Python access to project root directory

Question

Below folder structure of my application:

rootfolder
          /subfolder1/
          /subfolder2
          /subfolder3/test.py

my code inside of the subfolder3. But I want to write output of the code to subfolder1.

script_dir = os.path.dirname(__file__)

full_path = os.path.join(script_dir,'/subfolder1/')

I would like to know how can I do this wihout importing full path to directory.

"wihout importing full path to directory". Barring the typo, where are you importing anything? — Mad Physicist, Dec 12 '17 at 22:47
I mean os.path.join('/etc/test/tesfdfd/rootfolder/', '/subfolder1/'). I don't want to do like this. — Qaqa Leveo, Dec 12 '17 at 22:49
my second option was like this: path = os.path.abspath(__file__) dir_path = os.path.dirname(path) sys.path.append(dir_path) — Qaqa Leveo, Dec 12 '17 at 22:51
So you're OK with the way you have it now, with `script_dir` coming from `__file__`? — Mad Physicist, Dec 12 '17 at 22:51
Why would you append anything to `sys.path`? That won't change your script output, just where you import modules from. — Mad Physicist, Dec 12 '17 at 22:52
I have done above. I don't know why it can't find folder in sys.path — Qaqa Leveo, Dec 12 '17 at 22:52
`full_path = os.path.join(script_dir, os.pardir, 'subfolder1')` if you're running the script directly from `rootfolder/subfolder3/`. — zwer, Dec 12 '17 at 22:56
What do you get when you do: `print(os.path.abspath(__file__))` instead? — zwer, Dec 12 '17 at 23:01
@QaqaLeveo. Unless you show some concrete examples of the paths that you are dealing with, it is very difficult to solve your problem for you. Given the structure that you have shown here, I am unable to reproduce the issue. Also, please show how you run the script. Is it from the same directory or from another directory? — Mad Physicist, Dec 12 '17 at 23:06
Does this answer your question? [Python - Get path of root project structure](https://stackoverflow.com/questions/25389095/python-get-path-of-root-project-structure) — Давид Шико, Nov 12 '22 at 09:26

Mad Physicist · Accepted Answer · 2017-12-13T23:26:31.267

12

It sounds like you want something along the lines of

project_root = os.path.dirname(os.path.dirname(__file__))
output_path = os.path.join(project_root, 'subfolder1')

The project_root is set to the folder above your script's parent folder, which matches your description. The output folder then goes to subfolder1 under that.

I would also rephrase my import as

from os.path import dirname, join

That shortens your code to

project_root = dirname(dirname(__file__))
output_path = join(project_root, 'subfolder1')

I find this version to be easier to read.

edited Dec 13 '17 at 23:26

answered Dec 12 '17 at 22:55

Mad Physicist

107,652
25
181
264

Print the intermediate results: `__file__`, `project_root` and `output_path`. Make sure they are correct. – Mad Physicist Dec 12 '17 at 23:00
all of them under my hand. Really can't understand what is wrong with this. All of them seems true – Qaqa Leveo Dec 12 '17 at 23:01
I mean show me the printouts. I want to be able to diagnose this. – Mad Physicist Dec 12 '17 at 23:01
`project_root = os.path.dirname(os.path.dirname(__file__))` only returns `NameError: name '__file__' is not defined`. Am I missing something here? – Jakob Mar 14 '23 at 20:03
@Jakob. What version of python are you using, and where is the module stored? – Mad Physicist Mar 14 '23 at 23:46
@MadPhysicist. Python version is 3.9.16 and if you with module mean the python path, it's in a Conda environment under `/Users/username/miniconda3/envs/conda_env/bin/python`. The project (and where I run the code from) is located under `/Users/username/Desktop/Projects/project_name` though. – Jakob Mar 15 '23 at 06:54
Hmm, okay. I tested it in a jupyter notebook (`.ipynb`) and that didn't work. It works in a `.py` file though. – Jakob Mar 15 '23 at 08:15

score -2 · Answer 2 · answered Dec 13 '17 at 01:56

-2

The best way to get this done is to turn your project into a module. Python uses an __init__.py file to recognize this setup. So we can simply create an empty __init__.py file at the root directory. The structure would look like:

rootfolder
          /subfolder1/
          /subfolder2
          /subfolder3/test.py
          __init__.py

Once that is done, you can reference any subfolders like the following:

subfolder1/output.txt

Therefore, your script would look something like this:

f = open("subfolder1/output.txt", "w+")
f.write("works!")
f.close()

answered Dec 13 '17 at 01:56

ZooM SiX

75
4

1

`__init__` makes it a package, not a module. You will then need an `__init__` in `subfolder3` to mark it as a sub-package. None of this, however, will affect where you put your output data. Also, use `with` to open files. – Mad Physicist Dec 13 '17 at 21:32
You're right, my fault! Marking subfolder3 as a sub-package isn't needed for this though. – ZooM SiX Dec 13 '17 at 23:10
The reference won't magically resolve itself unless you run from a very specific directory. That is why OP is using `__file__` and why having an `__init__` or not having it is totally irrelevant. – Mad Physicist Dec 13 '17 at 23:26

Python access to project root directory

2 Answers2