Getting the caller function name inside another function in Python?

If you have 2 functions like:

def A
def B

and A calls B, can you get who is calling B inside B, like:

def A () :
    B ()

def B () :
    this.caller.name

You can use the inspect module to get the info you want. Its stack method returns a list of frame records.

• For Python 2 each frame record is a list. The third element in each record is the caller name. What you want is this:

   >>> import inspect
   >>> def f():
   ...     print inspect.stack()[1][3]
   ...
   >>> def g():
   ...     f()
   ...
   >>> g()
   g
  

---

• For Python 3.5+, each frame record is a named tuple so you need to replace

   print inspect.stack()[1][3]
  

  with

   print(inspect.stack()[1].function)
  

  on the above code.

There are two ways, using sys and inspect modules:

• sys._getframe(1).f_code.co_name
• inspect.stack()[1][3]

The stack() form is less readable and is implementation dependent since it calls sys._getframe(), see extract from inspect.py:

def stack(context=1):
    """Return a list of records for the stack above the caller's frame."""
    return getouterframes(sys._getframe(1), context)

Note (June 2018): today, I would probably use inspect module, see other answers

sys._getframe(1).f_code.co_name like in the example below:

>>> def foo():
...  global x
...  x = sys._getframe(1)
...
>>> def y(): foo()
...
>>> y()
>>> x.f_code.co_name
'y'
>>>  

Important note: as it's obvious from the _getframe method name (hey, it starts with an underscore), it's not an API method one should be thoughtlessly rely on.

This works for me! :D

>>> def a():
...     import sys
...     print sys._getframe(1).f_code.co_name
...
>>> def b():
...     a()
...
...
>>> b()
b
>>>

you can user the logging module and specify the %(funcName)s option in BaseConfig()

import logging
logging.basicConfig(filename='/tmp/test.log', level=logging.DEBUG, format='%(asctime)s | %(levelname)s | %(funcName)s |%(message)s')

def A():
    logging.info('info')