How can I get the function of a parent class (as opposed to the unbound method)? In the example below, I would like to get the function and the doc string of the parent class. We have a list of methods (post, get) and there should be a matching function name within either the parent or child class. We need to get the function object and see its attributes. For the child class, this is easy, but I don't see how to do it for the parent class.
I've simplified the example below, but in our more complex case, we have a large flask app that we rewrote to use a common set of base/parent classes (with decorated functions). Our third party library uses __dict__.get to generate swagger documentation.
class A():
def get(self):
"""geta"""
desc = 'geta'
print('got it')
class B(A):
def post(self):
"""postb"""
desc = 'postb'
print('posted')
# this is the part we need to change:
methods = ['post', 'get']
for method in methods:
f = B.__dict__.get(method, None)
print(f)
print(f.__doc__)
The results will be:
<function post at 0x1054f96e0>
postb
None
None
I have an idea of iterating B.bases looking for matching method names. I am worried about a long and deep set of nested if and for loops, and am hoping for a more pythonic and clean solution. dir(B) lists all of the functions, but i don't know how to grab a function except through the dict.
You can use the class __mro__ magic method:
In [3]: class A: #(object) if using 2.x
...: def post(s):
...: """foo"""
...: pass
...:
...:
In [4]: class B(A):
...: def get(s):
...: """boo"""
...: pass
...:
In [8]: methods = ['get', 'post']
In [10]: for m in methods:
...: for c in B.__mro__:
...: f = c.__dict__.get(m)
...: if f:
...: print(f)
...: print(f.__doc__)
break # if you don't want duplicate functions
# (with subclass listed first)
...:
<function B.get at 0x102ece378>
boo
<function A.post at 0x102ee8730>
foo
However, this may print the function twice if one of your subclasses is overriding a method from its parent (not sure if you care or it is what you want)