How to get the caller's method name in the called method?

Question

Python: How to get the caller's method name in the called method?

Assume I have 2 methods:

def method1(self):
    ...
    a = A.method2()

def method2(self):
    ...

If I don't want to do any change for method1, how to get the name of the caller (in this example, the name is method1) in method2?

Answer 1

Python has just made it better in 3.11, with the addition of codeobject.co_qualname, which return the function fully qualified name.

If you were using the solution from @Asclepius above, consider doing this instead, as it also includes the class name where the method is defined:

import inspect

def who_is_calling():
    print(inspect.currentframe().f_back.f_code.co_qualname)
    

class Klass:
    def method(self): 
        who_is_calling()      # prints 'Klass.method'

    @classmethod
    def class_method(cls): 
        who_is_calling()      # prints 'Klass.class_method'

    @staticmethod
    def static_method():
        who_is_calling()      # prints 'Klass.static_method'

    who_is_calling()          # prints 'Klass'

    class InnerKlass:
        def inner_method(self): 
            who_is_calling()  # prints 'Klass.InnerKlass.inner_method'


def unbound_function(): 
    who_is_calling()          # prints 'unbound_function'

What is good about this? It works for any functions, methods, class methods, static methods... and even a class itself (and nested classes)!

Answer 2

Shorter version:

import inspect

def f1(): f2()

def f2():
    print ('caller name:', inspect.stack()[1][3])

f1()

(with thanks to @Alex, and Stefaan Lippen)

Answer 3

Here is how to get the calling method and its class:

import sys
frame = sys._getframe().f_back
calling_method = frame.f_code.co_name
calling_class = frame.f_locals.get("self").__class__.__name__

Answer 4

Here is a convenient method using @user47's answer.

def trace_caller():
    try:
        raise Exception
    except Exception:
        frame = sys.exc_info()[2].tb_frame.f_back.f_back
        print(" >> invoked by:", frame.f_code.co_name)

Answer 5

For multilevel calling with the purpose of only printing until before the module name, the following function is a simple solution:

import inspect

def caller_name():
    frames = inspect.stack()
    caller_name = ''
    for f in frames:
        if f.function == '<module>':
            return caller_name
        caller_name = f.function

def a():
    caller = caller_name()
    print(f"'a' was called from '{caller}'")

def b():
    a()

def c():
    b()

c()

Output:

'a' was called from 'c'

Answer 6

An alternative to sys._getframe() is used by Python's Logging library to find caller information. Here's the idea:

1. raise an Exception

2. immediately catch it in an Except clause

3. use sys.exc_info to get Traceback frame (tb_frame).

4. from tb_frame get last caller's frame using f_back.

5. from last caller's frame get the code object that was being executed in that frame.

   In our sample code it would be method1 (not method2) being executed.

6. From code object obtained, get the object's name -- this is caller method's name in our sample.

Here's the sample code to solve example in the question:

def method1():
    method2()

def method2():
    try:
        raise Exception
    except Exception:
        frame = sys.exc_info()[2].tb_frame.f_back

    print("method2 invoked by: ", frame.f_code.co_name)

# Invoking method1
method1()

Output:

method2 invoked by: method1

Frame has all sorts of details, including line number, file name, argument counts, argument type and so on. The solution works across classes and modules too.

Answer 7

You can use decorators, and do not have to use stacktrace

If you want to decorate a method inside a class

import functools

# outside ur class
def printOuterFunctionName(func):
@functools.wraps(func)
def wrapper(self):
    print(f'Function Name is: {func.__name__}')
    func(self)    
return wrapper 

class A:
  @printOuterFunctionName
  def foo():
    pass

you may remove functools, self if it is procedural

Answer 8

Code:

#!/usr/bin/env python
import inspect

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

def caller1():
    print "inside: ",called()

def caller2():
    print "inside: ",called()
    
if __name__=='__main__':
    caller1()
    caller2()

Output:

shahid@shahid-VirtualBox:~/Documents$ python test_func.py 
inside:  caller1
inside:  caller2
shahid@shahid-VirtualBox:~/Documents$

Answer 9

I would use inspect.currentframe().f_back.f_code.co_name. Its use hasn't been covered in any of the prior answers which are mainly of one of three types:

• Some prior answers use inspect.stack but it's known to be too slow.
• Some prior answers use sys._getframe which is an internal private function given its leading underscore, and so its use is implicitly discouraged.
• One prior answer uses inspect.getouterframes(inspect.currentframe(), 2)[1][3] but it's entirely unclear what [1][3] is accessing.

import inspect
from types import FrameType
from typing import cast


def demo_the_caller_name() -> str:
    """Return the calling function's name."""
    # Ref: https://stackoverflow.com/a/57712700/
    return cast(FrameType, cast(FrameType, inspect.currentframe()).f_back).f_code.co_name


if __name__ == '__main__':
    def _test_caller_name() -> None:
        assert demo_the_caller_name() == '_test_caller_name'
    _test_caller_name()

Note that cast(FrameType, frame) is used to satisfy mypy.

---

Acknowlegement: comment by 1313e for an answer.

Answer 10

inspect.getframeinfo and other related functions in inspect can help:

>>> import inspect
>>> def f1(): f2()
... 
>>> def f2():
...   curframe = inspect.currentframe()
...   calframe = inspect.getouterframes(curframe, 2)
...   print('caller name:', calframe[1][3])
... 
>>> f1()
caller name: f1

this introspection is intended to help debugging and development; it's not advisable to rely on it for production-functionality purposes.

Answer 11

I found a way if you're going across classes and want the class the method belongs to AND the method. It takes a bit of extraction work but it makes its point. This works in Python 2.7.13.

import inspect, os

class ClassOne:
    def method1(self):
        classtwoObj.method2()

class ClassTwo:
    def method2(self):
        curframe = inspect.currentframe()
        calframe = inspect.getouterframes(curframe, 4)
        print '\nI was called from', calframe[1][3], \
        'in', calframe[1][4][0][6: -2]

# create objects to access class methods
classoneObj = ClassOne()
classtwoObj = ClassTwo()

# start the program
os.system('cls')
classoneObj.method1()

Answer 12

Bit of an amalgamation of the stuff above. But here's my crack at it.

def print_caller_name(stack_size=3):
    def wrapper(fn):
        def inner(*args, **kwargs):
            import inspect
            stack = inspect.stack()

            modules = [(index, inspect.getmodule(stack[index][0]))
                       for index in reversed(range(1, stack_size))]
            module_name_lengths = [len(module.__name__)
                                   for _, module in modules]

            s = '{index:>5} : {module:^%i} : {name}' % (max(module_name_lengths) + 4)
            callers = ['',
                       s.format(index='level', module='module', name='name'),
                       '-' * 50]

            for index, module in modules:
                callers.append(s.format(index=index,
                                        module=module.__name__,
                                        name=stack[index][3]))

            callers.append(s.format(index=0,
                                    module=fn.__module__,
                                    name=fn.__name__))
            callers.append('')
            print('\n'.join(callers))

            fn(*args, **kwargs)
        return inner
    return wrapper

Use:

@print_caller_name(4)
def foo():
    return 'foobar'

def bar():
    return foo()

def baz():
    return bar()

def fizz():
    return baz()

fizz()

output is

level :             module             : name
--------------------------------------------------
    3 :              None              : fizz
    2 :              None              : baz
    1 :              None              : bar
    0 :            __main__            : foo

Answer 13

I've come up with a slightly longer version that tries to build a full method name including module and class.

https://gist.github.com/2151727 (rev 9cccbf)

# Public Domain, i.e. feel free to copy/paste
# Considered a hack in Python 2

import inspect

def caller_name(skip=2):
    """Get a name of a caller in the format module.class.method

       `skip` specifies how many levels of stack to skip while getting caller
       name. skip=1 means "who calls me", skip=2 "who calls my caller" etc.

       An empty string is returned if skipped levels exceed stack height
    """
    stack = inspect.stack()
    start = 0 + skip
    if len(stack) < start + 1:
      return ''
    parentframe = stack[start][0]    

    name = []
    module = inspect.getmodule(parentframe)
    # `modname` can be None when frame is executed directly in console
    # TODO(techtonik): consider using __main__
    if module:
        name.append(module.__name__)
    # detect classname
    if 'self' in parentframe.f_locals:
        # I don't know any way to detect call from the object method
        # XXX: there seems to be no way to detect static method call - it will
        #      be just a function call
        name.append(parentframe.f_locals['self'].__class__.__name__)
    codename = parentframe.f_code.co_name
    if codename != '<module>':  # top level usually
        name.append( codename ) # function or a method

    ## Avoid circular refs and frame leaks
    #  https://docs.python.org/2.7/library/inspect.html#the-interpreter-stack
    del parentframe, stack

    return ".".join(name)

Answer 14

This seems to work just fine:

import sys
print sys._getframe().f_back.f_code.co_name