In the file that has the script, you want to do something like this:

import os
dirname = os.path.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')

This will give you the absolute path to the file you're looking for. Note that if you're using setuptools, you should probably use its package resources API instead.

UPDATE: I'm responding to a comment here so I can paste a code sample. :-)

Am I correct in thinking that __file__ is not always available (e.g. when you run the file directly rather than importing it)?

I'm assuming you mean the __main__ script when you mention running the file directly. If so, that doesn't appear to be the case on my system (python 2.5.1 on OS X 10.5.7):

#foo.py
import os
print os.getcwd()
print __file__

#in the interactive interpreter
>>> import foo
/Users/jason
foo.py

#and finally, at the shell:
~ % python foo.py
/Users/jason
foo.py

However, I do know that there are some quirks with __file__ on C extensions. For example, I can do this on my Mac:

>>> import collections #note that collections is a C extension in Python 2.5
>>> collections.__file__
'/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/lib-
dynload/collections.so'

However, this raises an exception on my Windows machine.

It's 2018 now, and Python have already evolve to the __future__ long time ago. So how about using the amazing pathlib coming with Python 3.4 to accomplish the task instead of struggling with os, os.path, glob , shutil, etc.

So we have 3 paths here (possibly duplicated):

• mod_path: which is the path of the simple helper script
• src_path: which contains a couple of template files waiting to be copied.
• cwd: current directory, the destination of those template files.

and the problem is: we don't have the full path of src_path, only know it's relative path to the mod_path.

Now let's solve this with the the amazing pathlib:

# Hope you don't be imprisoned by legacy Python code :)
from pathlib import Path

# `cwd`: current directory is straightforward
cwd = Path.cwd()

# `mod_path`: According to the accepted answer and combine with future power
# if we are in the `helper_script.py`
mod_path = Path(__file__).parent
# OR if we are `import helper_script`
mod_path = Path(helper_script.__file__).parent

# `src_path`: with the future power, it's just so straightforward
relative_path_1 = 'same/parent/with/helper/script/'
relative_path_2 = '../../or/any/level/up/'
src_path_1 = (mod_path / relative_path_1).resolve()
src_path_2 = (mod_path / relative_path_2).resolve()

In the future, it's just that simple.

Moreover, we can select and check and copy/move those template files with pathlib:

if src_path != cwd:
    # When we have different types of files in the `src_path`
    for template_path in src_path.glob('*.ini'):
        fname = template_path.name
        target = cwd / fname
        if not target.exists():
            # This is the COPY action
            with target.open(mode='wb') as fd:
                fd.write(template_path.read_bytes())
            # If we want MOVE action, we could use:
            # template_path.replace(target)

you need os.path.realpath (sample below adds the parent directory to your path)

import sys,os
sys.path.append(os.path.realpath('..'))

See sys.path As initialized upon program startup, the first item of this list, path[0], is the directory containing the script that was used to invoke the Python interpreter.

Use this path as the root folder from which you apply your relative path

>>> import sys
>>> import os.path
>>> sys.path[0]
'C:\\Python25\\Lib\\idlelib'
>>> os.path.relpath(sys.path[0], "path_to_libs") # if you have python 2.6
>>> os.path.join(sys.path[0], "path_to_libs")
'C:\\Python25\\Lib\\idlelib\\path_to_libs'

From what suggest others and from pathlib documentation, a simple (but not ideal) solution is the following (suppose the file we need to refer to is Test/data/users.csv):

# Current file location: Tests/src/long/module/subdir/some_script.py
from pathlib import Path

# back to Tests/
PROJECT_ROOT = Path(__file__).parents[4]
# then down to Test/data/users.csv
CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'  

with CSV_USERS_PATH.open() as users:
    print(users.read())

This works but looks a bit odd because if you move some_script.py around, the path to the root of our project may change (and we would therefore need to change the parents[4] part).

I think I found a better solution that, based on the same idea. We will use a file paths.py to store where the root of the project is, this file will remain at the same location compared to the root directory.

Tests
├── data
│  └── users.csv
└── src
   ├── long
   │  └── module
   │     └── subdir
   │        └── some_script.py
   ├── main.py
   └── paths.py

Where paths.py's only responsability is to provide PROJECT_ROOT:

from pathlib import Path

PROJECT_ROOT = Path(__file__).parents[1]

All scripts can now use paths.PROJECT_ROOT to express absolute paths from the root of the project. For example in src/long/module/subdir/some_script.py we could have:

from paths import PROJECT_ROOT

CSV_USERS_PATH = PROJECT_ROOT / 'data' / 'users.csv'

def hello():
    with CSV_USERS_PATH.open() as f:
        print(f.read())

And everything goes as expected:

~/Tests/src/$ python main.py

/Users/cglacet/Tests/data/users.csv
hello, user

~/Tests/$ python src/main.py

/Users/cglacet/Tests/data/users.csv
hello, user

The main.py script simply is:

from long.module.subdir import some_script

some_script.hello()

summary of the most important commands

>>> import os
>>> os.path.join('/home/user/tmp', 'subfolder')
'/home/user/tmp/subfolder'
>>> os.path.normpath('/home/user/tmp/../test/..')
'/home/user'
>>> os.path.relpath('/home/user/tmp', '/home/user')
'tmp'
>>> os.path.isabs('/home/user/tmp')
True
>>> os.path.isabs('/tmp')
True
>>> os.path.isabs('tmp')
False
>>> os.path.isabs('./../tmp')
False
>>> os.path.realpath('/home/user/tmp/../test/..') # follows symbolic links
'/home/user'

A detailed description is found in the docs. These are linux paths. Windows should work analogous.

Example

Here's an example, tested in Python '3.9.5`:

your current directory: 'c:\project1\code\'
and you want to access the following folder: 'c:\project1\dataset\train\'.
Then you can access the folder using the following address: '../dataset/train/'

References

If you want some more information about path in Python, read this:

• PEP - 355
• PEP - 519

I think to work with all systems use "ntpath" instead of "os.path". Today, it works well with Windows, Linux and Mac OSX.

import ntpath
import os
dirname = ntpath.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')

A simple solution would be

import os
os.chdir(os.path.dirname(__file__))

From C:\Users\xyz\myFolder to C:\Users\xyz\testdata :

import os
working_dir = os.path.abspath(os.path.dirname(os.path.dirname(__file__)))
# C:\Users\xyz\myFolder
print(working_dir)
updated_working_dir = os.path.join(os.path.realpath(working_dir + '/../'), 'testdata')
# C:\Users\xyz\testdata
print(updated_working_dir)

Output

C:\Users\xyz\myFolder
C:\Users\xyz\testdata

Here is my sumup:

First, define the tool function named relpath, which convert a relative path to current file into a relative path to cwd

import os
relpath = lambda p: os.path.normpath(os.path.join(os.path.dirname(__file__), p))

Then we use it to wrap paths which is relative to current file

path1 = relpath('../src/main.py')

And you can also call sys.path.append() to import file relative to current file position

sys.path.append(relpath('..')) # so that you can import from upper dir

The full example code : https://gist.github.com/luochen1990/9b1ffa30f5c4a721dab5991e040e3eb1

Say the current archive named "Helper" and the upper directory named "Workshop", and the template files are in \Workshop\Templates, then the relative path in Python is "..\Templates".

This a simple way to add a relative path to the system path set . For example, for frequent case when the target directory is one level above (thus, '/../') the working directory:

import os
import sys
workingDir = os.getcwd()
targetDir = os.path.join(os.path.relpath(workingDir + '/../'),'target_directory')
sys.path.insert(0,targetDir)

This solution was tested for:

Python 3.9.6 | packaged by conda-forge | (default, Jul 11 2021, 03:37:25) [MSC v.1916 64 bit (AMD64)]