This is a scenario where a module contains a Logger and it is being imported into another module. When main.py is called however, no LogRecords are written to the log. What can be revised in order for the log to be called?
#objects.py
import logging
logger = logging.getLogger(__name__)
logger.setLevel(logging.INFO)
formatter = logging.Formatter()
log_handler = logging.FileHandler('log_objects.log')
log_handler.setFormatter(formatter)
logger.addHandler(log_handler)
class Person:
def __init__(self):
pass
logger.info(f"A new {self.__class__.__name__} is created.")
#main.py
from my_objects import objects
if __name__ == '__main__':
objects.Person()
-dir/
-my_objects/
__init__.py
log_objects.log
objects.py
-main.py
I'm expecting to see this in the log:
"A new Person is created."
Upon execution nothing shows in the log
Since you do not specify a path name for log_objects.log, the file will be created under the directory where the Python interpreter is run, which in this case, is where main.py is located, rather than under the my_objects sub-directory.
If you want log_objects.log to be created under the my_objects directory, you can specify a relative path:
log_handler = logging.FileHandler('my_objects/log_objects.log')
It works as expected. 'log_objects.log' is got created with valid line.
Are you expecting the text to be printed on the console as well? Once specific file handler is added the default handler will be overwritten.
(or) You possibly missing that how module load happens in python
$ python main.py
$ tree .
.
├── log_objects.log
├── main.py
└── my_objects
├── __init__.py
└── objects.py
$ cat log_objects.log
A new Person is created.
I am currently having difficulties import some functions which are located in a python file which is in the parent directory of the working directory of the main flask application script. Here's how the structure looks like
project_folder
- public
--app.py
-scripts.py
here's a replica code for app.py:
def some_function():
from scripts import func_one, func_two
func_one()
func_two()
print('done')
if __name__ == "__main__":
some_function()
scripts.py contain the function as such:
def func_one():
print('function one successfully imported')
def func_two():
print('function two successfully imported')
What is the pythonic way of importing those functions in my app.py?
Precede it with a dot so that it searches the current directory (project_folder) instead of your python path:
from .scripts import func_one, func_two
The details of relative imports are described in PEP 328
Edit: I assumed you were working with a package. Consider adding an __init__.py file.
Anyways, you can import anything in python by altering the system path:
import sys
sys.path.append("/path/to/directory")
from x import y
1.
import importlib.util
def loadbasic():
spec = importlib.util.spec_from_file_location("basic", os.path.join(os.path.split(__file__)[0], 'basic.py'))
basic = importlib.util.module_from_spec(spec)
spec.loader.exec_module(basic)
return basic #returns a module
Or create an empty file __init__.py in the directory.
And
do not pollute your path with appends.
I'm trying to create an auto_import function which is part of a library: the purpose of this to avoid listing from .x import y many times in __init__ files, only do something this import lib; lib.auto_import(__file__) <- this would search for python files in that folder where the __init__ is present and would import all stuff by exec statement (i.e. exec('from .x import abc')).
My problem is that, somehow the 'from' statement always tries to import .x from lib directory, even if I change the cwd to the directory where the actual __init__ file is placed... How should I solve this? How should I change the search dir for from . statement?
Structure:
$ ls -R
.:
app.py lib x
./lib:
__init__.py auto_import.py
./x:
__init__.py y
./x/y:
__init__.py y.py
e.g.: ./x/y/__init__.py contains import lib; lib.auto_import(__file__)
auto_import is checking for files in dir of __file__ and import them with exec('from .{} import *') (but this from . is always the lib folder and not the dir of __file__, and that is my question, how to change this to dir of __file__
Of course the whole stuff is imported in app.py like:
import x
print(x.y)
Thanks
EDIT1: final auto_import (globals() / gns cannot be avoided )
import os, sys, inspect
def auto_import(gns):
current_frame = inspect.currentframe()
caller_frame = inspect.getouterframes(current_frame)[1]
src_file = caller_frame[1]
for item in os.listdir(os.path.dirname(src_file)):
item = item.split('.py')[0]
if item in ['__init__', '__pycache__']:
continue
gns.update(__import__(item, gns, locals(), ['*'], 1).__dict__)
The problem of your approach is that auto_import is defined in lib/auto_import.py so the context for exec('from .x import *') is always lib/. Even though you manage to fix the path problem, lib.auto_import(__file__) will not import anything to the namespace of lib.x.y, because the function locates in another module.
Use the built-in function __import__
Here is the auto_import script:
myimporter.py
# myimporter.py
def __import_siblings__(gns, lns={}):
for name in find_sibling_names(gns['__file__']):
gns.update((k,v) for k,v in __import__(name, gns,lns).__dict__.items() if not k.startswith('_'))
import re,os
def find_sibling_names(filename):
pyfp = re.compile(r'([a-zA-Z]\w*)\.py$')
files = (pyfp.match(f) for f in os.listdir(os.path.dirname(filename)))
return set(f.group(1) for f in files if f)
Inside your lib/x/y/__init__.py
#lib/x/y/__init__.py
from myimporter import __import_siblings__
__import_siblings__(globals())
Let's say you have a dummy module that need to be imported to y:
#lib/x/y/dummy.py
def hello():
print 'hello'
Test it:
import x.y
x.y.hello()
Please be aware that from lib import * is usually a bad habit because of namespace pollution. Use it with caution.
Refs:
1
2
Use sys module
With this folder structure:
RootDir
|-- module.py
|--ChildDir
|-- main.py
Now in main.py you can do
import sys
sys.path.append('..')
import module
A believe there are other hacks, but this is the one I know of and works for my purpose. I am not sure, whether it is the best option to go for some kind of auto_import stuff though.
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)
Below Code Returns the path until your project root
import sys
print(sys.path[1])
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()
Try:
ROOT_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
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.
Simple and Dynamic!
this solution works on any OS and in any level of directory:
Assuming your project folder name is my_project
from pathlib import Path
current_dir = Path(__file__)
project_dir = [p for p in current_dir.parents if p.parts[-1]=='my_project'][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
I decided for myself as follows.
Need to get the path to 'MyProject/drivers' from the main file.
MyProject/
├─── RootPackge/
│ ├── __init__.py
│ ├── main.py
│ └── definitions.py
│
├─── drivers/
│ └── geckodriver.exe
│
├── requirements.txt
└── setup.py
definitions.py
Put not in the root of the project, but in the root of the main package
from pathlib import Path
ROOT_DIR = Path(__file__).parent.parent
Use ROOT_DIR:
main.py
# imports must be relative,
# not from the root of the project,
# but from the root of the main package.
# Not this way:
# from RootPackge.definitions import ROOT_DIR
# But like this:
from definitions import ROOT_DIR
# Here we use ROOT_DIR
# get path to MyProject/drivers
drivers_dir = ROOT_DIR / 'drivers'
# Thus, you can get the path to any directory
# or file from the project root
driver = webdriver.Firefox(drivers_dir)
driver.get('http://www.google.com')
Then PYTHON_PATH will not be used to access the 'definitions.py' file.
Works in PyCharm:
run file 'main.py' (ctrl + shift + F10 in Windows)
Works in CLI from project root:
$ py RootPackge/main.py
Works in CLI from RootPackge:
$ cd RootPackge
$ py main.py
Works from directories above project:
$ cd ../../../../
$ py MyWork/PythoProjects/MyProject/RootPackge/main.py
Works from anywhere if you give an absolute path to the main file.
Doesn't depend on venv.
Here is a package that solves that problem: from-root
pip install from-root
from from_root import from_root, from_here
# path to config file at the root of your project
# (no matter from what file of the project the function is called!)
config_path = from_root('config.json')
# path to the data.csv file at the same directory where the callee script is located
# (has nothing to do with the current working directory)
data_path = from_here('data.csv')
Check out the link above and read the readme to see more use cases
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')
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])
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)
I used the ../ method to fetch the current project path.
Example:
Project1 -- D:\projects
src
ConfigurationFiles
Configuration.cfg
Path="../src/ConfigurationFiles/Configuration.cfg"
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
Here's my take on this issue.
I have a simple use-case that bugged me for a while. Tried a few solutions, but I didn't like either of them flexible enough.
So here's what I figured out.
create a blank python file in the root dir -> I call this beacon.py
(assuming that the project root is in the PYTHONPATH so it can be imported)
add a few lines to my module/class which I call here not_in_root.py.
This will import the beacon.py module and get the path to that
module
Here's an example project structure
this_project
├── beacon.py
├── lv1
│ ├── __init__.py
│ └── lv2
│ ├── __init__.py
│ └── not_in_root.py
...
The content of the not_in_root.py
import os
from pathlib import Path
class Config:
try:
import beacon
print(f"'import beacon' -> {os.path.dirname(os.path.abspath(beacon.__file__))}") # only for demo purposes
print(f"'import beacon' -> {Path(beacon.__file__).parent.resolve()}") # only for demo purposes
except ModuleNotFoundError as e:
print(f"ModuleNotFoundError: import beacon failed with {e}. "
f"Please. create a file called beacon.py and place it to the project root directory.")
project_root = Path(beacon.__file__).parent.resolve()
input_dir = project_root / 'input'
output_dir = project_root / 'output'
if __name__ == '__main__':
c = Config()
print(f"Config.project_root: {c.project_root}")
print(f"Config.input_dir: {c.input_dir}")
print(f"Config.output_dir: {c.output_dir}")
The output would be
/home/xyz/projects/this_project/venv/bin/python /home/xyz/projects/this_project/lv1/lv2/not_in_root.py
'import beacon' -> /home/xyz/projects/this_project
'import beacon' -> /home/xyz/projects/this_project
Config.project_root: /home/xyz/projects/this_project
Config.input_dir: /home/xyz/projects/this_project/input
Config.output_dir: /home/xyz/projects/this_project/output
Of course, it doesn't need to be called beacon.py nor need to be empty, essentially any python file (importable) file would do as long as it's in the root directory.
Using an empty .py file sort of guarantees that it will not be moved elsewhere due to some future refactoring.
Cheers
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 ended up needing to do this in various different situations where different answers worked correctly, others didn't, or either with various modifications, so I made this package to work for most situations
pip install get-project-root
from get_project_root import root_path
project_root = root_path(ignore_cwd=False)
# >> "C:/Users/person/source/some_project/"
https://pypi.org/project/get-project-root/
This is not exactly the answer to this question; But it might help someone. In fact, if you know the names of the folders, you can do this.
import os
import sys
TMP_DEL = '×'
PTH_DEL = '\\'
def cleanPath(pth):
pth = pth.replace('/', TMP_DEL)
pth = pth.replace('\\', TMP_DEL)
return pth
def listPath():
return sys.path
def getPath(__file__):
return os.path.abspath(os.path.dirname(__file__))
def getRootByName(__file__, dirName):
return getSpecificParentDir(__file__, dirName)
def getSpecificParentDir(__file__, dirName):
pth = cleanPath(getPath(__file__))
dirName = cleanPath(dirName)
candidate = f'{TMP_DEL}{dirName}{TMP_DEL}'
if candidate in pth:
pth = (pth.split(candidate)[0]+TMP_DEL +
dirName).replace(TMP_DEL*2, TMP_DEL)
return pth.replace(TMP_DEL, PTH_DEL)
return None
def getSpecificChildDir(__file__, dirName):
for x in [x[0] for x in os.walk(getPath(__file__))]:
dirName = cleanPath(dirName)
x = cleanPath(x)
if TMP_DEL in x:
if x.split(TMP_DEL)[-1] == dirName:
return x.replace(TMP_DEL, PTH_DEL)
return None
List available folders:
print(listPath())
Usage:
#Directories
#ProjectRootFolder/.../CurrentFolder/.../SubFolder
print(getPath(__file__))
# c:\ProjectRootFolder\...\CurrentFolder
print(getRootByName(__file__, 'ProjectRootFolder'))
# c:\ProjectRootFolder
print(getSpecificParentDir(__file__, 'ProjectRootFolder'))
# c:\ProjectRootFolder
print(getSpecificParentDir(__file__, 'CurrentFolder'))
# None
print(getSpecificChildDir(__file__, 'SubFolder'))
# c:\ProjectRootFolder\...\CurrentFolder\...\SubFolder
One-line solution
Hi all! I have been having this issue for ever as well and none of the solutions worked for me, so I used a similar approach that here::here() uses in R.
Install the groo package: pip install groo-ozika
Place a hidden file in your root directory, e.g. .my_hidden_root_file.
Then from anywhere lower in the directory hierarchy (i.e. within
the root) run the following:
from groo.groo import get_root
root_folder = get_root(".my_hidden_root_file")
That's it!
It just executes the following function:
def get_root(rootfile):
import os
from pathlib import Path
d = Path(os.getcwd())
found = 0
while found == 0:
if os.path.isfile(os.path.join(d, rootfile)):
found = 1
else:
d=d.parent
return d
The project root directory does not have __init__.py.
I solved this problem by looking for an ancestor directory that does not have __init__.py.
from functools import lru_cache
from pathlib import Path
#lru_cache()
def get_root_dir() -> str:
path = Path().cwd()
while Path(path, "__init__.py").exists():
path = path.parent
return str(path)
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
Important: This solution requires you to run the file as a module with python -m pkg.file and not as a script like python file.py.
import sys
import os.path as op
root_pkg_dirname = op.dirname(sys.modules[__name__.partition('.')[0]].__file__)
Other answers have requirements like depending on an environment variable or the position of another module in the package structure.
As long as you run the script as python -m pkg.file (with the -m), this approach is self-contained and will work in any module of the package, including in the top-level __init__.py file.
import sys
import os.path as op
root_pkg_name, _, _ = __name__.partition('.')
root_pkg_module = sys.modules[root_pkg_name]
root_pkg_dirname = op.dirname(root_pkg_module.__file__)
config_path = os.path.join(root_pkg_dirname, '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().