Get fully qualified class name of an object in Python

Question

For logging purposes I want to retrieve the fully qualified class name of a Python object. (With fully qualified I mean the class name including the package and module name.)

I know about x.__class__.__name__, but is there a simple method to get the package and module?

Answer 1

With the following program

#! /usr/bin/env python

import foo

def fullname(o):
  # o.__module__ + "." + o.__class__.__qualname__ is an example in
  # this context of H.L. Mencken's "neat, plausible, and wrong."
  # Python makes no guarantees as to whether the __module__ special
  # attribute is defined, so we take a more circumspect approach.
  # Alas, the module name is explicitly excluded from __qualname__
  # in Python 3.

  module = o.__class__.__module__
  if module is None or module == str.__class__.__module__:
    return o.__class__.__name__  # Avoid reporting __builtin__
  else:
    return module + '.' + o.__class__.__name__

bar = foo.Bar()
print fullname(bar)

and Bar defined as

class Bar(object):
  def __init__(self, v=42):
    self.val = v

the output is

$ ./prog.py
foo.Bar

Answer 2

The provided answers don't deal with nested classes. Though it's not available until Python 3.3 (PEP 3155), you really want to use __qualname__ of the class. Eventually (3.4? PEP 395), __qualname__ will also exist for modules to deal with cases where the module is renamed (i.e. when it is renamed to __main__).

Answer 3

Consider using the inspect module which has functions like getmodule which might be what are looking for:

>>>import inspect
>>>import xml.etree.ElementTree
>>>et = xml.etree.ElementTree.ElementTree()
>>>inspect.getmodule(et)
<module 'xml.etree.ElementTree' from 
        'D:\tools\python2.5.2\lib\xml\etree\ElementTree.pyc'>

Answer 4

Here's one based on Greg Bacon's excellent answer, but with a couple of extra checks:

__module__ can be None (according to the docs), and also for a type like str it can be __builtin__ (which you might not want appearing in logs or whatever). The following checks for both those possibilities:

def fullname(o):
    module = o.__class__.__module__
    if module is None or module == str.__class__.__module__:
        return o.__class__.__name__
    return module + '.' + o.__class__.__name__

(There might be a better way to check for __builtin__. The above just relies on the fact that str is always available, and its module is always __builtin__)

Answer 5

__module__ would do the trick.

Try:

>>> import re
>>> print re.compile.__module__
re

This site suggests that __package__ might work for Python 3.0; However, the examples given there won't work under my Python 2.5.2 console.

Answer 6

This is a hack but I'm supporting 2.6 and just need something simple:

>>> from logging.handlers import MemoryHandler as MH
>>> str(MH).split("'")[1]

'logging.handlers.MemoryHandler'

Answer 7

For python3.7 I use:

".".join([obj.__module__, obj.__name__])

Getting:

package.subpackage.ClassName

Answer 8

Some people (e.g. https://stackoverflow.com/a/16763814/5766934) arguing that __qualname__ is better than __name__. Here is an example that shows the difference:

$ cat dummy.py 
class One:
    class Two:
        pass

$ python3.6
>>> import dummy
>>> print(dummy.One)
<class 'dummy.One'>
>>> print(dummy.One.Two)
<class 'dummy.One.Two'>
>>> def full_name_with_name(klass):
...     return f'{klass.__module__}.{klass.__name__}'
>>> def full_name_with_qualname(klass):
...     return f'{klass.__module__}.{klass.__qualname__}'
>>> print(full_name_with_name(dummy.One))  # Correct
dummy.One
>>> print(full_name_with_name(dummy.One.Two))  # Wrong
dummy.Two
>>> print(full_name_with_qualname(dummy.One))  # Correct
dummy.One
>>> print(full_name_with_qualname(dummy.One.Two))  # Correct
dummy.One.Two

Note, it also works correctly for buildins:

>>> print(full_name_with_qualname(print))
builtins.print
>>> import builtins
>>> builtins.print
<built-in function print>

Answer 9

Since the interest of this topic is to get fully qualified names, here is a pitfall that occurs when using relative imports along with the main module existing in the same package. E.g., with the below module setup:

$ cat /tmp/fqname/foo/__init__.py
$ cat /tmp/fqname/foo/bar.py
from baz import Baz
print Baz.__module__
$ cat /tmp/fqname/foo/baz.py
class Baz: pass
$ cat /tmp/fqname/main.py
import foo.bar
from foo.baz import Baz
print Baz.__module__
$ cat /tmp/fqname/foo/hum.py
import bar
import foo.bar

Here is the output showing the result of importing the same module differently:

$ export PYTHONPATH=/tmp/fqname
$ python /tmp/fqname/main.py
foo.baz
foo.baz
$ python /tmp/fqname/foo/bar.py
baz
$ python /tmp/fqname/foo/hum.py
baz
foo.baz

When hum imports bar using relative path, bar sees Baz.__module__ as just "baz", but in the second import that uses full name, bar sees the same as "foo.baz".

If you are persisting the fully-qualified names somewhere, it is better to avoid relative imports for those classes.

Answer 10

None of the answers here worked for me. In my case, I was using Python 2.7 and knew that I would only be working with newstyle object classes.

def get_qualified_python_name_from_class(model):
    c = model.__class__.__mro__[0]
    name = c.__module__ + "." + c.__name__
    return name