问题:在python中从字典设置属性
是否可以通过python中的字典创建对象,使得每个键都是该对象的属性?
像这样:
d = { 'name': 'Oscar', 'lastName': 'Reyes', 'age':32 }
e = Employee(d)
print e.name # Oscar
print e.age + 10 # 42
我认为这与该问题几乎完全相反:来自对象字段的Python字典
Is it possible to create an object from a dictionary in python in such a way that each key is an attribute of that object?
Something like this:
d = { 'name': 'Oscar', 'lastName': 'Reyes', 'age':32 }
e = Employee(d)
print e.name # Oscar
print e.age + 10 # 42
I think it would be pretty much the inverse of this question: Python dictionary from an object’s fields
回答 0
当然,是这样的:
class Employee(object):
def __init__(self, initial_data):
for key in initial_data:
setattr(self, key, initial_data[key])
更新资料
正如布伦特·纳什(Brent Nash)所建议的,您还可以通过允许使用关键字参数来使其更加灵活:
class Employee(object):
def __init__(self, *initial_data, **kwargs):
for dictionary in initial_data:
for key in dictionary:
setattr(self, key, dictionary[key])
for key in kwargs:
setattr(self, key, kwargs[key])
然后您可以这样称呼它:
e = Employee({"name": "abc", "age": 32})
或像这样:
e = Employee(name="abc", age=32)
甚至像这样:
employee_template = {"role": "minion"}
e = Employee(employee_template, name="abc", age=32)
Sure, something like this:
class Employee(object):
def __init__(self, initial_data):
for key in initial_data:
setattr(self, key, initial_data[key])
Update
As Brent Nash suggests, you can make this more flexible by allowing keyword arguments as well:
class Employee(object):
def __init__(self, *initial_data, **kwargs):
for dictionary in initial_data:
for key in dictionary:
setattr(self, key, dictionary[key])
for key in kwargs:
setattr(self, key, kwargs[key])
Then you can call it like this:
e = Employee({"name": "abc", "age": 32})
or like this:
e = Employee(name="abc", age=32)
or even like this:
employee_template = {"role": "minion"}
e = Employee(employee_template, name="abc", age=32)
回答 1
以这种方式设置属性几乎肯定不是解决问题的最佳方法。要么:
您知道所有字段都应该提前。在这种情况下,您可以显式设置所有属性。这看起来像
class Employee(object):
def __init__(self, name, last_name, age):
self.name = name
self.last_name = last_name
self.age = age
d = {'name': 'Oscar', 'last_name': 'Reyes', 'age':32 }
e = Employee(**d)
print e.name # Oscar
print e.age + 10 # 42
要么
您不知道所有字段都应该提前。在这种情况下,您应该将数据存储为dict,而不是污染对象命名空间。这些属性用于静态访问。这种情况看起来像
class Employee(object):
def __init__(self, data):
self.data = data
d = {'name': 'Oscar', 'last_name': 'Reyes', 'age':32 }
e = Employee(d)
print e.data['name'] # Oscar
print e.data['age'] + 10 # 42
与情况1基本等效的另一种解决方案是使用collections.namedtuple
。有关如何实现的信息,请参见van的答案。
Setting attributes in this way is almost certainly not the best way to solve a problem. Either:
You know what all the fields should be ahead of time. In that case, you can set all the attributes explicitly. This would look like
class Employee(object):
def __init__(self, name, last_name, age):
self.name = name
self.last_name = last_name
self.age = age
d = {'name': 'Oscar', 'last_name': 'Reyes', 'age':32 }
e = Employee(**d)
print e.name # Oscar
print e.age + 10 # 42
or
You don’t know what all the fields should be ahead of time. In this case, you should store the data as a dict instead of polluting an objects namespace. Attributes are for static access. This case would look like
class Employee(object):
def __init__(self, data):
self.data = data
d = {'name': 'Oscar', 'last_name': 'Reyes', 'age':32 }
e = Employee(d)
print e.data['name'] # Oscar
print e.data['age'] + 10 # 42
Another solution that is basically equivalent to case 1 is to use a collections.namedtuple
. See van’s answer for how to implement that.
回答 2
您可以使用访问对象的属性__dict__
,并对其调用update方法:
>>> class Employee(object):
... def __init__(self, _dict):
... self.__dict__.update(_dict)
...
>>> dict = { 'name': 'Oscar', 'lastName': 'Reyes', 'age':32 }
>>> e = Employee(dict)
>>> e.name
'Oscar'
>>> e.age
32
You can access the attributes of an object with __dict__
, and call the update method on it:
>>> class Employee(object):
... def __init__(self, _dict):
... self.__dict__.update(_dict)
...
>>> dict = { 'name': 'Oscar', 'lastName': 'Reyes', 'age':32 }
>>> e = Employee(dict)
>>> e.name
'Oscar'
>>> e.age
32
回答 3
为什么不只使用属性名称作为字典的键?
class StructMyDict(dict):
def __getattr__(self, name):
try:
return self[name]
except KeyError as e:
raise AttributeError(e)
def __setattr__(self, name, value):
self[name] = value
您可以使用命名参数,元组列表,字典或单独的属性分配进行初始化,例如:
nautical = StructMyDict(left = "Port", right = "Starboard") # named args
nautical2 = StructMyDict({"left":"Port","right":"Starboard"}) # dictionary
nautical3 = StructMyDict([("left","Port"),("right","Starboard")]) # tuples list
nautical4 = StructMyDict() # fields TBD
nautical4.left = "Port"
nautical4.right = "Starboard"
for x in [nautical, nautical2, nautical3, nautical4]:
print "%s <--> %s" % (x.left,x.right)
或者,您可以为未知值返回None,而不是引发属性错误。(web2py存储类中使用的一个技巧)
Why not just use attribute names as keys to a dictionary?
class StructMyDict(dict):
def __getattr__(self, name):
try:
return self[name]
except KeyError as e:
raise AttributeError(e)
def __setattr__(self, name, value):
self[name] = value
You can initialize with named arguments, a list of tuples, or a dictionary, or
individual attribute assignments, e.g.:
nautical = StructMyDict(left = "Port", right = "Starboard") # named args
nautical2 = StructMyDict({"left":"Port","right":"Starboard"}) # dictionary
nautical3 = StructMyDict([("left","Port"),("right","Starboard")]) # tuples list
nautical4 = StructMyDict() # fields TBD
nautical4.left = "Port"
nautical4.right = "Starboard"
for x in [nautical, nautical2, nautical3, nautical4]:
print "%s <--> %s" % (x.left,x.right)
Alternatively, instead of raising the attribute error, you can return None for unknown values. (A trick used in the web2py storage class)
回答 4
我认为,settattr
如果您确实需要支持,那么使用答案是可行的方法dict
。
但是,如果Employee
object只是可以使用点语法(.name
)而不是dict语法(['name']
)访问的结构,则可以使用namedtuple,如下所示:
from collections import namedtuple
Employee = namedtuple('Employee', 'name age')
e = Employee('noname01', 6)
print e
#>> Employee(name='noname01', age=6)
# create Employee from dictionary
d = {'name': 'noname02', 'age': 7}
e = Employee(**d)
print e
#>> Employee(name='noname02', age=7)
print e._asdict()
#>> {'age': 7, 'name': 'noname02'}
您确实具有_asdict()
将所有属性作为字典访问的方法,但是以后只能在构造过程中才能添加其他属性。
I think that answer using settattr
are the way to go if you really need to support dict
.
But if Employee
object is just a structure which you can access with dot syntax (.name
) instead of dict syntax (['name']
), you can use namedtuple like this:
from collections import namedtuple
Employee = namedtuple('Employee', 'name age')
e = Employee('noname01', 6)
print e
#>> Employee(name='noname01', age=6)
# create Employee from dictionary
d = {'name': 'noname02', 'age': 7}
e = Employee(**d)
print e
#>> Employee(name='noname02', age=7)
print e._asdict()
#>> {'age': 7, 'name': 'noname02'}
You do have _asdict()
method to access all properties as dictionary, but you cannot add additional attributes later, only during the construction.
回答 5
例如说
class A():
def __init__(self):
self.x=7
self.y=8
self.z="name"
如果您想一次设置属性
d = {'x':100,'y':300,'z':"blah"}
a = A()
a.__dict__.update(d)
say for example
class A():
def __init__(self):
self.x=7
self.y=8
self.z="name"
if you want to set the attributes at once
d = {'x':100,'y':300,'z':"blah"}
a = A()
a.__dict__.update(d)
回答 6
与使用dict类似,您可以像这样使用kwargs:
class Person:
def __init__(self, **kwargs):
self.properties = kwargs
def get_property(self, key):
return self.properties.get(key, None)
def main():
timmy = Person(color = 'red')
print(timmy.get_property('color')) #prints 'red'
similar to using a dict, you could just use kwargs like so:
class Person:
def __init__(self, **kwargs):
self.properties = kwargs
def get_property(self, key):
return self.properties.get(key, None)
def main():
timmy = Person(color = 'red')
print(timmy.get_property('color')) #prints 'red'