How can I get a list of all classes within current module in Python?


I’ve seen plenty of examples of people extracting all of the classes from a module, usually something like:

class Foo:

import inspect
import foo

for name, obj in inspect.getmembers(foo):
    if inspect.isclass(obj):
        print obj


But I can’t find out how to get all of the classes from the current module.

import inspect

class Foo:

def print_classes():
    for name, obj in inspect.getmembers(???): # what do I do here?
        if inspect.isclass(obj):
            print obj

import foo


This is probably something really obvious, but I haven’t been able to find anything. Can anyone help me out?

Asked By: mcccclean



I don’t know if there’s a ‘proper’ way to do it, but your snippet is on the right track: just add import foo to, do inspect.getmembers(foo), and it should work fine.

Answered By: int3

Try this:

import sys
current_module = sys.modules[__name__]

In your context:

import sys, inspect
def print_classes():
    for name, obj in inspect.getmembers(sys.modules[__name__]):
        if inspect.isclass(obj):

And even better:

clsmembers = inspect.getmembers(sys.modules[__name__], inspect.isclass)

Because inspect.getmembers() takes a predicate.

Answered By: Nadia Alramli

What about

g = globals().copy()
for name, obj in g.iteritems():


Answered By: Krab
import pyclbr

Note that the stdlib’s Python class browser module uses static source analysis, so it only works for modules that are backed by a real .py file.

Answered By: ncoghlan

I was able to get all I needed from the dir built in plus getattr.

# Works on pretty much everything, but be mindful that 
# you get lists of strings back

print dir(myproject)
print dir(myproject.mymodule)
print dir(myproject.mymodule.myfile)
print dir(myproject.mymodule.myfile.myclass)

# But, the string names can be resolved with getattr, (as seen below)

Though, it does come out looking like a hairball:

def list_supported_platforms():
        List supported platforms (to match sys.platform)

            list str: platform names
    return list(itertools.chain(
            # Get the class's constant
                # Get the module's first class, which we wrote
                    # Get the module
                    getattr(platforms, item),
                        getattr(platforms, item)
            # For each include in platforms/ 
            for item in dir(platforms)
            # Ignore magic, ourselves ( and a base class.
            if not item.startswith('__') and item not in ['index', 'base']
Answered By: ThorSummoner

Another solution which works in Python 2 and 3:
import sys

class Foo(object):

def print_classes():
    current_module = sys.modules[__name__]
    for key in dir(current_module):
        if isinstance( getattr(current_module, key), type ):

import foo
Answered By: Florian

If you want to have all the classes, that belong to the current module, you could use this :

import sys, inspect
def print_classes():
    is_class_member = lambda member: inspect.isclass(member) and member.__module__ == __name__
    clsmembers = inspect.getmembers(sys.modules[__name__], is_class_member)

If you use Nadia’s answer and you were importing other classes on your module, that classes will be being imported too.

So that’s why member.__module__ == __name__ is being added to the predicate used on is_class_member. This statement checks that the class really belongs to the module.

A predicate is a function (callable), that returns a boolean value.

Answered By: Benjy Malca

I think that you can do something like this.

class custom(object):
    __custom__ = True
class Alpha(custom):
    something = 3
def GetClasses():
    return [x for x in globals() if hasattr(globals()[str(x)], '__custom__')]

if you need own classes

Answered By: user10181506

This is the line that I use to get all of the classes that have been defined in the current module (ie not imported). It’s a little long according to PEP-8 but you can change it as you see fit.

import sys
import inspect

classes = [name for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass) 
          if obj.__module__ is __name__]

This gives you a list of the class names. If you want the class objects themselves just keep obj instead.

classes = [obj for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass)
          if obj.__module__ is __name__]

This is has been more useful in my experience.

Answered By: Austin Mackillop

I frequently find myself writing command line utilities wherein the first argument is meant to refer to one of many different classes. For example ./ feature command —-arguments, where Feature is a class and command is a method on that class. Here’s a base class that makes this easy.

The assumption is that this base class resides in a directory alongside all of its subclasses. You can then call ArgBaseClass(foo = bar).load_subclasses() which will return a dictionary. For example, if the directory looks like this:


Assuming implements class Feature(ArgBaseClass), then the above invocation of load_subclasses will return { 'feature' : <Feature object> }. The same kwargs (foo = bar) will be passed into the Feature class.

#!/usr/bin/env python3
import os, pkgutil, importlib, inspect

class ArgBaseClass():
    # Assign all keyword arguments as properties on self, and keep the kwargs for later.
    def __init__(self, **kwargs):
        self._kwargs = kwargs
        for (k, v) in kwargs.items():
            setattr(self, k, v)
        ms = inspect.getmembers(self, predicate=inspect.ismethod)
        self.methods = dict([(n, m) for (n, m) in ms if not n.startswith('_')])

    # Add the names of the methods to a parser object.
    def _parse_arguments(self, parser):
        parser.add_argument('method', choices=list(self.methods))
        return parser

    # Instantiate one of each of the subclasses of this class.
    def load_subclasses(self):
        module_dir = os.path.dirname(__file__)
        module_name = os.path.basename(os.path.normpath(module_dir))
        parent_class = self.__class__
        modules = {}
        # Load all the modules it the package:
        for (module_loader, name, ispkg) in pkgutil.iter_modules([module_dir]):
            modules[name] = importlib.import_module('.' + name, module_name)

        # Instantiate one of each class, passing the keyword arguments.
        ret = {}
        for cls in parent_class.__subclasses__():
            path = cls.__module__.split('.')
            ret[path[-1]] = cls(**self._kwargs)
        return ret
Answered By: Zane Claes
import Foo 

import collections
Answered By: Avinash Koneru

Go to Python Interpreter. type help (‘module_name’) , then press Enter.
e.g. help(‘os’) .
Here, I’ve pasted one part of the output below:

class statvfs_result(__builtin__.object)
     |  statvfs_result: Result from statvfs or fstatvfs.
     |  This object may be accessed either as a tuple of
     |    (bsize, frsize, blocks, bfree, bavail, files, ffree, favail, flag, namemax),
     |  or via the attributes f_bsize, f_frsize, f_blocks, f_bfree, and so on.
     |  See os.statvfs for more information.
     |  Methods defined here:
     |  __add__(...)
     |      x.__add__(y) <==> x+y
     |  __contains__(...)
     |      x.__contains__(y) <==> y in x
Answered By: sipreeti

The following can be placed at the top of the file:

def get_classes():
    import inspect, sys
    return dict(inspect.getmembers(
        lambda member: inspect.isclass(member) and member.__module__ == __name__

Note, this can be placed at the top of the module because we’ve wrapped the logic in a function definition. If you want the dictionary to exist as a top-level object you will need to place the definition at the bottom of the file to ensure all classes are included.

Answered By: cmarkle
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