get class name in decorator python 2.7

Question

How can I get the class of the method that is wrapped by a decorator with Python 2.7?

With this code:

def timeit(method): def timed(*args, **kw): result = method(*args, **kw) className = ??? print '%s.%s'%(className, method.func_name) #want printed: Test.run return result return timed class Test(object): @timeit def run(self, ): print 'run' a = Test() a.run()

I'm not sure what to put for className = ???. I want it to print Test.run.

I've seen similar solutions that use im_class and __ self__ but I don't have either.

skyking · Accepted Answer · 2015-08-31 19:16:46Z

You could just analyze the self argument to the method.

def timeit(method): def timed(self, *args, **kw): result = method(self, *args, **kw) className = type(self).__name__ print '%s.%s'%(className, method.func_name) #want printed: Test.run return result return timed class Test(object): @timeit def run(self): print 'run' a = Test() a.run()

If it's a classmethod then you could use two decorators (ie @classmethod @timeit) and get the name via cls.__name__ (where cls is the first argument to timed instead of self).

If it's a staticmethod you will have to turn it into a classmethod in order to be able to get a parameter with class info available.

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