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Get python class's namespace parent type

Is it possible to get the the namespace parent, or encapsulating type, of a class?
class base:
class sub:
def __init__(self):
# self is "__main__.extra.sub"
# want to create object of type "__main__.extra" from this
pass
class extra(base):
class sub(base.sub):
pass
o = extra.sub()
The problem in base.sub.__init__ is getting extra from the extra.sub.
The only solutions I can think of at the moment involve having all subclasses of base provide some link to their encapsulating class type or turning the type of self in base.sub.__init__ into a string an manipulating it into a new type string. Both a bit ughly.
It's clearly possible to go the other way, type(self()).sub would give you extra.sub from inside base.sub.__init__ for a extra type object, but how do I do .. instead of .sub ? :)

The real answer is that there is no general way to do this. Python classes are normal objects, but they are created a bit differently. A class does not exist until well after its entire body has been executed. Once a class is created, it can be bound to many different names. The only reference it has to where it was created are the __module__ and __qualname__ attributes, but both of these are mutable.
In practice, it is possible to write your example like this:
class Sub:
def __init__(self):
pass
class Base:
Sub = Sub
Sub.__qualname__ = 'Base.Sub'
class Sub(Sub):
pass
class Extra(Base):
Sub = Sub
Sub.__qualname__ = 'Extra.Sub'
del Sub # Unlink from global namespace
Barring the capitalization, this behaves exactly as your original example. Hopefully this clarifies which code has access to what, and shows that the most robust way to determine the enclosing scope of a class is to explicitly assign it somewhere. You can do this in any number of ways. The trivial way is just to assign it. Going back to your original notation:
class Base:
class Sub:
def __init__(self):
print(self.enclosing)
Base.Sub.enclosing = Base
class Extra(Base):
class Sub(Base.Sub):
pass
Extra.Sub.enclosing = Extra
Notice that since Base does not exist when it body is being executed, the assignment has to happen after the classes are both created. You can bypass this by using a metaclass or a decorator. That will allow you to mess with the namespace before the class object is assigned to a name, making the change more transparent.
class NestedMeta(type):
def __init__(cls, name, bases, namespace):
for name, obj in namespace.items():
if isinstance(obj, type):
obj.enclosing = cls
class Base(metaclass=NestedMeta):
class Sub:
def __init__(self):
print(self.enclosing)
class Extra(Base):
class Sub(Base.Sub):
pass
But this is again somewhat unreliable because not all metaclasses are an instance of type, which takes us back to the first statement in this answer.

In many cases, you can use the __qualname__ and __module__ attributes to get the name of the surrounding class:
import sys
cls = type(o)
getattr(sys.modules[cls.__module__], '.'.join(cls.__qualname__.split('.')[:-1]))
This is a very literal answer to your question. It just shows one way of getting the class in the enclosing scope without addressing the probably design flaws that lead to this being necessary in the first place, or any of the many possible corner cases that this would not cover.

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.

Find class in which a method is defined

I want to figure out the type of the class in which a certain method is defined (in essence, the enclosing static scope of the method), from within the method itself, and without specifying it explicitly, e.g.
class SomeClass:
def do_it(self):
cls = enclosing_class() # <-- I need this.
print(cls)
class DerivedClass(SomeClass):
pass
obj = DerivedClass()
# I want this to print 'SomeClass'.
obj.do_it()
Is this possible?

If you need this in Python 3.x, please see my other answer—the closure cell __class__ is all you need.
If you need to do this in CPython 2.6-2.7, RickyA's answer is close, but it doesn't work, because it relies on the fact that this method is not overriding any other method of the same name. Try adding a Foo.do_it method in his answer, and it will print out Foo, not SomeClass
The way to solve that is to find the method whose code object is identical to the current frame's code object:
def do_it(self):
mro = inspect.getmro(self.__class__)
method_code = inspect.currentframe().f_code
method_name = method_code.co_name
for base in reversed(mro):
try:
if getattr(base, method_name).func_code is method_code:
print(base.__name__)
break
except AttributeError:
pass
(Note that the AttributeError could be raised either by base not having something named do_it, or by base having something named do_it that isn't a function, and therefore doesn't have a func_code. But we don't care which; either way, base is not the match we're looking for.)
This may work in other Python 2.6+ implementations. Python does not require frame objects to exist, and if they don't, inspect.currentframe() will return None. And I'm pretty sure it doesn't require code objects to exist either, which means func_code could be None.
Meanwhile, if you want to use this in both 2.7+ and 3.0+, change that func_code to __code__, but that will break compatibility with earlier 2.x.
If you need CPython 2.5 or earlier, you can just replace the inpsect calls with the implementation-specific CPython attributes:
def do_it(self):
mro = self.__class__.mro()
method_code = sys._getframe().f_code
method_name = method_code.co_name
for base in reversed(mro):
try:
if getattr(base, method_name).func_code is method_code:
print(base.__name__)
break
except AttributeError:
pass
Note that this use of mro() will not work on classic classes; if you really want to handle those (which you really shouldn't want to…), you'll have to write your own mro function that just walks the hierarchy old-school… or just copy it from the 2.6 inspect source.
This will only work in Python 2.x implementations that bend over backward to be CPython-compatible… but that includes at least PyPy. inspect should be more portable, but then if an implementation is going to define frame and code objects with the same attributes as CPython's so it can support all of inspect, there's not much good reason not to make them attributes and provide sys._getframe in the first place…

First, this is almost certainly a bad idea, and not the way you want to solve whatever you're trying to solve but refuse to tell us about…
That being said, there is a very easy way to do it, at least in Python 3.0+. (If you need 2.x, see my other answer.)
Notice that Python 3.x's super pretty much has to be able to do this somehow. How else could super() mean super(THISCLASS, self), where that THISCLASS is exactly what you're asking for?*
Now, there are lots of ways that super could be implemented… but PEP 3135 spells out a specification for how to implement it:
Every function will have a cell named __class__ that contains the class object that the function is defined in.
This isn't part of the Python reference docs, so some other Python 3.x implementation could do it a different way… but at least as of 3.2+, they still have to have __class__ on functions, because Creating the class object explicitly says:
This class object is the one that will be referenced by the zero-argument form of super(). __class__ is an implicit closure reference created by the compiler if any methods in a class body refer to either __class__ or super. This allows the zero argument form of super() to correctly identify the class being defined based on lexical scoping, while the class or instance that was used to make the current call is identified based on the first argument passed to the method.
(And, needless to say, this is exactly how at least CPython 3.0-3.5 and PyPy3 2.0-2.1 implement super anyway.)
In [1]: class C:
...: def f(self):
...: print(__class__)
In [2]: class D(C):
...: pass
In [3]: D().f()
<class '__main__.C'>
Of course this gets the actual class object, not the name of the class, which is apparently what you were after. But that's easy; you just need to decide whether you mean __class__.__name__ or __class__.__qualname__ (in this simple case they're identical) and print that.
* In fact, this was one of the arguments against it: that the only plausible way to do this without changing the language syntax was to add a new closure cell to every function, or to require some horrible frame hacks which may not even be doable in other implementations of Python. You can't just use compiler magic, because there's no way the compiler can tell that some arbitrary expression will evaluate to the super function at runtime…

If you can use #abarnert's method, do it.
Otherwise, you can use some hardcore introspection (for python2.7):
import inspect
from http://stackoverflow.com/a/22898743/2096752 import getMethodClass
def enclosing_class():
frame = inspect.currentframe().f_back
caller_self = frame.f_locals['self']
caller_method_name = frame.f_code.co_name
return getMethodClass(caller_self.__class__, caller_method_name)
class SomeClass:
def do_it(self):
print(enclosing_class())
class DerivedClass(SomeClass):
pass
DerivedClass().do_it() # prints 'SomeClass'
Obviously, this is likely to raise an error if:
called from a regular function / staticmethod / classmethod
the calling function has a different name for self (as aptly pointed out by #abarnert, this can be solved by using frame.f_code.co_varnames[0])

Sorry for writing yet another answer, but here's how to do what you actually want to do, rather than what you asked for:
this is about adding instrumentation to a code base to be able to generate reports of method invocation counts, for the purpose of checking certain approximate runtime invariants (e.g. "the number of times that method ClassA.x() is executed is approximately equal to the number of times that method ClassB.y() is executed in the course of a run of a complicated program).
The way to do that is to make your instrumentation function inject the information statically. After all, it has to know the class and method it's injecting code into.
I will have to instrument many classes by hand, and to prevent mistakes I want to avoid typing the class names everywhere. In essence, it's the same reason why typing super() is preferable to typing super(ClassX, self).
If your instrumentation function is "do it manually", the very first thing you want to turn it into an actual function instead of doing it manually. Since you obviously only need static injection, using a decorator, either on the class (if you want to instrument every method) or on each method (if you don't) would make this nice and readable. (Or, if you want to instrument every method of every class, you might want to define a metaclass and have your root classes use it, instead of decorating every class.)
For example, here's an easy way to instrument every method of a class:
import collections
import functools
import inspect
_calls = {}
def inject(cls):
cls._calls = collections.Counter()
_calls[cls.__name__] = cls._calls
for name, method in cls.__dict__.items():
if inspect.isfunction(method):
#functools.wraps(method)
def wrapper(*args, **kwargs):
cls._calls[name] += 1
return method(*args, **kwargs)
setattr(cls, name, wrapper)
return cls
#inject
class A(object):
def f(self):
print('A.f here')
#inject
class B(A):
def f(self):
print('B.f here')
#inject
class C(B):
pass
#inject
class D(C):
def f(self):
print('D.f here')
d = D()
d.f()
B.f(d)
print(_calls)
The output:
{'A': Counter(),
'C': Counter(),
'B': Counter({'f': 1}),
'D': Counter({'f': 1})}
Exactly what you wanted, right?

You can either do what #mgilson suggested or take another approach.
class SomeClass:
pass
class DerivedClass(SomeClass):
pass
This makes SomeClass the base class for DerivedClass.
When you normally try to get the __class__.name__ then it will refer to derived class rather than the parent.
When you call do_it(), it's really passing DerivedClass as self, which is why you are most likely getting DerivedClass being printed.
Instead, try this:
class SomeClass:
pass
class DerivedClass(SomeClass):
def do_it(self):
for base in self.__class__.__bases__:
print base.__name__
obj = DerivedClass()
obj.do_it() # Prints SomeClass
Edit:
After reading your question a few more times I think I understand what you want.
class SomeClass:
def do_it(self):
cls = self.__class__.__bases__[0].__name__
print cls
class DerivedClass(SomeClass):
pass
obj = DerivedClass()
obj.do_it() # prints SomeClass

[Edited]
A somewhat more generic solution:
import inspect
class Foo:
pass
class SomeClass(Foo):
def do_it(self):
mro = inspect.getmro(self.__class__)
method_name = inspect.currentframe().f_code.co_name
for base in reversed(mro):
if hasattr(base, method_name):
print(base.__name__)
break
class DerivedClass(SomeClass):
pass
class DerivedClass2(DerivedClass):
pass
DerivedClass().do_it()
>> 'SomeClass'
DerivedClass2().do_it()
>> 'SomeClass'
SomeClass().do_it()
>> 'SomeClass'
This fails when some other class in the stack has attribute "do_it", since this is the signal name for stop walking the mro.

Is there a method like '__getattribute__' for class (not instance) variables?

I have a class sysprops in which I'd like to have a number of constants. However, I'd like to pull the values for those constants from the database, so I'd like some sort of hook any time one of these class constants are accessed (something like the getattribute method for instance variables).
class sysprops(object):
SOME_CONSTANT = 'SOME_VALUE'
sysprops.SOME_CONSTANT # this statement would not return 'SOME_VALUE' but instead a dynamic value pulled from the database.

Although I think it is a very bad idea to do this, it is possible:
class GetAttributeMetaClass(type):
def __getattribute__(self, key):
print 'Getting attribute', key
class sysprops(object):
__metaclass__ = GetAttributeMetaClass

While the other two answers have a valid method. I like to take the route of 'least-magic'.
You can do something similar to the metaclass approach without actually using them. Simply by using a decorator.
def instancer(cls):
return cls()
#instancer
class SysProps(object):
def __getattribute__(self, key):
return key # dummy
This will create an instance of SysProps and then assign it back to the SysProps name. Effectively shadowing the actual class definition and allowing a constant instance.
Since decorators are more common in Python I find this way easier to grasp for other people that have to read your code.

sysprops.SOME_CONSTANT can be the return value of a function if SOME_CONSTANT were a property defined on type(sysprops).
In other words, what you are talking about is commonly done if sysprops were an instance instead of a class.
But here is the kicker -- classes are instances of metaclasses. So everything you know about controlling the behavior of instances through the use of classes applies equally well to controlling the behavior of classes through the use of metaclasses.
Usually the metaclass is type, but you are free to define other metaclasses by subclassing type. If you place a property SOME_CONSTANT in the metaclass, then the instance of that metaclass, e.g. sysprops will have the desired behavior when Python evaluates sysprops.SOME_CONSTANT.
class MetaSysProps(type):
#property
def SOME_CONSTANT(cls):
return 'SOME_VALUE'
class SysProps(object):
__metaclass__ = MetaSysProps
print(SysProps.SOME_CONSTANT)
yields
SOME_VALUE

In nested classes, how to access outer class's elements from nested class in Python?

I have this scenario, where I need to ask a nested class to append items to a list in the outer class. Heres pseudocode thats similar to what Im trying to do. How would I go about getting it to work?
class Outer(object):
outerlist = []
class Inner(object):
def __call__(self, arg1):
outerlist.append(arg1)
if __name__ == "__main__":
f = Outer()
f.Inner("apple")
f.Inner("orange")
print f.outerlist()
This is what I hope to see - apple, orange
Details:
OS X, Python 2.7

This will have the desired result.
Notice the use of __new__ (line A). This means that instead of doing any construction stuff properly Inner just appends the list. Also, to access class properties of a containing class just use the outer class's name (line B).
Lastly see line C. the list is not a function. You had some extra brackets in your original code.
class Outer(object):
outerlist = []
class Inner(object):
def __new__(self, arg1): #A
Outer.outerlist.append(arg1) #B
f = Outer()
f.Inner("apple")
f.Inner("orange")
print f.outerlist #C

You can only access the outer class with the full global name:
class Outer(object):
outerlist = []
class Inner(object):
def __call__(self, arg1):
Outer.outerlist.append(arg1)
In python, there generally is no need to nest classes in any case. There are use-cases where it makes sense to define a class in a function (to use scoped variabels), but rarely is there a need for nested classes.

Now that I understand your design, you're going about things all wrong.
First, Outer is a class; it doesn't get __call__ed. Your line 17 is just going to construct an empty Outer object and do nothing with it. If you want to "call" the Outer object, you can define an __init__ method—or, as Sheena suggests, define a __new__ and intercept the initialization, since you don't actually need the initialized object anyway.
Honestly, I don't think someone who doesn't understand how __call__ works yet should be trying to build something tricky like this yet. But if you insist, read on.
It's a very odd, and probably bad, design to collect this kind of stuff in a class instead of an instance. Keep in mind that class variables are effectively globals, with all that entails. If you try to use Outer reentrantly, or from multiple threads/event handlers/greenlets/whatever, the uses will end up stomping all over each other. Even if you think that isn't possibly going to be a problem now, it likely will at some point in the future.
You could create an Outer instance, and use its members as decorators. For example:
from outer_library import Outer
outer = Outer()
#outer.get("/")
…
But I'm not sure that's much better. The entire design here seems to involve performing actions at the module level, even though it looks like you're just defining normal functions and calling a function at the end. The fact that you've managed to confuse yourself should be evidence of how confusing a design this is.
But if you do want to do that, what you probably want to do is define classes inside the Outer.__init__ method, and assign them to instance members. Classes are first-class values, and can be assigned to variables just like any other values. Then, use the __init__ (or __new__) methods of those classes to do the work you wanted, making the classes simulate functions.
This may seem confusing or misleading. But remember that the whole point of what you're trying to do is to use a class in a way that it looks like a method, so that kind of confusion is inherent in the problem. But if you prefer, you can write decorators as functions (in this case, as normal instance methods of Outer); it just makes a different part of the problem harder instead of this part.
A more normal way to design something like this would be to make Outer a perfectly normal class that people can either subclass or create instances of, and provide an explicit non-fancy way to attach handler methods or functions to URLs. Then, once that's working, design a way to simplify that handler registration with a decorator.

Why don't you pass in the outer class instance as a parameter to the inner class method? You can access the members of the outer class instance that way.
class Outer(object):
outerlist = []
class Inner(object):
def __init__(self, outer_self):
self.outer = outer_self
def __call__(self, arg1):
self.outer.outerlist.append(arg1)
if __name__ == "__main__":
f = Outer()
i = f.Inner(f)
i("apple")
i("orange")
print f.outerlist

I think the option pointed out by picmate (passing the outer class as an input of the inner one) is a good choice for this particular problem. Note that it could also work without having to 'nest' the classes (in my opinion, something to avoid here since it is unnecessary). Note that I have also embedded the initialization of the inner class inside the outer one (this step is not strictly needed, but I include it here for illustrative purposes):
class Outer(object):
outerlist = []
def __init__(self):
self.Inner=Inner(Outer=self)
def get_basket(self):
print "Outer basket: %s" %(self.outerlist)
class Inner(object):
def __init__(self,Outer):
self.Outer=Outer
def __call__(self, arg1):
self.Outer.outerlist.append(arg1)
def get_basket(self):
print "Inner basket: %s" %(self.Outer.outerlist)
if __name__ == "__main__":
f = Outer()
print "Initial state"
f.get_basket()
f.Inner.get_basket()
f.Inner("apple")
f.Inner("orange")
print "Final state"
f.get_basket()
f.Inner.get_basket()