我想检查是最好的方法。例如:

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

我用自己的方法在多重继承场景中安全地调用super(我把所有代码都放在这里)

def safe_super(_class, _inst):
    """safe super call"""
    try:
        return getattr(super(_class, _inst), _inst.__fname__)
    except:
        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

示例用法:

class A(object):

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

    @with_name
    def test(self):
        print 'called from A\n'
        safe_super(A, self)()

class B(object):

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

    @with_name
    def test(self):
        print 'called from B\n'
        safe_super(B, self)()

class C(A, B):

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

    @with_name
    def test(self):
        print 'called from C\n'
        safe_super(C, self)()

测试:

a = C()
a.test()

输出:

called from C
called from A
called from B

在每个@with_name修饰的方法中，您可以访问self。__fname__作为当前函数名。

我把这个实用工具放在附近:

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

用法:

myself()

print(inspect.stack()[0].function)似乎也可以工作(Python 3.5)。

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())
    bar()

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

foo()
bar()

在IDE中，代码输出

你好，我是foo，爸爸是 你好，我是bar，爸爸是foo 你好，我是bar，爸爸是

import sys

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

class A(object):
    def __init__(self):
        pass
    def test_class_func_name(self):
        print(func_name())

def test_func_name():
    print(func_name())

测试:

a = A()
a.test_class_func_name()
test_func_name()

输出:

test_class_func_name
test_func_name