Python - Get Path of Root Project Structure

Python access to project root directory

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.

Finding File Relative to Project Root in Python

This would be a different answer if it were and the question were "How do you import that from here," but "How do you load that" is pretty straightforward.

Remember that __file__ is the relative path from your curdir to the python file it's being executed in. This lets us calculate the path in two ways: one using the modern pathlib stdlib library and one with os commands.

# using pathlib
from pathlib import Path

thisfile = Path(__file__)
programs = thisfile.parent
project1 = programs.parent
data = project1 / 'data' # yep, we're dividing by a string. pathlib is awesome
resultscsv = data / 'results.csv'

# or

resultscsv = Path(__file__).parent.parent / 'data' / 'results.csv'
# using os
import os.path

thisfile = os.path.abspath(__file__)
programs = os.path.dirname(thisfile)
project1 = os.path.dirname(programs)
data = os.path.join(project1, 'data')
resultscsv = os.path.join(data, 'results.csv')

# or

resultscsv = os.path.join(os.path.dirname(os.path.dirname(os.path.abspath(__file__))), 'data', 'results.csv')

# or POSSIBLY, but this might not work in all places

resultscsv = os.path.join(__file__, '..', '..', 'data', 'results.csv')

The pathlib approach looks massively more readable to me, and is compounded with the fact that then opening the file becomes:

with'r') as f:

rather than the (slightly) more obtuse

with open(resultscsv, mode='r') as f:

Why is python assuming my path is the project root, which is two directory levels up?

According to :



Indicates whether to run a file in the file's directory instead of the current folder.

So presumably just set python.terminal.executeInFileDir to true.

Getting absolute path when trying to get root directory

cd ../ is change directory command which can be used to go back in directory tree.

If you are in /home/, then command
cd ../../../../ will change you current working directory to /home/

In string notation, if ROOT_DIR is giving you /home/, then ROOT_DIR/../../../../ should correspond to /home/, which you want.

Python - Loading files relative from project root

I like to create some sort of configuration file in the project root that has an aboslute path to the root. I do this only because the frameworks i usually use (django, scrapy) have some sort of convention like this

├ ...
├── pve
│ ├── blahblah
│ │ ├──
│ │ ├── TestDefinition.pyc
│ │ ├──
│ │ └── __init__.pyc
│ └──
├── src
│ └── definitions
│ └── THISFILE.yml


import os
PROJECT_ROOT = os.path.abspath(os.path.dirname(__file__))
DEFINITIONS_ROOT = os.path.join(PROJECT_ROOT, 'src', 'definitions')

from myproject import settings

How do I get the parent directory in Python?

Python 3.4

Use the pathlib module.

from pathlib import Path
path = Path("/here/your/path/file.txt")
Old answer

Try this:

import os
print os.path.abspath(os.path.join(yourpath, os.pardir))

where yourpath is the path you want the parent for.

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