How do you map a fully qualified class name to its class object in Python?

Question

You can get the fully qualified class name of a Python object like this (see this question):

>>> import Queue
>>> q = Queue.PriorityQueue()
>>> def fullname(o):
    return o.__module__ + "." + o.__class__.__name__  
...   
>>> fullname(q)
'Queue.PriorityQueue'
>>>

How do you do the inverse, ie, map a fully qualified class name like 'Queue.PriorityQueue' to its associated class object (Queue.PriorityQueue)?

I don't think you can, because if I `from Queue import PriorityQueue` then `Queue.PriorityQueue` isn't a class object in my program, it's just `PriorityQueue`. — agf, Aug 28 '11 at 03:31
A long text but I think that's what you need: [Link](http://blog.garlicsim.org/post/2958629526/address-tools-more-powerful-replacements-for-eval) — JBernardo, Aug 28 '11 at 03:39

Eryk Sun · Accepted Answer · 2014-05-13T05:19:32.783

11

You can use importlib in 2.7:

from importlib import import_module

name = 'xml.etree.ElementTree.ElementTree'
parts = name.rsplit('.', 1)
ElementTree = getattr(import_module(parts[0]), parts[1])
tree = ElementTree()

In older versions you can use the __import__ function. It defaults to returning the top level of a package import (e.g. xml). However, if you pass it a non-empty fromlist, it returns the named module instead:

name = 'xml.etree.ElementTree.ElementTree'
parts = name.rsplit('.', 1)    
ElementTree = getattr(__import__(parts[0], fromlist=['']), parts[1])
tree = ElementTree()

edited May 13 '14 at 05:19

answered Aug 28 '11 at 06:10

Eryk Sun

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Thanks @eryksun, this is exactly what we needed (we're using Python 2.6). – Jim Ferrans Aug 30 '11 at 16:20

score 1 · Answer 2 · edited Aug 30 '11 at 02:32

For Python 2.6/2.7



    import sys
    def hasModule(moduleName):
        return moduleName in sys.modules

    def getModule(moduleName):
        if hasModule(moduleName):
            return sys.modules[moduleName]

    def loadModule(moduleName):
        if not hasModule(moduleName):
            return __import__(moduleName)
        return getModule(moduleName)

    def createInstance(fqcn, *args):
        paths = fqcn.split('.')
        moduleName = '.'.join(paths[:-1])
        className = paths[-1]
        module = loadModule(moduleName)
        if module is not None:
            return getattr(module, className)(*args)

    pq = "Queue.PriorityQueue"
    pqObj = createInstance(pq)
    pqObj.put(1)
    print pqObj.get() #1

Jeremy · Answer 3 · 2011-08-28T03:55:20.267

Identifiers in Python aren't really meant to be used like this, but with sys.modules and __import__ you can achive what you want.

import sys

def get_by_qualified_name(name):
    parts = name.split(".")
    module_name = parts[0]
    attribute_names = parts[1:]

    if module_name not in sys.modules:
        __import__(module_name)

    result = sys.modules[module_name]

    for attribute_name in attribute_names:
        result = getattr(result, attribute_name)

    return result

Example Use

my_queue = get_by_qualified_name("Queue.Queue")()
my_queue.put("Hello World")
print my_queue.get() # prints "Hello World"

This will fail if the module is in a package which doesn't import it as part of __ALL__. This is possible to fix, but it requires more elaborate use of __import__.

This will also fail if your string doesn't mean `module.attr1.attr2`, but `package1.package2.module.attr1.attr2`. — glglgl, Aug 28 '11 at 05:43

How do you map a fully qualified class name to its class object in Python?

3 Answers3