如何在Python中获取调用方法的名称（动态添加）？

Question

我有类：

class SimpleClass: 
    def __init__(self): 
     pass 

    def init_levels(self): 
     levels = get_levels_list() 
     for level in levels: 
      Transplant(foo, Simplelog, method_name=level) 

移植是动态的添加方法类的类：

class Transplant: 
    def __init__(self, method, host, method_name=None): 
     self.host = host 
     self.method = method 
     self.method_name = method_name 
     setattr(host, method_name or method.__name__, self) 

    def __call__(self, *args, **kwargs): 
     nargs = [self.host] 
     nargs.extend(args) 
     return apply(self.method, nargs, kwargs) 

富是 “移植” 功能：

def foo(self): 
    return 

我怎样才能在foo里面调用方法名称？

比如我执行：

simpleinst = SimpleClass() 
simpleinst.init_levels() 

如何修改我的代码为FOO定义体越来越被调用的方法的名字吗？

Answer 1

你必须明确传递：

class Transplant: 
    def __init__(self, method, host, method_name=None): 
     self.host = host 
     self.method = method 
     self.method_name = method_name or method.__name__ 
     setattr(host, method_name or method.__name__, self) 

    def __call__(self, *args, **kwargs): 
     nargs = [self.host, self.method_name] 
     nargs.extend(args) 
     return apply(self.method, nargs, kwargs) 

并延长foo接受这个参数。

Answer 2

你看过getattr？

getattr(self, "method") 

Answer 3

你可以用一个函数工厂做，make_foo：

的关键步骤是重新定义函数的func_name属性：

foo.func_name = name 

---

class SimpleClass: 
    def init_levels(self): 
     levels = ['foo', 'bar'] 
     for level in levels: 
      # Your original code defined this on `SimpleLog`. Did you mean `SimpleClass`? 
      setattr(SimpleClass, level, make_foo(level)) 

def make_foo(name): 
    def foo(self): 
     print('{n} has been called'.format(n = foo.func_name)) 

    foo.func_name = name 
    return foo 

simpleinst = SimpleClass() 
simpleinst.init_levels() 
simpleinst.foo() 
# foo has been called 
simpleinst.bar() 
# bar has been called