Python - inspect - Get full caller name (package.module.function)

caller_name.py

# 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
    """
    name  = []
    limit = iter(xrange(1, skip + 1))
    frame = sys._getframe(1)

    try:
        while frame and next(limit):
            frame = frame.f_back
            
    except StopIteration:
        parentframe = frame
    else:
        return ''   
    
    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
    del parentframe
    return ".".join(name)