How do I get the filepath for a class in Python?

Question

Given a class C in Python, how can I determine which file the class was defined in? I need something that can work from either the class C, or from an instance of C.

I am doing this because I am generally a fan of putting files that belong together in the same folder. I want to create a class that uses a Django template to render itself as HTML. The base implementation should infer the filename for the template based on the filename that the class is defined in.

Say I put a class LocationArtifact in the file "base/artifacts.py", then I want the default behaviour to be that the template name is "base/LocationArtifact.html".

Duplicate: stackoverflow.com/questions/269795/…, stackoverflow.com/questions/602846/… — S.Lott
– S.Lott, Commented Mar 30, 2009 at 14:06
Those assume you know the module you are looking up the file for, I will just have the module string as I work with implementations off a class. — Staale
– Staale, Commented Mar 31, 2009 at 5:51
Does this answer your question? How do I find the location of Python module sources? — mkrieger1
– mkrieger1, Commented Jul 30, 2023 at 21:46
@mkrieger1 that doesn't work, because the input here is a class, not a module. It's easy enough to check the __file__ of a module, but classes don't generally have that, nor do they have a direct link to the module in which they were defined. It takes a couple of extra steps, as in Jarret Hardie's answer. — Karl Knechtel
– Karl Knechtel, Commented Jun 29, 2024 at 23:51

DNS · Accepted Answer · 2009-03-30 17:03:40Z

195

You can use the inspect module, like this:

import inspect
inspect.getfile(C.__class__)

edited Mar 30, 2009 at 17:03

answered Mar 30, 2009 at 14:14

DNS

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5 Comments

AJP Over a year ago

Not sure what I was doing different but instead of getfile I had to use: inspect.getmodule(C.__class__)

Daniel Braun Over a year ago

Note: does not work on a classes created by the user

cheshirekow Over a year ago

This should probably be inspect.getfile(C). If C is a class, then C.__class__ refers to object, which will raise an exception TypeError: <module 'builtins' (built-in)> is a built-in class. I think that only for an instance c do you want to use inspect.getfile(c.__class__).

martian111 Over a year ago

This did not for a class that extends an abstract base class (metaclass=abc.ABCMeta), as it returns /usr/local/lib/python3.7/abc.py instead of the appropriate file. The solution by @JarretHardie (below) worked better.

Cecil Curry Over a year ago

The accepted answer: guaranteed to be the wrong answer. It's a StackOverflow tradition by now.

Jarret Hardie · Accepted Answer · 2009-03-30 14:05:42Z

55

try:

import sys, os
os.path.abspath(sys.modules[LocationArtifact.__module__].__file__)

answered Mar 30, 2009 at 14:05

Jarret Hardie

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5 Comments

martian111 Over a year ago

To get the path from an instance of C (as desired by the OP), replace LocationArtifact with obj.__class__ where obj is an instance of LocationArtifact.

Vidar Over a year ago

This also works for meta classes!

Micah Over a year ago

inspect take alot resources, this is perfect

Cecil Curry Over a year ago

This is the way. inspect should generally be avoided in production code. It's slow. It's fragile. It fails in common edge cases. If you can't write a one-liner that does it, you probably shouldn't do it.

Abbas Gadhia Over a year ago

yuck...........

Soviut · Accepted Answer · 2009-03-30 14:17:09Z

5

This is the wrong approach for Django and really forcing things.

The typical Django app pattern is:

/project
- /appname
  - models.py
  - views.py
  - /templates
    - index.html
    - etc.

answered Mar 30, 2009 at 14:17

Soviut

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2 Comments

S.Lott Over a year ago

+1: Do what Django does naturally and life is so much simpler.

Soviut Over a year ago

Agreed. Django is one of the frameworks with the least amount of "magic", but templates, template tags and apps have some expectations as part of their pattern. If you're having to do wacky class inference you're probably going in the wrong direction.

mustapha_1234 · Accepted Answer · 2023-06-15 12:05:48Z

-1

You can use the error to get the path of your class

import os
import traceback
class C:
  def getClassPath(self):
    #get foldet path
    try:
        raise ValueError("Example error")
    except Exception as e:
        tb = traceback.extract_tb(e.__traceback__)
       
        return os.path.dirname(tb[-1].filename)
c = C()
print('path is',c.getClassPath())

answered Jun 15, 2023 at 12:05

mustapha_1234

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