how to get the caller's filename, method name in python

Question

for example, a.boo method calls b.foo method. In b.foo method, how can I get a\'s file name (I don\'t want to pass __file__

Answer 1

You can use the inspect module to achieve this:

frame = inspect.stack()[1]
module = inspect.getmodule(frame[0])
filename = module.__file__