Getting the caller function name inside another function in Python? [duplicate]

Solution 1:

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.

Solution 2:

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)

Solution 3:

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.

Solution 4:

This works for me! :D

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

Solution 5:

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')