I’ve got a python project with a configuration file in the project root. The configuration file needs to be accessed in a few different files throughout the project.

So it looks something like: <ROOT>/configuration.conf <ROOT>/A/a.py, <ROOT>/A/B/b.py (when b,a.py access the configuration file).

What’s the best / easiest way to get the path to the project root and the configuration file without depending on which file inside the project I’m in? i.e without using ../../? It’s okay to assume that we know the project root’s name.

You can do this how Django does it: define a variable to the Project Root from a file that is in the top-level of the project. For example, if this is what your project structure looks like:

project/
    configuration.conf
    definitions.py
    main.py
    utils.py

In definitions.py you can define (this requires import os):

ROOT_DIR = os.path.dirname(os.path.abspath(__file__)) # This is your Project Root

Thus, with the Project Root known, you can create a variable that points to the location of the configuration (this can be defined anywhere, but a logical place would be to put it in a location where constants are defined – e.g. definitions.py):

CONFIG_PATH = os.path.join(ROOT_DIR, 'configuration.conf')  # requires `import os`

Then, you can easily access the constant (in any of the other files) with the import statement (e.g. in utils.py): from definitions import CONFIG_PATH.

Other answers advice to use a file in the top-level of the project. This is not necessary if you use pathlib.Path and parent (Python 3.4 and up). Consider the following directory structure where all files except README.md and utils.py have been omitted.

project
│   README.md
|
└───src
│   │   utils.py
|   |   ...
|   ...

In utils.py we define the following function.

from pathlib import Path

def get_project_root() -> Path:
    return Path(__file__).parent.parent

In any module in the project we can now get the project root as follows.

from src.utils import get_project_root

root = get_project_root()

Benefits: Any module which calls get_project_root can be moved without changing program behavior. Only when the module utils.py is moved we have to update get_project_root and the imports (refactoring tools can be used to automate this).

All the previous solutions seem to be overly complicated for what I think you need, and often didn’t work for me. The following one-line command does what you want:

import os
ROOT_DIR = os.path.abspath(os.curdir)

To get the path of the “root” module, you can use:

import os
import sys
os.path.dirname(sys.modules['__main__'].__file__)

But more interestingly if you have an config “object” in your top-most module you could -read- from it like so:

app = sys.modules['__main__']
stuff = app.config.somefunc()

A standard way to achieve this would be to use the pkg_resources module which is part of the setuptools package. setuptools is used to create an install-able python package.

You can use pkg_resources to return the contents of your desired file as a string and you can use pkg_resources to get the actual path of the desired file on your system.

Let’s say that you have a package called stackoverflow.

stackoverflow/
|-- app
|   `-- __init__.py
`-- resources
    |-- bands
    |   |-- Dream\ Theater
    |   |-- __init__.py
    |   |-- King's\ X
    |   |-- Megadeth
    |   `-- Rush
    `-- __init__.py

3 directories, 7 files

Now let’s say that you want to access the file Rush from a module app.run. Use pkg_resources.resouces_filename to get the path to Rush and pkg_resources.resource_string to get the contents of Rush; thusly:

import pkg_resources

if __name__ == "__main__":
    print pkg_resources.resource_filename('resources.bands', 'Rush')
    print pkg_resources.resource_string('resources.bands', 'Rush')

The output:

/home/sri/workspace/stackoverflow/resources/bands/Rush
Base: Geddy Lee
Vocals: Geddy Lee
Guitar: Alex Lifeson
Drums: Neil Peart

This works for all packages in your python path. So if you want to know where lxml.etree exists on your system:

import pkg_resources

if __name__ == "__main__":
    print pkg_resources.resource_filename('lxml', 'etree')

output:

/usr/lib64/python2.7/site-packages/lxml/etree

The point is that you can use this standard method to access files that are installed on your system (e.g pip install xxx or yum -y install python-xxx) and files that are within the module that you’re currently working on.

Try:

ROOT_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))

Below Code Returns the path until your project root

import sys
print(sys.path[1])

I struggled with this problem too until I came to this solution. This is the cleanest solution in my opinion.

In your setup.py add “packages”

setup(
name='package_name'
version='0.0.1'
.
.
.
packages=['package_name']
.
.
.
)

In your python_script.py

import pkg_resources
import os

resource_package = pkg_resources.get_distribution(
    'package_name').location
config_path = os.path.join(resource_package,'configuration.conf')

Just an example: I want to run runio.py from within helper1.py

Project tree example:

myproject_root
- modules_dir/helpers_dir/helper1.py
- tools_dir/runio.py

Get project root:

import os
rootdir = os.path.dirname(os.path.realpath(__file__)).rsplit(os.sep, 2)[0]

Build path to script:

runme = os.path.join(rootdir, "tools_dir", "runio.py")
execfile(runme)

This worked for me using a standard PyCharm project with my virtual environment (venv) under the project root directory.

Code below isnt the prettiest, but consistently gets the project root. It returns the full directory path to venv from the VIRTUAL_ENV environment variable e.g. /Users/NAME/documents/PROJECT/venv

It then splits the path at the last /, giving an array with two elements. The first element will be the project path e.g. /Users/NAME/documents/PROJECT

import os

print(os.path.split(os.environ['VIRTUAL_ENV'])[0])

I’ve recently been trying to do something similar and I have found these answers inadequate for my use cases (a distributed library that needs to detect project root). Mainly I’ve been battling different environments and platforms, and still haven’t found something perfectly universal.

Code local to project

I’ve seen this example mentioned and used in a few places, Django, etc.

import os
print(os.path.dirname(os.path.abspath(__file__)))

Simple as this is, it only works when the file that the snippet is in is actually part of the project. We do not retrieve the project directory, but instead the snippet’s directory

Similarly, the sys.modules approach breaks down when called from outside the entrypoint of the application, specifically I’ve observed a child thread cannot determine this without relation back to the ‘main‘ module. I’ve explicitly put the import inside a function to demonstrate an import from a child thread, moving it to top level of app.py would fix it.

app/
|-- config
|   `-- __init__.py
|   `-- settings.py
`-- app.py

app.py

#!/usr/bin/env python
import threading


def background_setup():
    # Explicitly importing this from the context of the child thread
    from config import settings
    print(settings.ROOT_DIR)


# Spawn a thread to background preparation tasks
t = threading.Thread(target=background_setup)
t.start()

# Do other things during initialization

t.join()

# Ready to take traffic

settings.py

import os
import sys


ROOT_DIR = None


def setup():
    global ROOT_DIR
    ROOT_DIR = os.path.dirname(sys.modules['__main__'].__file__)
    # Do something slow

Running this program produces an attribute error:

>>> import main
>>> Exception in thread Thread-1:
Traceback (most recent call last):
  File "C:\Python2714\lib\threading.py", line 801, in __bootstrap_inner
    self.run()
  File "C:\Python2714\lib\threading.py", line 754, in run
    self.__target(*self.__args, **self.__kwargs)
  File "main.py", line 6, in background_setup
    from config import settings
  File "config\settings.py", line 34, in <module>
    ROOT_DIR = get_root()
  File "config\settings.py", line 31, in get_root
    return os.path.dirname(sys.modules['__main__'].__file__)
AttributeError: 'module' object has no attribute '__file__'

…hence a threading-based solution

Location independent

Using the same application structure as before but modifying settings.py

import os
import sys
import inspect
import platform
import threading


ROOT_DIR = None


def setup():
    main_id = None
    for t in threading.enumerate():
        if t.name == 'MainThread':
            main_id = t.ident
            break

    if not main_id:
        raise RuntimeError("Main thread exited before execution")

    current_main_frame = sys._current_frames()[main_id]
    base_frame = inspect.getouterframes(current_main_frame)[-1]

    if platform.system() == 'Windows':
        filename = base_frame.filename
    else:
        filename = base_frame[0].f_code.co_filename

    global ROOT_DIR
    ROOT_DIR = os.path.dirname(os.path.abspath(filename))

Breaking this down: First we want to accurately find the thread ID of the main thread. In Python3.4+ the threading library has threading.main_thread() however, everybody doesn’t use 3.4+ so we search through all threads looking for the main thread save it’s ID. If the main thread has already exited, it won’t be listed in the threading.enumerate(). We raise a RuntimeError() in this case until I find a better solution.

main_id = None
for t in threading.enumerate():
    if t.name == 'MainThread':
        main_id = t.ident
        break

if not main_id:
    raise RuntimeError("Main thread exited before execution")

Next we find the very first stack frame of the main thread. Using the cPython specific function sys._current_frames() we get a dictionary of every thread’s current stack frame. Then utilizing inspect.getouterframes() we can retrieve the entire stack for the main thread and the very first frame. current_main_frame = sys._current_frames()[main_id] base_frame = inspect.getouterframes(current_main_frame)[-1] Finally, the differences between Windows and Linux implementations of inspect.getouterframes() need to be handled. Using the cleaned up filename, os.path.abspath() and os.path.dirname() clean things up.

if platform.system() == 'Windows':
    filename = base_frame.filename
else:
    filename = base_frame[0].f_code.co_filename

global ROOT_DIR
ROOT_DIR = os.path.dirname(os.path.abspath(filename))

So far I’ve tested this on Python2.7 and 3.6 on Windows as well as Python3.4 on WSL

If you are working with anaconda-project, you can query the PROJECT_ROOT from the environment variable –> os.getenv(‘PROJECT_ROOT’). This works only if the script is executed via anaconda-project run .

If you do not want your script run by anaconda-project, you can query the absolute path of the executable binary of the Python interpreter you are using and extract the path string up to the envs directory exclusiv. For example: The python interpreter of my conda env is located at:

/home/user/project_root/envs/default/bin/python

# You can first retrieve the env variable PROJECT_DIR.
# If not set, get the python interpreter location and strip off the string till envs inclusiv...

if os.getenv('PROJECT_DIR'):
    PROJECT_DIR = os.getenv('PROJECT_DIR')
else:
    PYTHON_PATH = sys.executable
    path_rem = os.path.join('envs', 'default', 'bin', 'python')
    PROJECT_DIR = py_path.split(path_rem)[0]

This works only with conda-project with fixed project structure of a anaconda-project

I used the ../ method to fetch the current project path.

Example: Project1 — D:\projects

src

ConfigurationFiles

Configuration.cfg

Path=”../src/ConfigurationFiles/Configuration.cfg”

At the time of writing, none of the other solutions are very self-contained. They depend either on an environment variable or the position of the module in the package structure. The top answer with the ‘Django’ solution falls victim to the latter by requiring a relative import. It also has the disadvantage of having to modify a module at the top level.

This should be the correct approach for finding the directory path of the top-level package:

import sys
import os

root_name, _, _ = __name__.partition('.')
root_module = sys.modules[root_name]
root_dir = os.path.dirname(root_module.__file__)

config_path = os.path.join(root_dir, 'configuration.conf')

It works by taking the first component in the dotted string contained in __name__ and using it as a key in sys.modules which returns the module object of the top-level package. Its __file__ attribute contains the path we want after trimming off /__init__.py using os.path.dirname().

This solution is self-contained. It works anywhere in any module of the package, including in the top-level __init__.py file.

I had to implement a custom solution because it’s not as simple as you might think. My solution is based on stack trace inspection (inspect.stack()) + sys.path and is working fine no matter the location of the python module in which the function is invoked nor the interpreter (I tried by running it in PyCharm, in a poetry shell and other…). This is the full implementation with comments:

def get_project_root_dir() -> str:
    """
    Returns the name of the project root directory.

    :return: Project root directory name
    """

    # stack trace history related to the call of this function
    frame_stack: [FrameInfo] = inspect.stack()

    # get info about the module that has invoked this function
    # (index=0 is always this very module, index=1 is fine as long this function is not called by some other
    # function in this module)
    frame_info: FrameInfo = frame_stack[1]

    # if there are multiple calls in the stacktrace of this very module, we have to skip those and take the first
    # one which comes from another module
    if frame_info.filename == __file__:
        for frame in frame_stack:
            if frame.filename != __file__:
                frame_info = frame
                break

    # path of the module that has invoked this function
    caller_path: str = frame_info.filename

    # absolute path of the of the module that has invoked this function
    caller_absolute_path: str = os.path.abspath(caller_path)

    # get the top most directory path which contains the invoker module
    paths: [str] = [p for p in sys.path if p in caller_absolute_path]
    paths.sort(key=lambda p: len(p))
    caller_root_path: str = paths[0]

    if not os.path.isabs(caller_path):
        # file name of the invoker module (eg: "mymodule.py")
        caller_module_name: str = Path(caller_path).name

        # this piece represents a subpath in the project directory
        # (eg. if the root folder is "myproject" and this function has ben called from myproject/foo/bar/mymodule.py
        # this will be "foo/bar")
        project_related_folders: str = caller_path.replace(os.sep + caller_module_name, '')

        # fix root path by removing the undesired subpath
        caller_root_path = caller_root_path.replace(project_related_folders, '')

    dir_name: str = Path(caller_root_path).name

    return dir_name

There are many answers here but I couldn’t find something simple that covers all cases so allow me to suggest my solution too:

import pathlib
import os

def get_project_root():
    """
    There is no way in python to get project root. This function uses a trick.
    We know that the function that is currently running is in the project.
    We know that the root project path is in the list of PYTHONPATH
    look for any path in PYTHONPATH list that is contained in this function's path
    Lastly we filter and take the shortest path because we are looking for the root.
    :return: path to project root
    """
    apth = str(pathlib.Path().absolute())
    ppth = os.environ['PYTHONPATH'].split(':')
    matches = [x for x in ppth if x in apth]
    project_root = min(matches, key=len)
    return project_root