my_dictionary = pd.DataFrame({
'key1': {
'inner_key1': 'value1'
},
'key2': {
'inner_key2': 'value2'
}
})
print(my_dictionary.key1.inner_key1)
# Output: value1
请注意,这可能在规范化数据结构(其中每个字典条目都具有相同的结构)下工作得更好。在上面的第二个例子中,得到的DataFrame是:
key1 key2
inner_key1 value1 NaN
inner_key2 NaN value2