How to find all the subclasses of a class given its name?

Question

I need a working approach of getting all classes that are inherited from a base class in Python.

Answer 1

New-style classes (i.e. subclassed from object, which is the default in Python 3) have a __subclasses__ method which returns the subclasses:

class Foo(object): pass
class Bar(Foo): pass
class Baz(Foo): pass
class Bing(Bar): pass

Here are the names of the subclasses:

print([cls.__name__ for cls in Foo.__subclasses__()])
# ['Bar', 'Baz']

Here are the subclasses themselves:

print(Foo.__subclasses__())
# [<class '__main__.Bar'>, <class '__main__.Baz'>]

Confirmation that the subclasses do indeed list Foo as their base:

for cls in Foo.__subclasses__():
    print(cls.__base__)
# <class '__main__.Foo'>
# <class '__main__.Foo'>

Note if you want subsubclasses, you'll have to recurse:

def all_subclasses(cls):
    return set(cls.__subclasses__()).union(
        [s for c in cls.__subclasses__() for s in all_subclasses(c)])

print(all_subclasses(Foo))
# {<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>}

Note that if the class definition of a subclass hasn't been executed yet - for example, if the subclass's module hasn't been imported yet - then that subclass doesn't exist yet, and __subclasses__ won't find it.

---

You mentioned "given its name". Since Python classes are first-class objects, you don't need to use a string with the class's name in place of the class or anything like that. You can just use the class directly, and you probably should.

If you do have a string representing the name of a class and you want to find that class's subclasses, then there are two steps: find the class given its name, and then find the subclasses with __subclasses__ as above.

How to find the class from the name depends on where you're expecting to find it. If you're expecting to find it in the same module as the code that's trying to locate the class, then

cls = globals()[name]

would do the job, or in the unlikely case that you're expecting to find it in locals,

cls = locals()[name]

If the class could be in any module, then your name string should contain the fully-qualified name - something like 'pkg.module.Foo' instead of just 'Foo'. Use importlib to load the class's module, then retrieve the corresponding attribute:

import importlib
modname, _, clsname = name.rpartition('.')
mod = importlib.import_module(modname)
cls = getattr(mod, clsname)

However you find the class, cls.__subclasses__() would then return a list of its subclasses.

Answer 2

If you just want direct subclasses then .__subclasses__() works fine. If you want all subclasses, subclasses of subclasses, and so on, you'll need a function to do that for you.

Here's a simple, readable function that recursively finds all subclasses of a given class:

def get_all_subclasses(cls):
    all_subclasses = []

    for subclass in cls.__subclasses__():
        all_subclasses.append(subclass)
        all_subclasses.extend(get_all_subclasses(subclass))

    return all_subclasses

Answer 3

The simplest solution in general form:

def get_subclasses(cls):
    for subclass in cls.__subclasses__():
        yield from get_subclasses(subclass)
        yield subclass

And a classmethod in case you have a single class where you inherit from:

@classmethod
def get_subclasses(cls):
    for subclass in cls.__subclasses__():
        yield from subclass.get_subclasses()
        yield subclass

Answer 4

Python 3.6 - __init_subclass__

As other answer mentioned you can check the __subclasses__ attribute to get the list of subclasses, since python 3.6 you can modify this attribute creation by overriding the __init_subclass__ method.

class PluginBase:
    subclasses = []

    def __init_subclass__(cls, **kwargs):
        super().__init_subclass__(**kwargs)
        cls.subclasses.append(cls)

class Plugin1(PluginBase):
    pass

class Plugin2(PluginBase):
    pass

This way, if you know what you're doing, you can override the behavior of of __subclasses__ and omit/add subclasses from this list.

Answer 5

Note: I see that someone (not @unutbu) changed the referenced answer so that it no longer uses vars()['Foo'] — so the primary point of my post no longer applies.

FWIW, here's what I meant about @unutbu's answer only working with locally defined classes — and that using eval() instead of vars() would make it work with any accessible class, not only those defined in the current scope.

For those who dislike using eval(), a way is also shown to avoid it.

First here's a concrete example demonstrating the potential problem with using vars():

class Foo(object): pass
class Bar(Foo): pass
class Baz(Foo): pass
class Bing(Bar): pass

# unutbu's approach
def all_subclasses(cls):
    return cls.__subclasses__() + [g for s in cls.__subclasses__()
                                       for g in all_subclasses(s)]

print(all_subclasses(vars()['Foo']))  # Fine because  Foo is in scope
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

def func():  # won't work because Foo class is not locally defined
    print(all_subclasses(vars()['Foo']))

try:
    func()  # not OK because Foo is not local to func()
except Exception as e:
    print('calling func() raised exception: {!r}'.format(e))
    # -> calling func() raised exception: KeyError('Foo',)

print(all_subclasses(eval('Foo')))  # OK
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

# using eval('xxx') instead of vars()['xxx']
def func2():
    print(all_subclasses(eval('Foo')))

func2()  # Works
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

This could be improved by moving the eval('ClassName') down into the function defined, which makes using it easier without loss of the additional generality gained by using eval() which unlike vars() is not context-sensitive:

# easier to use version
def all_subclasses2(classname):
    direct_subclasses = eval(classname).__subclasses__()
    return direct_subclasses + [g for s in direct_subclasses
                                    for g in all_subclasses2(s.__name__)]

# pass 'xxx' instead of eval('xxx')
def func_ez():
    print(all_subclasses2('Foo'))  # simpler

func_ez()
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

Lastly, it's possible, and perhaps even important in some cases, to avoid using eval() for security reasons, so here's a version without it:

def get_all_subclasses(cls):
    """ Generator of all a class's subclasses. """
    try:
        for subclass in cls.__subclasses__():
            yield subclass
            for subclass in get_all_subclasses(subclass):
                yield subclass
    except TypeError:
        return

def all_subclasses3(classname):
    for cls in get_all_subclasses(object):  # object is base of all new-style classes.
        if cls.__name__.split('.')[-1] == classname:
            break
    else:
        raise ValueError('class %s not found' % classname)
    direct_subclasses = cls.__subclasses__()
    return direct_subclasses + [g for s in direct_subclasses
                                    for g in all_subclasses3(s.__name__)]

# no eval('xxx')
def func3():
    print(all_subclasses3('Foo'))

func3()  # Also works
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

Answer 6

Here is a simple but efficient version of code:

def get_all_subclasses(cls):
    subclass_list = []

    def recurse(klass):
        for subclass in klass.__subclasses__():
            subclass_list.append(subclass)
            recurse(subclass)

    recurse(cls)

    return set(subclass_list)

Its time complexity is O(n) where n is the number of all subclasses if there's no multiple inheritance. It's more efficient than the functions that recursively create lists or yield classes with generators, whose complexity could be (1) O(nlogn) when the class hierarchy is a balanced tree or (2) O(n^2) when the class hierarchy is a biased tree.

Answer 7

A much shorter version for getting a list of all subclasses:

from itertools import chain

def subclasses(cls):
    return list(
        chain.from_iterable(
            [list(chain.from_iterable([[x], subclasses(x)])) for x in cls.__subclasses__()]
        )
    )

Answer 8

This isn't as good an answer as using the special built-in __subclasses__() class method which @unutbu mentions, so I present it merely as an exercise. The subclasses() function defined returns a dictionary which maps all the subclass names to the subclasses themselves.

def traced_subclass(baseclass):
    class _SubclassTracer(type):
        def __new__(cls, classname, bases, classdict):
            obj = type(classname, bases, classdict)
            if baseclass in bases: # sanity check
                attrname = '_%s__derived' % baseclass.__name__
                derived = getattr(baseclass, attrname, {})
                derived.update( {classname:obj} )
                setattr(baseclass, attrname, derived)
             return obj
    return _SubclassTracer

def subclasses(baseclass):
    attrname = '_%s__derived' % baseclass.__name__
    return getattr(baseclass, attrname, None)


class BaseClass(object):
    pass

class SubclassA(BaseClass):
    __metaclass__ = traced_subclass(BaseClass)

class SubclassB(BaseClass):
    __metaclass__ = traced_subclass(BaseClass)

print subclasses(BaseClass)

Output:

{'SubclassB': <class '__main__.SubclassB'>,
 'SubclassA': <class '__main__.SubclassA'>}

Answer 9

Here's a version without recursion:

def get_subclasses_gen(cls):

    def _subclasses(classes, seen):
        while True:
            subclasses = sum((x.__subclasses__() for x in classes), [])
            yield from classes
            yield from seen
            found = []
            if not subclasses:
                return

            classes = subclasses
            seen = found

    return _subclasses([cls], [])

This differs from other implementations in that it returns the original class. This is because it makes the code simpler and:

class Ham(object):
    pass

assert(issubclass(Ham, Ham)) # True

If get_subclasses_gen looks a bit weird that's because it was created by converting a tail-recursive implementation into a looping generator:

def get_subclasses(cls):

    def _subclasses(classes, seen):
        subclasses = sum(*(frozenset(x.__subclasses__()) for x in classes))
        found = classes + seen
        if not subclasses:
            return found

        return _subclasses(subclasses, found)

    return _subclasses([cls], [])

Answer 10

How can I find all subclasses of a class given its name?

We can certainly easily do this given access to the object itself, yes.

Simply given its name is a poor idea, as there can be multiple classes of the same name, even defined in the same module.

I created an implementation for another answer, and since it answers this question and it's a little more elegant than the other solutions here, here it is:

def get_subclasses(cls):
    """returns all subclasses of argument, cls"""
    if issubclass(cls, type):
        subclasses = cls.__subclasses__(cls)
    else:
        subclasses = cls.__subclasses__()
    for subclass in subclasses:
        subclasses.extend(get_subclasses(subclass))
    return subclasses

Usage:

>>> import pprint
>>> list_of_classes = get_subclasses(int)
>>> pprint.pprint(list_of_classes)
[<class 'bool'>,
 <enum 'IntEnum'>,
 <enum 'IntFlag'>,
 <class 'sre_constants._NamedIntConstant'>,
 <class 'subprocess.Handle'>,
 <enum '_ParameterKind'>,
 <enum 'Signals'>,
 <enum 'Handlers'>,
 <enum 'RegexFlag'>]

Answer 11

While I'm very partial to the __init_subclass__ approach, this will preserve definition order, and avoid combinatorial order of growth if you have a very dense hierarchy with multiple inheritance everywhere:

def descendents(cls):
    '''Does not return the class itself'''
    R = {}
    def visit(cls):
        for subCls in cls.__subclasses__():
            R[subCls] = True
            visit(subCls)
    visit(cls)
    return list(R.keys())

This works because dictionaries remember the insertion order of their keys. A list approach would also work.