Inherit docstrings in Python class inheritance

You're not the only one! There was a discussion on comp.lang.python about this a while ago, and a recipe was created. Check it out here.

"""
doc_inherit decorator

Usage:

class Foo(object):
    def foo(self):
        "Frobber"
        pass

class Bar(Foo):
    @doc_inherit
    def foo(self):
        pass 

Now, Bar.foo.__doc__ == Bar().foo.__doc__ == Foo.foo.__doc__ == "Frobber"
"""

from functools import wraps

class DocInherit(object):
    """
    Docstring inheriting method descriptor

    The class itself is also used as a decorator
    """

    def __init__(self, mthd):
        self.mthd = mthd
        self.name = mthd.__name__

    def __get__(self, obj, cls):
        if obj:
            return self.get_with_inst(obj, cls)
        else:
            return self.get_no_inst(cls)

    def get_with_inst(self, obj, cls):

        overridden = getattr(super(cls, obj), self.name, None)

        @wraps(self.mthd, assigned=('__name__','__module__'))
        def f(*args, **kwargs):
            return self.mthd(obj, *args, **kwargs)

        return self.use_parent_doc(f, overridden)

    def get_no_inst(self, cls):

        for parent in cls.__mro__[1:]:
            overridden = getattr(parent, self.name, None)
            if overridden: break

        @wraps(self.mthd, assigned=('__name__','__module__'))
        def f(*args, **kwargs):
            return self.mthd(*args, **kwargs)

        return self.use_parent_doc(f, overridden)

    def use_parent_doc(self, func, source):
        if source is None:
            raise NameError, ("Can't find '%s' in parents"%self.name)
        func.__doc__ = source.__doc__
        return func

doc_inherit = DocInherit 

You can concatenate the docstrings easily:

class Foo(object):
    """
    Foo Class.
    This class foos around.
    """
    pass

class Bar(Foo):
    """
    Bar class, children of Foo
    Use this when you want to Bar around.
    parent:
    """ 
    __doc__ += Foo.__doc__
    pass

However, that is useless. Most documentation generation tool (Sphinx and Epydoc included) will already pull parent docstring, including for methods. So you don't have to do anything.

Not particularly elegant, but simple and direct:

class X(object):
  """This class has a method foo()."""
  def foo(): pass

class Y(X):
  __doc__ = X.__doc__ + ' Also bar().'
  def bar(): pass

Now:

>>> print Y.__doc__
This class has a method foo(). Also bar().

Starting with Python 3.5, docstrings are now inherited if a subclass doesn't have any documentation.

You can use inspect.getdoc() accordingly:

import inspect

class A:
  """Hello"""

class B(A):
  pass

class C(B):
  __doc__ = inspect.getdoc(B) + " World"

Do note that the builtin help function pulls recursively only for overridden functions but not for class docstring for some odd reason:

class A:
  """Hello"""
  def test():
    """Testdoc"""

class B(A):
  def test():
    pass

>>> help(B.test)
... Testdoc ...
>>> help(B)
...  # No "Hello".

It is a very old thread. But If anyone is looking for a simple way, you can do this with __init_subclass__ which is called whenever you inherit that class.

def __init_subclass__(cls, **kwargs):
    super().__init_subclass__(**kwargs)
    parent_method_docstr = {}
    for i, v in ParentClass.__dict__.items():
        if v and callable(v) and v.__doc__ is not None:
            parent_method_docstr[i] = v.__doc__

    for i, v in cls.__dict__.items():
        if v and callable(v) and v.__doc__ is None and i in parent_method_docstr:
            v.__doc__ = parent_method_docstr[i]