How to make a class property?
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Chapters
00:00 Question
00:57 Accepted answer (Score 122)
02:44 Answer 2 (Score 60)
03:16 Answer 3 (Score 46)
06:05 Answer 4 (Score 33)
06:27 Thank you
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Full question
https://stackoverflow.com/questions/5189...
Answer 2 links:
[descriptor documentation]: https://docs.python.org/3/howto/descript...
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https://meta.stackexchange.com/help/lice...
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Tags
#python #properties #classmethod
#avk47
ACCEPTED ANSWER
Score 133
Here's how I would do this:
class ClassPropertyDescriptor(object):
def __init__(self, fget, fset=None):
self.fget = fget
self.fset = fset
def __get__(self, obj, klass=None):
if klass is None:
klass = type(obj)
return self.fget.__get__(obj, klass)()
def __set__(self, obj, value):
if not self.fset:
raise AttributeError("can't set attribute")
type_ = type(obj)
return self.fset.__get__(obj, type_)(value)
def setter(self, func):
if not isinstance(func, (classmethod, staticmethod)):
func = classmethod(func)
self.fset = func
return self
def classproperty(func):
if not isinstance(func, (classmethod, staticmethod)):
func = classmethod(func)
return ClassPropertyDescriptor(func)
class Bar(object):
_bar = 1
@classproperty
def bar(cls):
return cls._bar
@bar.setter
def bar(cls, value):
cls._bar = value
# test instance instantiation
foo = Bar()
assert foo.bar == 1
baz = Bar()
assert baz.bar == 1
# test static variable
baz.bar = 5
assert foo.bar == 5
# test setting variable on the class
Bar.bar = 50
assert baz.bar == 50
assert foo.bar == 50
The setter didn't work at the time we call Bar.bar, because we are calling
TypeOfBar.bar.__set__, which is not Bar.bar.__set__.
Adding a metaclass definition solves this:
class ClassPropertyMetaClass(type):
def __setattr__(self, key, value):
if key in self.__dict__:
obj = self.__dict__.get(key)
if obj and type(obj) is ClassPropertyDescriptor:
return obj.__set__(self, value)
return super(ClassPropertyMetaClass, self).__setattr__(key, value)
# and update class define:
# class Bar(object):
# __metaclass__ = ClassPropertyMetaClass
# _bar = 1
# and update ClassPropertyDescriptor.__set__
# def __set__(self, obj, value):
# if not self.fset:
# raise AttributeError("can't set attribute")
# if inspect.isclass(obj):
# type_ = obj
# obj = None
# else:
# type_ = type(obj)
# return self.fset.__get__(obj, type_)(value)
Now all will be fine.
ANSWER 2
Score 68
If you define classproperty as follows, then your example works exactly as you requested.
class classproperty(object):
def __init__(self, f):
self.f = f
def __get__(self, obj, owner):
return self.f(owner)
The caveat is that you can't use this for writable properties. While e.I = 20 will raise an AttributeError, Example.I = 20 will overwrite the property object itself.
ANSWER 3
Score 51
[answer written based on python 3.4; the metaclass syntax differs in 2 but I think the technique will still work]
You can do this with a metaclass...mostly. Dappawit's almost works, but I think it has a flaw:
class MetaFoo(type):
@property
def thingy(cls):
return cls._thingy
class Foo(object, metaclass=MetaFoo):
_thingy = 23
This gets you a classproperty on Foo, but there's a problem...
print("Foo.thingy is {}".format(Foo.thingy))
# Foo.thingy is 23
# Yay, the classmethod-property is working as intended!
foo = Foo()
if hasattr(foo, "thingy"):
print("Foo().thingy is {}".format(foo.thingy))
else:
print("Foo instance has no attribute 'thingy'")
# Foo instance has no attribute 'thingy'
# Wha....?
What the hell is going on here? Why can't I reach the class property from an instance?
I was beating my head on this for quite a while before finding what I believe is the answer. Python @properties are a subset of descriptors, and, from the descriptor documentation (emphasis mine):
The default behavior for attribute access is to get, set, or delete the attribute from an object’s dictionary. For instance,
a.xhas a lookup chain starting witha.__dict__['x'], thentype(a).__dict__['x'], and continuing through the base classes oftype(a)excluding metaclasses.
So the method resolution order doesn't include our class properties (or anything else defined in the metaclass). It is possible to make a subclass of the built-in property decorator that behaves differently, but (citation needed) I've gotten the impression googling that the developers had a good reason (which I do not understand) for doing it that way.
That doesn't mean we're out of luck; we can access the properties on the class itself just fine...and we can get the class from type(self) within the instance, which we can use to make @property dispatchers:
class Foo(object, metaclass=MetaFoo):
_thingy = 23
@property
def thingy(self):
return type(self).thingy
Now Foo().thingy works as intended for both the class and the instances! It will also continue to do the right thing if a derived class replaces its underlying _thingy (which is the use case that got me on this hunt originally).
This isn't 100% satisfying to me -- having to do setup in both the metaclass and object class feels like it violates the DRY principle. But the latter is just a one-line dispatcher; I'm mostly okay with it existing, and you could probably compact it down to a lambda or something if you really wanted.
ANSWER 4
Score 30
I think you may be able to do this with the metaclass. Since the metaclass can be like a class for the class (if that makes sense). I know you can assign a __call__() method to the metaclass to override calling the class, MyClass(). I wonder if using the property decorator on the metaclass operates similarly.
Wow, it works:
class MetaClass(type):
def getfoo(self):
return self._foo
foo = property(getfoo)
@property
def bar(self):
return self._bar
class MyClass(object):
__metaclass__ = MetaClass
_foo = 'abc'
_bar = 'def'
print MyClass.foo
print MyClass.bar
Note: This is in Python 2.7. Python 3+ uses a different technique to declare a metaclass. Use: class MyClass(metaclass=MetaClass):, remove __metaclass__, and the rest is the same.