Python Calls logging. Getting the base class name
<h2>I have a Python source:</h2>

<pre><code>import inspect  
import logging  

logging.basicConfig(level=logging.DEBUG)  

class base_class:  
    def base_method(self):  
      logging.debug("  %s:%s() {" % (self.__class__.__name__, (inspect.stack()[0][3])))
      # . . .
      logging.debug("  } // %s:%s()" % (self.__class__.__name__, (inspect.stack()[0][3])))

class derived_class(base_class):
  def derived_method(self):
    logging.debug("%s:%s()." % (self.__class__.__name__, (inspect.stack()[0][3])))
    self.base_method()
    logging.debug("} // %s:%s()" % (self.__class__.__name__, (inspect.stack()[0][3])))

derived_class().derived_method()
</code></pre>

<hr>

<h2>It prints:</h2>

<pre><code>DEBUG:root:derived_class:derived_method().
DEBUG:root:  derived_class:base_method() {
DEBUG:root:  } // derived_class:base_method()
DEBUG:root:} // derived_class:derived_method()
</code></pre>

<hr>

<h2>I want:</h2>

<pre><code>DEBUG:root:derived_class:derived_method().
DEBUG:root:  base_class:base_method() {
DEBUG:root:  } // base_class:base_method()
DEBUG:root:} // derived_class:derived_method()
</code></pre>

<hr>

<p>Is that possible?
I failed to find an answer in several hours.</p>
