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Function to behave differently on class vs on instance

I'd like a particular function to be callable as a classmethod, and to behave differently when it's called on an instance.
For example, if I have a class Thing, I want Thing.get_other_thing() to work, but also thing = Thing(); thing.get_other_thing() to behave differently.
I think overwriting the get_other_thing method on initialization should work (see below), but that seems a bit hacky. Is there a better way?
class Thing:
def __init__(self):
self.get_other_thing = self._get_other_thing_inst()
#classmethod
def get_other_thing(cls):
# do something...
def _get_other_thing_inst(self):
# do something else

Great question! What you seek can be easily done using descriptors.
Descriptors are Python objects which implement the descriptor protocol, usually starting with __get__().
They exist, mostly, to be set as a class attribute on different classes. Upon accessing them, their __get__() method is called, with the instance and owner class passed in.
class DifferentFunc:
"""Deploys a different function accroding to attribute access
I am a descriptor.
"""
def __init__(self, clsfunc, instfunc):
# Set our functions
self.clsfunc = clsfunc
self.instfunc = instfunc
def __get__(self, inst, owner):
# Accessed from class
if inst is None:
return self.clsfunc.__get__(None, owner)
# Accessed from instance
return self.instfunc.__get__(inst, owner)
class Test:
#classmethod
def _get_other_thing(cls):
print("Accessed through class")
def _get_other_thing_inst(inst):
print("Accessed through instance")
get_other_thing = DifferentFunc(_get_other_thing,
_get_other_thing_inst)
And now for the result:
>>> Test.get_other_thing()
Accessed through class
>>> Test().get_other_thing()
Accessed through instance
That was easy!
By the way, did you notice me using __get__ on the class and instance function? Guess what? Functions are also descriptors, and that's the way they work!
>>> def func(self):
... pass
...
>>> func.__get__(object(), object)
<bound method func of <object object at 0x000000000046E100>>
Upon accessing a function attribute, it's __get__ is called, and that's how you get function binding.
For more information, I highly suggest reading the Python manual and the "How-To" linked above. Descriptors are one of Python's most powerful features and are barely even known.
Why not set the function on instantiation?
Or Why not set self.func = self._func inside __init__?
Setting the function on instantiation comes with quite a few problems:
self.func = self._funccauses a circular reference. The instance is stored inside the function object returned by self._func. This on the other hand is stored upon the instance during the assignment. The end result is that the instance references itself and will clean up in a much slower and heavier manner.
Other code interacting with your class might attempt to take the function straight out of the class, and use __get__(), which is the usual expected method, to bind it. They will receive the wrong function.
Will not work with __slots__.
Although with descriptors you need to understand the mechanism, setting it on __init__ isn't as clean and requires setting multiple functions on __init__.
Takes more memory. Instead of storing one single function, you store a bound function for each and every instance.
Will not work with properties.
There are many more that I didn't add as the list goes on and on.

Here is a bit hacky solution:
class Thing(object):
#staticmethod
def get_other_thing():
return 1
def __getattribute__(self, name):
if name == 'get_other_thing':
return lambda: 2
return super(Thing, self).__getattribute__(name)
print Thing.get_other_thing() # 1
print Thing().get_other_thing() # 2
If we are on class, staticmethod is executed. If we are on instance, __getattribute__ is first to be executed, so we can return not Thing.get_other_thing but some other function (lambda in my case)

Why differences in class.__slots__ assignment via decorator vs. class body?

I'm working on a decorator to implement some behaviors for an immutable class. I'd like a class to inherit from namedtuple (to have attribute immutability) and also want to add some new methods. Like this ... but correctly preventing new attributes being assigned to the new class.
When inheriting from namedtuple, you should define __new__ and set __slots__ to be an empty tuple (to maintain immutability):
def define_new(clz):
def __new(cls, *args, **kwargs):
return super(clz, cls).__new__(cls, *args, **kwargs)
clz.__new__ = staticmethod(__new) # delegate namedtuple.__new__ to namedtuple
return clz
#define_new
class C(namedtuple('Foo', "a b c")):
__slots__ = () # Prevent assignment of new vars
def foo(self): return "foo"
C(1,2,3).x = 123 # Fails, correctly
... great. But now I'd like to move the __slots__ assignment into the decorator:
def define_new(clz):
def __new(cls, *args, **kwargs):
return super(clz, cls).__new__(cls, *args, **kwargs)
#clz.__slots__ = ()
clz.__slots__ = (123) # just for testing
clz.__new__ = staticmethod(__new)
return clz
#define_new
class C(namedtuple('Foo', "a b c")):
def foo(self): return "foo"
c = C(1,2,3)
print c.__slots__ # Is the (123) I assigned!
c.x = 456 # Assignment succeeds! Not immutable.
print c.__slots__ # Is still (123)
Which is a little surprising.
Why has moving the assignment of __slots__ into the decorator caused a change in behavior?
If I print C.__slots__, I get the object I assigned. What do the x get stored?

The code doesn't work because __slots__ is not a normal class property consulted at run-time. It is a fundamental property of the class that affects the memory layout of each of its instances, and as such must be known when the class is created and remain static throughout the its lifetime. While Python (presumably for backward compatibility) allows assigning to __slots__ later, the assignment has no effect on the behavior of existing or future instances.
How __slots__ is set
The value of __slots__ determined by the class author is passed to the class constructor when the class object is being created. This is done when the class statement is executed; for example:
class X:
__slots__ = ()
The above statement is equivalent1 to creating a class object and assigning it to X:
X = type('X', (), {'__slots__': ()})
The type object is the metaclass, the factory that creates and returns a class when called. The metaclass invocation accepts the name of the type, its superclasses, and its definition dict. Most of the contents of the definition dict can also be assigned later The definition dict contains directives that affect low-level layour of the class instances. As you discovered, later assignment to __slots__ simply has no effect.
Setting __slots__ from the outside
To modify __slots__ so that it is picked up by Python, one must specify it when the class is being created. This can be accomplished with a metaclass, the type responsible for instantiating types. The metaclass drives the creation of the class object and it can make sure __slots__ makes its way into the class definition dict before the constructor is invoked:
class DefineNew(type):
def __new__(metacls, name, bases, dct):
def __new__(cls, *new_args, **new_kwargs):
return super(defcls, cls).__new__(cls, *new_args, **new_kwargs)
dct['__slots__'] = ()
dct['__new__'] = __new__
defcls = super().__new__(metacls, name, bases, dct)
return defcls
class C(namedtuple('Foo', "a b c"), metaclass=DefineNew):
def foo(self):
return "foo"
Testing results in the expected:
>>> c = C(1, 2, 3)
>>> c.foo()
'foo'
>>> c.bar = 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'C' object has no attribute 'bar'
Metaclass mixing pitfall
Note that the C type object will itself be an instance of DefineMeta - which is not surprising, since that follows from the definition of a metaclass. But this might cause an error if you ever inherit from both C and a type that specifies a metaclass other than type or DefineMeta. Since we only need the metaclass to hook into class creation, but are not using it later, it is not strictly necessary for C to be created as an instance of DefineMeta - we can instead make it an instance of type, just like any other class. This is achieved by changing the line:
defcls = super().__new__(metacls, name, bases, dct)
to:
defcls = type.__new__(type, name, bases, dct)
The injection of __new__ and __slots__ will remain, but C will be a most ordinary type with the default metaclass.
In conclusion...
Defining a __new__ which simply calls the superclass __new__ is always superfluous - presumably the real code will also do something different in the injected __new__, e.g. provide the default values for the namedtuple.
1
In the actual class definition the compiler adds a couple of additional items to the class dict, such as the module name. Those are useful, but they do not affect the class definition in the fundamental way that __slots__ does. If X had methods, their function objects would also be included in the dict keyed by function name - automatically inserted as a side effect of executing the def statement in the class definition namespace dict.

__slots__ has to be present during class creation. It affects the memory layout of a class's instances, which isn't something you can just change at will. (Imagine if you already had instances of the class and you tried to reassign the class's __slots__ at that point; instances would all break.) The processing that bases the memory layout on __slots__ only happens during class creation.
Assigning __slots__ in a decorator is too late to do anything. It has to happen before class creation, in the class body or a metaclass __new__.
Also, your define_new is pointless; namedtuple.__new__ already does what you need your __new__ to do.

Is it safe to create an instance of a class in its metaclass in Python?

How safe is it to create an instance of a class in the metaclass constructors (__new__ and __init__)? I'm specifically interested for Python 2.7, but what Python 3 does is also welcome.
The Python data model docs sound like they're written for the normal class instance creation case, and I'm not really sure how the rules might be subtly different when its occurring in a metaclass.
For example, lets say I have code like this:
class Meta(type):
NEWED = []
INITED_BEFORE = []
INITED_AFTER = []
def __new__(meta, name, bases, dict):
cls = super(Meta, meta).__new__(meta, name, bases, dict)
instance = cls()
Meta.NEWED.append(instance)
return cls
def __init__(cls, name, bases, dict):
Meta.INITED_BEFORE.append(cls())
super(Meta, cls).__init__(name, bases, dict)
Meta.INITED_AFTER.append(cls())
class Foo(object):
__metaclass__ = Meta
At which points, if any, is it safe to construct an instance of Foo while the metaclass is constructing it, and what sort of caveats are there?
One suspicion I have is that, if Foo inherited other classes, or was subclassed, and those other classes had their own metaclass, then calling cls() in any of the metaclass methods would be calling it on a not-yet-finished class object. Is that true?

One suspicion I have is that, if Foo inherited other classes, or was subclassed, and those other classes had their own metaclass, then calling cls() in any of the metaclass methods would be calling it on a not-yet-finished class object. Is that true?
That is true, to the extent that in the metaclass's __new__, its __init__ won't have been called yet. Your metaclass has an __init__, as might any subclass, so you should make sure that gets called first. So I wouldn't try to make an instance of a class that hasn't been fully instantiated yet.
One thing you might do is manually call the metaclass's __init__ from within __new__. But you would have to set a flag or something in __init__ to make sure that the class doesn't get __init__'d twice. There might be a more clever way too that I'm not thinking of at the moment.

Python: make every instance have a different superclass

In Python, I currently have instances of a class like MyClass('name1'), MyClass('name2') and so on.
I want to make it so that each instance has its own superclass, i.e., I want MyClass('name1') to be an instance of Name1MyClass and MyClass('name2') to be an instance of Name2MyClass. Name1MyClass and Name2MyClass would be dynamically generated subclasses of MyClass. I can't figure out how to do this, because it seems that Python always makes whatever is returned from __new__ an instance of that class. It isn't clear to me how to do it in a metaclass either.
The reason I want to do this is that I want to define __doc__ docstrings on the instances. But it seems that help completely ignores __doc__ on instances; it only looks on classes. So to put a different docstring on each instance, I need to make each instance have its own custom class.

I could be wrong, but I don't think you want a metaclass here. __metaclass__es are used when the class is created, not when you call the class to construct a new instance of the class (or something else).
Here's an answer using __new__ without a metaclass. It feels a bit hacky, but it seems to work:
_sentinel = Ellipsis
class MyClass(object):
def __new__(cls, name):
if name is _sentinel:
return object.__new__(cls)
else:
instance = type(name + cls.__name__, (MyClass,), {})(_sentinel)
# Initialization goes here.
return instance
print type(MyClass('name1'))
print type(MyClass('name2'))
There's a catch here -- All the business logic of initializing then new instance must be done in __new__. Since __new__ is returning a different type than the class it is bound to, __init__ won't get called.
Another option is to create a class factory:
class MyClass(object):
pass
def class_factory(name):
new_cls = type(name + MyClass.__name__, (MyClass,), {})
return new_cls() # Or pass whatever you want in here...
print type(class_factory('name1'))
print type(class_factory('name2'))
Finally, you could even create a non-__new__ class method:
class MyClass(object):
#classmethod
def class_factory(cls, name):
new_cls = type(name + cls.__name__, (cls,), {})
return new_cls() # Or pass whatever you want in here...
print type(MyClass.class_factory('name1'))
print type(MyClass.class_factory('name2'))

How to auto register a class when it's defined

I want to have an instance of class registered when the class is defined. Ideally the code below would do the trick.
registry = {}
def register( cls ):
registry[cls.__name__] = cls() #problem here
return cls
#register
class MyClass( Base ):
def __init__(self):
super( MyClass, self ).__init__()
Unfortunately, this code generates the error NameError: global name 'MyClass' is not defined.
What's going on is at the #problem here line I'm trying to instantiate a MyClass but the decorator hasn't returned yet so it doesn't exist.
Is the someway around this using metaclasses or something?

Yes, meta classes can do this. A meta class' __new__ method returns the class, so just register that class before returning it.
class MetaClass(type):
def __new__(cls, clsname, bases, attrs):
newclass = super(MetaClass, cls).__new__(cls, clsname, bases, attrs)
register(newclass) # here is your register function
return newclass
class MyClass(object):
__metaclass__ = MetaClass
The previous example works in Python 2.x. In Python 3.x, the definition of MyClass is slightly different (while MetaClass is not shown because it is unchanged - except that super(MetaClass, cls) can become super() if you want):
#Python 3.x
class MyClass(metaclass=MetaClass):
pass
As of Python 3.6 there is also a new __init_subclass__ method (see PEP 487) that can be used instead of a meta class (thanks to #matusko for his answer below):
class ParentClass:
def __init_subclass__(cls, **kwargs):
super().__init_subclass__(**kwargs)
register(cls)
class MyClass(ParentClass):
pass
[edit: fixed missing cls argument to super().__new__()]
[edit: added Python 3.x example]
[edit: corrected order of args to super(), and improved description of 3.x differences]
[edit: add Python 3.6 __init_subclass__ example]

Since python 3.6 you don't need metaclasses to solve this
In python 3.6 simpler customization of class creation was introduced (PEP 487).
An __init_subclass__ hook that initializes all subclasses of a given class.
Proposal includes following example of subclass registration
class PluginBase:
subclasses = []
def __init_subclass__(cls, **kwargs):
super().__init_subclass__(**kwargs)
cls.subclasses.append(cls)
In this example, PluginBase.subclasses will contain a plain list of
all subclasses in the entire inheritance tree. One should note that
this also works nicely as a mixin class.

The problem isn't actually caused by the line you've indicated, but by the super call in the __init__ method. The problem remains if you use a metaclass as suggested by dappawit; the reason the example from that answer works is simply that dappawit has simplified your example by omitting the Base class and therefore the super call. In the following example, neither ClassWithMeta nor DecoratedClass work:
registry = {}
def register(cls):
registry[cls.__name__] = cls()
return cls
class MetaClass(type):
def __new__(cls, clsname, bases, attrs):
newclass = super(cls, MetaClass).__new__(cls, clsname, bases, attrs)
register(newclass) # here is your register function
return newclass
class Base(object):
pass
class ClassWithMeta(Base):
__metaclass__ = MetaClass
def __init__(self):
super(ClassWithMeta, self).__init__()
#register
class DecoratedClass(Base):
def __init__(self):
super(DecoratedClass, self).__init__()
The problem is the same in both cases; the register function is called (either by the metaclass or directly as a decorator) after the class object is created, but before it has been bound to a name. This is where super gets gnarly (in Python 2.x), because it requires you to refer to the class in the super call, which you can only reasonably do by using the global name and trusting that it will have been bound to that name by the time the super call is invoked. In this case, that trust is misplaced.
I think a metaclass is the wrong solution here. Metaclasses are for making a family of classes that have some custom behaviour in common, exactly as classes are for making a family of instances that have some custom behavior in common. All you're doing is calling a function on a class. You wouldn't define a class to call a function on a string, neither should you define a metaclass to call a function on a class.
So, the problem is a fundamental incompatibility between: (1) using hooks in the class creation process to create instances of the class, and (2) using super.
One way to resolve this is to not use super. super solves a hard problem, but it introduces others (this is one of them). If you're using a complex multiple inheritance scheme, super's problems are better than the problems of not using super, and if you're inheriting from third-party classes that use super then you have to use super. If neither of those conditions are true, then just replacing your super calls with direct base class calls may actually be a reasonable solution.
Another way is to not hook register into class creation. Adding register(MyClass) after each of your class definitions is pretty equivalent to adding #register before them or __metaclass__ = Registered (or whatever you call the metaclass) into them. A line down the bottom is much less self-documenting than a nice declaration up the top of the class though, so this doesn't feel great, but again it may actually be a reasonable solution.
Finally, you can turn to hacks that are unpleasant, but will probably work. The problem is that a name is being looked up in a module's global scope just before it's been bound there. So you could cheat, as follows:
def register(cls):
name = cls.__name__
force_bound = False
if '__init__' in cls.__dict__:
cls.__init__.func_globals[name] = cls
force_bound = True
try:
registry[name] = cls()
finally:
if force_bound:
del cls.__init__.func_globals[name]
return cls
Here's how this works:
We first check to see whether __init__ is in cls.__dict__ (as opposed to whether it has an __init__ attribute, which will always be true). If it's inherited an __init__ method from another class we're probably fine (because the superclass will already be bound to its name in the usual way), and the magic we're about to do doesn't work on object.__init__ so we want to avoid trying that if the class is using a default __init__.
We lookup the __init__ method and grab it's func_globals dictionary, which is where global lookups (such as to find the class referred to in a super call) will go. This is normally the global dictionary of the module where the __init__ method was originally defined. Such a dictionary is about to have the cls.__name__ inserted into it as soon as register returns, so we just insert it ourselves early.
We finally create an instance and insert it into the registry. This is in a try/finally block to make sure we remove the binding we created whether or not creating an instance throws an exception; this is very unlikely to be necessary (since 99.999% of the time the name is about to be rebound anyway), but it's best to keep weird magic like this as insulated as possible to minimise the chance that someday some other weird magic interacts badly with it.
This version of register will work whether it's invoked as a decorator or by the metaclass (which I still think is not a good use of a metaclass). There are some obscure cases where it will fail though:
I can imagine a weird class that doesn't have an __init__ method but inherits one that calls self.someMethod, and someMethod is overridden in the class being defined and makes a super call. Probably unlikely.
The __init__ method might have been defined in another module originally and then used in the class by doing __init__ = externally_defined_function in the class block. The func_globals attribute of the other module though, which means our temporary binding would clobber any definition of this class' name in that module (oops). Again, unlikely.
Probably other weird cases I haven't thought of.
You could try to add more hacks to make it a little more robust in these situations, but the nature of Python is both that these kind of hacks are possible and that it's impossible to make them absolutely bullet proof.

The answers here didn't work for me in python3, because __metaclass__ didn't work.
Here's my code registering all subclasses of a class at their definition time:
registered_models = set()
class RegisteredModel(type):
def __new__(cls, clsname, superclasses, attributedict):
newclass = type.__new__(cls, clsname, superclasses, attributedict)
# condition to prevent base class registration
if superclasses:
registered_models.add(newclass)
return newclass
class CustomDBModel(metaclass=RegisteredModel):
pass
class BlogpostModel(CustomDBModel):
pass
class CommentModel(CustomDBModel):
pass
# prints out {<class '__main__.BlogpostModel'>, <class '__main__.CommentModel'>}
print(registered_models)

Calling the Base class directly should work (instead of using super()):
def __init__(self):
Base.__init__(self)

It can be also done with something like this (without a registry function)
_registry = {}
class MetaClass(type):
def __init__(cls, clsname, bases, methods):
super().__init__(clsname, bases, methods)
_registry[cls.__name__] = cls
class MyClass1(metaclass=MetaClass): pass
class MyClass2(metaclass=MetaClass): pass
print(_registry)
# {'MyClass1': <class '__main__.MyClass1'>, 'MyClass2': <class '__main__.MyClass2'>}
Additionally, if we need to use a base abstract class (e.g. Base() class), we can do it this way (notice the metacalss inherits from ABCMeta instead of type)
from abc import ABCMeta
_registry = {}
class MetaClass(ABCMeta):
def __init__(cls, clsname, bases, methods):
super().__init__(clsname, bases, methods)
_registry[cls.__name__] = cls
class Base(metaclass=MetaClass): pass
class MyClass1(Base): pass
class MyClass2(Base): pass
print(_registry)
# {'Base': <class '__main__.Base'>, 'MyClass1': <class '__main__.MyClass1'>, 'MyClass2': <class '__main__.MyClass2'>}