호출 된 메소드에서 호출자의 메소드 이름을 얻는 방법은 무엇입니까?

호출 된 메소드에서 호출자의 메소드 이름을 얻는 방법은 무엇입니까?

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파이썬 : 호출 된 메소드에서 호출자의 메소드 이름을 얻는 방법은 무엇입니까?

두 가지 방법이 있다고 가정합니다.

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

def method2(self):
    ...

method1을 변경하지 않으려면 method2에서 호출자의 이름을 얻는 방법 (이 예제에서 이름은 method1입니다)?

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inspect.getframeinfo 및 기타 관련 기능 inspect이 도움 이 될 수 있습니다.

>>> 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

이 내부 검사는 디버깅 및 개발을 돕기위한 것입니다. 생산 기능 목적으로 사용하는 것은 좋지 않습니다.

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더 짧은 버전 :

import inspect

def f1(): f2()

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

f1()

(@Alex 및 Stefaan Lippen 덕분에 )

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이것은 잘 작동하는 것 같습니다 :

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

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모듈과 클래스를 포함한 전체 메소드 이름을 작성하려고 시도하는 약간 더 긴 버전을 생각해 냈습니다.

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)

---

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

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

import inspect
import types
from typing import cast


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


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

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Acknowlegement: prior comment by 1313e for an answer.

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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()

---

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

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you can print a list of function like answered here https://stackoverflow.com/a/56897183/4039061

import inspect;print(*['\n\x1b[0;36;1m| \x1b[0;32;1m{:25}\x1b[0;36;1m| \x1b[0;35;1m{}'.format(str(x.function), x.filename+'\x1b[0;31;1m:'+str(x.lineno)+'\x1b[0m') for x in inspect.stack()])

참고URL : https://stackoverflow.com/questions/2654113/how-to-get-the-callers-method-name-in-the-called-method