Python, how to keep a non-class method as non-class method? - python

I am making a set of classes that call functions that were defined in a different module. To know which function they must call, the function is stored as a variable of the class (or at least that was what I tried). However, when I try to call it, it automatically assumes that the function is a class method and passes "self" as an argument, which logically causes an error because the function received too many arguments. Do you know how can I avoid the function becoming a class method.
The code would be like:
# Module A
def func1(a):
print a
def func2(a):
print a,a
# Module B
from A import *
class Parent:
def func():
self.sonFunc("Hiya!")
class Son1:
sonFunc = func1
class Son2:
sonFunc = func2
so = Son1()
s.func()
# Should print "Hiya!"
s = Son2()
s.func()
# Should print "Hiya! Hiya!"
Thanks

What you are doing is somewhat of a nonstandard/odd thing, but this should work:
class Son_1(object):
son_func = staticmethod(func_1)
class Son_2(object):
son_func = staticmethod(func_2)
Normally, staticmethod is used as a decorator, but since decorators are just syntactical sugar, you can use them this way too.
An arguably cleaner but also more advanced way would be with a metaclass:
class HasSonMeta(type):
def __new__(cls, name, bases, attrs):
attrs['son_func'] = staticmethod(attrs.pop('__son_func__'))
return type.__new__(cls, name, bases, attrs)
class Son1(object):
__metaclass__ = HasSonMeta
__son_func__ = func_1
class Son2(object):
__metaclass__ = HasSonMeta
__son_func__ = func_2
Using this form, you could also define the function directly in the class (though then it gets even more confusing to anyone reading this code):
class Son3(object):
__metaclass__ = HasSonMeta
def __son_func__():
pass
While there could be a very narrow/obscure scenario where this would be an optimal implementation, you would probably be better served by putting your functions in a base class and then referring to (or overridding) them as needed in the children.

Related

Python meta class - verifying the name of class where instance of other class with meta was initialized

I'm not sure if it is possible. I need to verify (with meta), the name of class which initializes the instance of class enhanced with meta in order to allow (or not) to use its class methods. I have prepared pseudo code to show what I need.
class meta:
__call__(cls, name, base, dict)
if some_class_name != "b":
del dict("some_func")
class a:
def some_func():
code code
class b:
a_instance = a()
a_instance.some_func()
a.some_func() can be only used when the class name's where it was called is b. Otherwise this method should be deleted from object.
This could be made with a metaclass, but it would not be needed, and might be a bit less effective
Normally people worry about metaclasses to modify a class as it is created, and then customize the __new__ or __init__ methods. By customizing the metaclass __getattribute__ you could, yes, control which methods are visible in each instance of the class - but that can also be made directly from your base class __getattribute__. At class creation time, we need to annotate the guarded methods and that can be done with the __init_subclass__ class method.
Now, to check the class name where one instance is bould as an attribute, that is more complicated: when code is run, it won't "know" where it was called from -
One could do a().some_func() at top level code: that creates an instance of "a", and runs the method without the instance ever being bound to any local or global variable.
Python do allow introspection of running code, so the called code (either the code in some_func or code in a.__getattribute__ can "see" the local variables where that code was called. Since the caller could "forge" local variables to whatever they'd like, this alone would not restrict code to be run - but it could be used as a simple mechanism to avoid someone that would unknowingly call the functions from "wrong places" at runtime.
Note that this code aditionally requieres that the calling method has a local self variable that points to the instance - and that is used to check the class name (as I said, this is easy to circunvent, but will avoid people calling your methods by mistake)
The restriction should be documented - and be cunted upon just as a safeguard.
So instead of a metaclass, there is a "RestrictorBase" class that should be inherited from any classes that have the protected methods. And them, for each method you want to protect by name in this way, you should apply the allowed_from decorator as bellow. As the comparison is with the in operator,
it will automatically work with single strings or with tuples of strings (say, you want it to work from "a" and "b").
import inspect
RestrictedRegistry = {}
class RestrictorBase:
def __init_subclass__(cls, **kwargs):
for attrname, attr in cls.__dict__.items():
if allowed:=getattr(attr, "_allowed_from", None):
RestrictedRegistry[attr] = allowed
super().__init_subclass__(**kwargs)
def __getattribute__(self, attrname):
attr = super().__getattribute__(attrname)
if (func:=getattr(attr, "__func__", None)) and func in RestrictedRegistry:
caller_vars = inspect.currentframe().f_back.f_locals
if "self" in caller_vars and caller_vars["self"].__class__.__name__ in RestrictedRegistry[func]:
return attr
raise AttributeError(f" {self.__class__.__name__!r} object has no attribute {attrname}")
return attr
def allowed_from(names):
def deco(func):
func._allowed_from = names
return func
return deco
def hashable(obj):
try:
hash(obj)
except TypeError:
return False
return True
class a(RestrictorBase):
#allowed_from('b')
def some_func(self):
print("Restricted function running")
class b:
def __init__(self):
self.a_instance = a()
def other_method(self):
self.a_instance.some_func()
And this is the code above running in the interactive interpreter:
In [121]: b().other_method()
Restricted function running
In [122]: class c(b): pass
In [123]: c().other_method()
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-123-6746b8f5138c> in <module>
----> 1 c().other_method()
[...]
AttributeError: 'a' object has no attribute some_func
In [124]:

How can I refer to the currently being defined class? [duplicate]

For a recursive function we can do:
def f(i):
if i<0: return
print i
f(i-1)
f(10)
However is there a way to do the following thing?
class A:
# do something
some_func(A)
# ...
If I understand your question correctly, you should be able to reference class A within class A by putting the type annotation in quotes. This is called forward reference.
class A:
# do something
def some_func(self, a: 'A')
# ...
See ref below
https://github.com/python/mypy/issues/3661
https://www.youtube.com/watch?v=AJsrxBkV3kc
In Python you cannot reference the class in the class body, although in languages like Ruby you can do it.
In Python instead you can use a class decorator but that will be called once the class has initialized. Another way could be to use metaclass but it depends on what you are trying to achieve.
You can't with the specific syntax you're describing due to the time at which they are evaluated. The reason the example function given works is that the call to f(i-1) within the function body is because the name resolution of f is not performed until the function is actually called. At this point f exists within the scope of execution since the function has already been evaluated. In the case of the class example, the reference to the class name is looked up during while the class definition is still being evaluated. As such, it does not yet exist in the local scope.
Alternatively, the desired behavior can be accomplished using a metaclass like such:
class MetaA(type):
def __init__(cls):
some_func(cls)
class A(object):
__metaclass__=MetaA
# do something
# ...
Using this approach you can perform arbitrary operations on the class object at the time that the class is evaluated.
Maybe you could try calling __class__.
Right now I'm writing a code that calls a class method from within the same class.
It is working well so far.
I'm creating the class methods using something like:
#classmethod
def my_class_method(cls):
return None
And calling then by using:
x = __class__.my_class_method()
It seems most of the answers here are outdated. From python3.7:
from __future__ import annotations
Example:
$ cat rec.py
from __future__ import annotations
class MyList:
def __init__(self,e):
self.data = [e]
def add(self, e):
self.data.append(e)
return self
def score(self, other:MyList):
return len([e
for e in self.data
if e in other.data])
print(MyList(8).add(3).add(4).score(MyList(4).add(9).add(3)))
$ python3.7 rec.py
2
Nope. It works in a function because the function contents are executed at call-time. But the class contents are executed at define-time, at which point the class doesn't exist yet.
It's not normally a problem because you can hack further members into the class after defining it, so you can split up a class definition into multiple parts:
class A(object):
spam= 1
some_func(A)
A.eggs= 2
def _A_scramble(self):
self.spam=self.eggs= 0
A.scramble= _A_scramble
It is, however, pretty unusual to want to call a function on the class in the middle of its own definition. It's not clear what you're trying to do, but chances are you'd be better off with decorators (or the relatively new class decorators).
There isn't a way to do that within the class scope, not unless A was defined to be something else first (and then some_func(A) will do something entirely different from what you expect)
Unless you're doing some sort of stack inspection to add bits to the class, it seems odd why you'd want to do that. Why not just:
class A:
# do something
pass
some_func(A)
That is, run some_func on A after it's been made. Alternately, you could use a class decorator (syntax for it was added in 2.6) or metaclass if you wanted to modify class A somehow. Could you clarify your use case?
If you want to do just a little hacky thing do
class A(object):
...
some_func(A)
If you want to do something more sophisticated you can use a metaclass. A metaclass is responsible for manipulating the class object before it gets fully created. A template would be:
class AType(type):
def __new__(meta, name, bases, dct):
cls = super(AType, meta).__new__(meta, name, bases, dct)
some_func(cls)
return cls
class A(object):
__metaclass__ = AType
...
type is the default metaclass. Instances of metaclasses are classes so __new__ returns a modified instance of (in this case) A.
For more on metaclasses, see http://docs.python.org/reference/datamodel.html#customizing-class-creation.
If the goal is to call a function some_func with the class as an argument, one answer is to declare some_func as a class decorator. Note that the class decorator is called after the class is initialized. It will be passed the class that is being decorated as an argument.
def some_func(cls):
# Do something
print(f"The answer is {cls.x}")
return cls # Don't forget to return the class
#some_func
class A:
x = 1
If you want to pass additional arguments to some_func you have to return a function from the decorator:
def some_other_func(prefix, suffix):
def inner(cls):
print(f"{prefix} {cls.__name__} {suffix}")
return cls
return inner
#some_other_func("Hello", " and goodbye!")
class B:
x = 2
Class decorators can be composed, which results in them being called in the reverse order they are declared:
#some_func
#some_other_func("Hello", "and goodbye!")
class C:
x = 42
The result of which is:
# Hello C and goodbye!
# The answer is 42
What do you want to achieve? It's possible to access a class to tweak its definition using a metaclass, but it's not recommended.
Your code sample can be written simply as:
class A(object):
pass
some_func(A)
If you want to refer to the same object, just use 'self':
class A:
def some_func(self):
another_func(self)
If you want to create a new object of the same class, just do it:
class A:
def some_func(self):
foo = A()
If you want to have access to the metaclass class object (most likely not what you want), again, just do it:
class A:
def some_func(self):
another_func(A) # note that it reads A, not A()
Do remember that in Python, type hinting is just for auto-code completion therefore it helps IDE to infer types and warn user before runtime. In runtime, type hints almost never used(except in some cases) so you can do something like this:
from typing import Any, Optional, NewType
LinkListType = NewType("LinkList", object)
class LinkList:
value: Any
_next: LinkListType
def set_next(self, ll: LinkListType):
self._next = ll
if __name__ == '__main__':
r = LinkList()
r.value = 1
r.set_next(ll=LinkList())
print(r.value)
And as you can see IDE successfully infers it's type as LinkList:
Note: Since the next can be None, hinting this in the type would be better, I just didn't want to confuse OP.
class LinkList:
value: Any
next: Optional[LinkListType]
It's ok to reference the name of the class inside its body (like inside method definitions) if it's actually in scope... Which it will be if it's defined at top level. (In other cases probably not, due to Python scoping quirks!).
For on illustration of the scoping gotcha, try to instantiate Foo:
class Foo(object):
class Bar(object):
def __init__(self):
self.baz = Bar.baz
baz = 15
def __init__(self):
self.bar = Foo.Bar()
(It's going to complain about the global name 'Bar' not being defined.)
Also, something tells me you may want to look into class methods: docs on the classmethod function (to be used as a decorator), a relevant SO question. Edit: Ok, so this suggestion may not be appropriate at all... It's just that the first thing I thought about when reading your question was stuff like alternative constructors etc. If something simpler suits your needs, steer clear of #classmethod weirdness. :-)
Most code in the class will be inside method definitions, in which case you can simply use the name A.

Unresolved reference in class method

I don't think it is because of the scope of the function, but I get a
Unresolved reference at get_all_predicates(examples).count(predicate_list[0])
inside get_entropy_of_attributes(examples, predicate_list) function in my class Tree:
class Tree:
def get_examples(examples, attributes):
for value in examples:
yield dict(zip(attributes, value.strip().replace(" ", "").split(',')))
def get_all_predicates(examples):
return [d['Predicate'] for d in examples]
def get_entropy_of_attributes(examples, predicate_list):
get_all_predicates(examples).count(predicate_list[0])
return 0
examples = list(get_examples(all_examples, name_of_attributes))
predicate_list = list(set(get_all_predicates(examples)))
get_entropy_of_attributes(examples, predicate_list)
all_examples is a list of dictionary and name_of_attributes is a list, that holds values imported from a text file.
all_examples = [{'P_Length': '1.4', 'P_Width': '0.2', 'Predicate': 'I-setosa', 'Sepal_Width': '3.5', 'S_Length': '5.1'}, ...]
name_of_attributes = ["Check","P-Width"]
Any help?
Classes do not have scopes, only namespaces. This means that functions defined within them cannot see other class variables automatically.
class Foo(object):
var = 1 # lets create a class variable
def foo():
print(var) # this doesn't work!
To access a class variable, you need use attribute syntax: either Foo.var (to access via the class) or, if you're writing an instance method, with self.var (to access via the current instance, which will be passed in as the first argument).
class Bar(object):
var = 1
def bar1():
print(Bar.var) # works
def bar2(self):
print(self.var) # also works, if called on an instance, e.g. `Bar().bar2()`
With this kind of setup you can almost fix your current code (but not quite).
def get_entropy_of_attributes(examples, predicate_list):
Tree.get_all_predicates(examples).count(predicate_list[0]) # name the class
return 0
If you call this after the class is fully initialized, it will work without any exceptions (though it's implementation seems a bit nonsensical). However, it doesn't work when you call it to define a class variable, as your current code does. That's because the class object is only created and bound to the class name after all of the class body has been run.
I think the fix for that is probably to redesign your class in a more conventional way. Rather than having class variables set up based on various globals (like all_examples), you should probably create instances of your class by passing in arguments to the constructor and making the other variables you calculate from them instance attributes. I'd try to write it out, but frankly I don't understand what you're doing well enough.
If you want to call class methods, you have to call them with self, e.g.
class myClass:
def __init__(self):
pass
def get_all_predicates(self):
print('asd')
def do_something(self):
self.get_all_predicates() # working
get_all_predicates() # → Unresolved reference
test = myClass()
test.do_something()
See this link for examples for Python classes.

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'>}

python decorator to modify variable in current scope

Goal: Make a decorator which can modify the scope that it is used in.
If it worked:
class Blah(): # or perhaps class Blah(ParentClassWhichMakesThisPossible)
def one(self):
pass
#decorated
def two(self):
pass
>>> Blah.decorated
["two"]
Why? I essentially want to write classes which can maintain specific dictionaries of methods, so that I can retrieve lists of available methods of different types on a per class basis. errr.....
I want to do this:
class RuleClass(ParentClass):
#rule
def blah(self):
pass
#rule
def kapow(self):
pass
def shazam(self):
class OtherRuleClass(ParentClass):
#rule
def foo(self):
pass
def bar(self):
pass
>>> RuleClass.rules.keys()
["blah", "kapow"]
>>> OtherRuleClass.rules.keys()
["foo"]
You can do what you want with a class decorator (in Python 2.6) or a metaclass. The class decorator version:
def rule(f):
f.rule = True
return f
def getRules(cls):
cls.rules = {}
for attr, value in cls.__dict__.iteritems():
if getattr(value, 'rule', False):
cls.rules[attr] = value
return cls
#getRules
class RuleClass:
#rule
def foo(self):
pass
The metaclass version would be:
def rule(f):
f.rule = True
return f
class RuleType(type):
def __init__(self, name, bases, attrs):
self.rules = {}
for attr, value in attrs.iteritems():
if getattr(value, 'rule', False):
self.rules[attr] = value
super(RuleType, self).__init__(name, bases, attrs)
class RuleBase(object):
__metaclass__ = RuleType
class RuleClass(RuleBase):
#rule
def foo(self):
pass
Notice that neither of these do what you ask for (modify the calling namespace) because it's fragile, hard and often impossible. Instead they both post-process the class -- through the class decorator or the metaclass's __init__ method -- by inspecting all the attributes and filling the rules attribute. The difference between the two is that the metaclass solution works in Python 2.5 and earlier (down to 2.2), and that the metaclass is inherited. With the decorator, subclasses have to each apply the decorator individually (if they want to set the rules attribute.)
Both solutions do not take inheritance into account -- they don't look at the parent class when looking for methods marked as rules, nor do they look at the parent class rules attribute. It's not hard to extend either to do that, if that's what you want.
Problem is, at the time the decorated decorator is called, there is no object Blah yet: the class object is built after the class body finishes executing. Simplest is to have decorated stash the info "somewhere else", e.g. a function attribute, then a final pass (a class decorator or metaclass) reaps that info into the dictionary you desire.
Class decorators are simpler, but they don't get inherited (so they wouldn't come from a parent class), while metaclasses are inherited -- so if you insist on inheritance, a metaclass it will have to be. Simplest-first, with a class decorator and the "list" variant you have at the start of your Q rather than the "dict" variant you have later:
import inspect
def classdecorator(aclass):
decorated = []
for name, value in inspect.getmembers(aclass, inspect.ismethod):
if hasattr(value, '_decorated'):
decorated.append(name)
del value._decorated
aclass.decorated = decorated
return aclass
def decorated(afun):
afun._decorated = True
return afun
now,
#classdecorator
class Blah(object):
def one(self):
pass
#decorated
def two(self):
pass
gives you the Blah.decorated list you request in the first part of your Q. Building a dict instead, as you request in the second part of your Q, just means changing decorated.append(name) to decorated[name] = value in the code above, and of course initializing decorated in the class decorator to an empty dict rather than an empty list.
The metaclass variant would use the metaclass's __init__ to perform essentially the same post-processing after the class body is built -- a metaclass's __init__ gets a dict corresponding to the class body as its last argument (but you'll have to support inheritance yourself by appropriately dealing with any base class's analogous dict or list). So the metaclass approach is only "somewhat" more complex in practice than a class decorator, but conceptually it's felt to be much more difficult by most people. I'll give all the details for the metaclass if you need them, but I'd recommend sticking with the simpler class decorator if feasible.

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