Determine function name from within that function (without using traceback)


In Python, without using the traceback module, is there a way to determine a function’s name from within that function?

Say I have a module foo with a function bar. When executing, is there a way for bar to know bar‘s name? Or better yet,‘s name?  
def bar():
    print "my name is", __myname__ # <== how do I calculate this at runtime?
Asked By: Rob



import inspect

def foo():
   print(inspect.stack()[1][3])  # will give the caller of foos name, if something called foo



Answered By: Andreas Jung

Python doesn’t have a feature to access the function or its name within the function itself. It has been proposed but rejected. If you don’t want to play with the stack yourself, you should either use "bar" or bar.__name__ depending on context.

The given rejection notice is:

This PEP is rejected. It is not clear how it should be implemented or what the precise semantics should be in edge cases, and there aren’t enough important use cases given. response has been lukewarm at best.

Answered By: Rosh Oxymoron

I guess inspect is the best way to do this. For example:

import inspect
def bar():
    print("My name is", inspect.stack()[0][3])
Answered By: Bjorn

You can get the name that it was defined with using the approach that @Andreas Jung shows, but that may not be the name that the function was called with:

import inspect

def Foo():
   print inspect.stack()[0][3]

Foo2 = Foo

>>> Foo()

>>> Foo2()

Whether that distinction is important to you or not I can’t say.

Answered By: bgporter
functionNameAsString = sys._getframe().f_code.co_name

I wanted a very similar thing because I wanted to put the function name in a log string that went in a number of places in my code. Probably not the best way to do that, but here’s a way to get the name of the current function.

Answered By: Ron Davis

There are few ways to get the same result:

import sys
import inspect

def what_is_my_name():

Note that the inspect.stack calls are thousands of times slower than the alternatives:

$ python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][0].f_code.co_name'
1000 loops, best of 3: 499 usec per loop
$ python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][3]'
1000 loops, best of 3: 497 usec per loop
$ python -m timeit -s 'import inspect, sys' 'inspect.currentframe().f_code.co_name'
10000000 loops, best of 3: 0.1 usec per loop
$ python -m timeit -s 'import inspect, sys' 'sys._getframe().f_code.co_name'
10000000 loops, best of 3: 0.135 usec per loop

Update 08/2021 (original post was written for Python2.7)

Python 3.9.1 (default, Dec 11 2020, 14:32:07)
[GCC 7.3.0] :: Anaconda, Inc. on linux

python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][0].f_code.co_name'
500 loops, best of 5: 390 usec per loop
python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][3]'
500 loops, best of 5: 398 usec per loop
python -m timeit -s 'import inspect, sys' 'inspect.currentframe().f_code.co_name'
2000000 loops, best of 5: 176 nsec per loop
python -m timeit -s 'import inspect, sys' 'sys._getframe().f_code.co_name'
5000000 loops, best of 5: 62.8 nsec per loop
Answered By: Alex Granovsky

I found a wrapper that will write the function name

from functools import wraps

def tmp_wrap(func):
    def tmp(*args, **kwargs):
        print func.__name__
        return func(*args, **kwargs)
    return tmp

def my_funky_name():
    print "STUB"


This will print



Answered By: cad106uk

I keep this handy utility nearby:

import inspect
myself = lambda: inspect.stack()[1][3]


Answered By: xxyzzy

Here’s a future-proof approach.

Combining @CamHart’s and @Yuval’s suggestions with @RoshOxymoron’s accepted answer has the benefit of avoiding:

  • _hidden and potentially deprecated methods
  • indexing into the stack (which could be reordered in future pythons)

So I think this plays nice with future python versions (tested on 2.7.3 and 3.3.2):

from __future__ import print_function
import inspect

def bar():
    print("my name is '{}'".format(inspect.currentframe().f_code.co_name))

Update: tested on 3.7.10, 3.8.10, and 3.9.5

Answered By: hobs

You can use a decorator:

def my_function(name=None):
    return name

def get_function_name(function):
    return function(name=function.__name__)

>>> get_function_name(my_function)
Answered By: Douglas Denhartog

This is actually derived from the other answers to the question.

Here’s my take:

import sys

# for current func name, specify 0 or no argument.
# for name of caller of current func, specify 1.
# for name of caller of caller of current func, specify 2. etc.
currentFuncName = lambda n=0: sys._getframe(n + 1).f_code.co_name

def testFunction():
    print "You are in function:", currentFuncName()
    print "This function's caller was:", currentFuncName(1)    

def invokeTest():


# end of file

The likely advantage of this version over using inspect.stack() is that it should be thousands of times faster [see Alex Melihoff’s post and timings regarding using sys._getframe() versus using inspect.stack() ].

Answered By: Gino
import inspect

def whoami():
    return inspect.stack()[1][3]

def whosdaddy():
    return inspect.stack()[2][3]

def foo():
    print "hello, I'm %s, daddy is %s" % (whoami(), whosdaddy())

def bar():
    print "hello, I'm %s, daddy is %s" % (whoami(), whosdaddy())


In IDE the code outputs

hello, I’m foo, daddy is

hello, I’m bar, daddy is foo

hello, I’m bar, daddy is

Answered By: Lee

I do my own approach used for calling super with safety inside multiple inheritance scenario (I put all the code)

def safe_super(_class, _inst):
    """safe super call"""
        return getattr(super(_class, _inst), _inst.__fname__)
        return (lambda *x,**kx: None)

def with_name(function):
    def wrap(self, *args, **kwargs):
        self.__fname__ = function.__name__
        return function(self, *args, **kwargs)
return wrap

sample usage:

class A(object):

    def __init__():
        super(A, self).__init__()

    def test(self):
        print 'called from An'
        safe_super(A, self)()

class B(object):

    def __init__():
        super(B, self).__init__()

    def test(self):
        print 'called from Bn'
        safe_super(B, self)()

class C(A, B):

    def __init__():
        super(C, self).__init__()

    def test(self):
        print 'called from Cn'
        safe_super(C, self)()

testing it :

a = C()


called from C
called from A
called from B

Inside each @with_name decorated method you have access to self.__fname__ as the current function name.

Answered By: Mel Viso Martinez

print(inspect.stack()[0].function) seems to work too (Python 3.5).

Answered By: Pierre Voisin
import sys

def func_name():
    :return: name of caller
    return sys._getframe(1).f_code.co_name

class A(object):
    def __init__(self):
    def test_class_func_name(self):

def test_func_name():


a = A()


Answered By: nordborn

I recently tried to use the above answers to access the docstring of a function from the context of that function but as the above questions were only returning the name string it did not work.

Fortunately I found a simple solution. If like me, you want to refer to the function rather than simply get the string representing the name you can apply eval() to the string of the function name.

import sys
def foo():
    """foo docstring"""
Answered By: John Forbes

I suggest not to rely on stack elements. If someone use your code within different contexts (python interpreter for instance) your stack will change and break your index ([0][3]).

I suggest you something like that:

class MyClass:

    def __init__(self):
        self.function_name = None

    def _Handler(self, **kwargs):
        print('Calling function {} with parameters {}'.format(self.function_name, kwargs))
        self.function_name = None

    def __getattr__(self, attr):
        self.function_name = attr
        return self._Handler

mc = MyClass()
mc.test(FirstParam='my', SecondParam='test')
Answered By: Genschi

I am not sure why people make it complicated:

import sys 
print("%s/%s" %(sys._getframe().f_code.co_filename, sys._getframe().f_code.co_name))
Answered By: karthik r

This is pretty easy to accomplish with a decorator.

>>> from functools import wraps

>>> def named(func):
...     @wraps(func)
...     def _(*args, **kwargs):
...         return func(func.__name__, *args, **kwargs)
...     return _

>>> @named
... def my_func(name, something_else):
...     return name, something_else

>>> my_func('hello, world')
('my_func', 'hello, world')
Answered By: Jeff Laughlin

Use __name__ attribute:

def bar():
    print(f"my name is {bar.__name__}")

You can easily access function’s name from within the function using __name__ attribute.

>>> def bar():
...     print(f"my name is {bar.__name__}")
>>> bar()
my name is bar

I’ve come across this question myself several times, looking for the ways to do it. Correct answer is contained in the Python’s documentation (see Callable types section).

Every function has a __name__ parameter that returns its name and even __qualname__ parameter that returns its full name, including which class it belongs to (see Qualified name).

Answered By: Vagiz Duseev
str(str(inspect.currentframe())).split(' ')[-1][:-1]
Answered By: Harry Kearney

Sincesys._getframe().f_back.f_code.co_name does not work at all in python 3.9, following could be used from now:

from inspect import currentframe

def testNameFunction() -> str:
    return currentframe().f_back.f_code.co_name

print(f'function name is {testNameFunction()}(...)')


function name is testNameFunction(...)
Answered By: Manifest Man

I like the idea of using a decorator but I’d prefer to avoid touching the function arguments. Hence, I’m providing yet another alternative:

import functools

def withname(f):
    def wrapper(*args, **kwargs):
        global __name
        __saved_name = globals().get("__name")
        __name = f.__name__
        ret = f(*args, **kwargs)
        __name = __saved_name
        return ret
    return wrapper

def f():
    print(f"in f: __name=={__name}")
    print(f"back in f: __name=={__name}")

def g():
    print(f"in g: __name=={__name}")

We need to save and restore __name when calling the function as consequence of it being a global variable. Calling f() above produces:

in f: __name==f
in g: __name==g
back in f: __name==f

Unfortunately, there is no alternative to the global variable if we don’t change the function arguments. Referencing a variable, that is not created in the context of the function, will generate code that would look for a global variable:

>>> def f(): print(__function__)
>>> from dis import dis
>>> dis(f)
  1           0 LOAD_GLOBAL              0 (print)
              2 LOAD_GLOBAL              1 (__function__)
              4 CALL_FUNCTION            1
              6 POP_TOP
              8 LOAD_CONST               0 (None)
             10 RETURN_VALUE
Answered By: olivecoder
import inspect

def method_name():
    return inspect.stack()[1][3]

def method_name_caller():
    return inspect.stack()[2][3]

def asdf():

def asdf2():
Answered By: seunggabi

@jeff-laughlin’s answer is beautiful. I have modified it slightly to achieve what I think is the intent: to trace out the execution of functions, and also to capture the list of arguments as well as the keyword arguments. Thank you @jeff-laughlin!

from functools import wraps                                                                                                                                                                                                     
import time                                                                                                                                                                                                                     
def named(func):                                                                                                                                                                                                                
    def _(*args, **kwargs):                                                                                                                                                                                                     
        print(f"From wrapper function: Executing function named: {func.__name__}, with arguments: {args}, and keyword arguments: {kwargs}.")                                                                                    
        print(f"From wrapper function: {func}")                                                                                                                                                                                 
        start_time = time.time()                                                                                                                                                                                                
        return_value = func(*args, **kwargs)                                                                                                                                                                                    
        end_time = time.time()                                                                                                                                                                                                  
        elapsed_time = end_time - start_time                                                                                                                                                                                    
        print(f"From wrapper function: Execution of {func.__name__} took {elapsed_time} seconds.")                                                                                                                              
        return return_value                                                                                                                                                                                                     
    return _                                                                                                                                                                                                                    
def thanks(message, concepts, username='@jeff-laughlin'):                                                                                                                                                                       
    print(f"From inner function: {message} {username} for teaching me about the {concepts} concepts of closures and decorators!")                                                                                               
thanks('Thank you', 'two', username='@jeff-laughlin')                                                                                                                                                                           
thanks('Thank you', 'two', username='stackoverflow')

From wrapper function: Executing function named: thanks, with arguments: (‘Thank you’, ‘two’), and keyword arguments: {‘username’: ‘@jeff-laughlin’}.

From wrapper function: <function thanks at 0x7f13e6ceaa60>

From inner function: Thank you @jeff-laughlin for teaching me about the two concepts of closures and decorators!

From wrapper function: Execution of thanks took 2.193450927734375e-05 seconds.


From wrapper function: Executing function named: thanks, with arguments: (‘Thank you’, ‘two’), and keyword arguments: {‘username’: ‘stackoverflow’}.

From wrapper function: <function thanks at 0x7f13e6ceaa60>

From inner function: Thank you stackoverflow for teaching me about the two concepts of closures and decorators!

From wrapper function: Execution of thanks took 7.152557373046875e-06 seconds.
<function thanks at 0x7f13e6ceaca0>

What is most surprising to me is that there is a way to intercept functions at runtime, inspect them, and take some actions based on this. The other surprising thing is the memory address of the inner function was the same both times. Does anyone know why this is? I have a ways to go before I can understand this decorator/closure magic.

Answered By: Pritam Dodeja

It seems from all the answer above that use the inspect library, all are writing something like:

import inspect


But, since the return of inspect.stack()[0] is a NamedTuple of the form:

FrameInfo(frame=<frame at 0x103578810, file '<stdin>', line 1, code <module>>, filename='<stdin>', lineno=1, function='<module>', code_context=None, index=None)

One can simply call by the name, i.e. inspect.stack()[0].function

A small dummy example can be seen here:

    def test_train_UGRIZY_noZ(self, architecture, dataset, hyperrun, wloss):
        log.warning(f"{inspect.stack()[0].function} -- Not Implemented Yet")

Which when run prints:

WARNING - test_train_UGRIZY_noZ -- Not Implemented Yet
Answered By: tallamjr