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'

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

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

Hi first of all you should understand functions os.path.abspath(path) and os.path.relpath(path)

In short os.path.abspath(path) makes a relative path to absolute path. And if the path provided is itself a absolute path then the function returns the same path.

similarly os.path.relpath(path) makes a absolute path to relative path. And if the path provided is itself a relative path then the function returns the same path.

Below example can let you understand the above concept properly:

suppose i have a file input_file_list.txt which contains list of input files to be processed by my python script.

D:concinput1.dic

D:concinput2.dic

D:Copyioconcinput_file_list.txt

If you see above folder structure, input_file_list.txt is present in Copyofconc folder and the files to be processed by the python script are present in conc folder

But the content of the file input_file_list.txt is as shown below:

..concinput1.dic

..concinput2.dic

And my python script is present in D: drive.

And the relative path provided in the input_file_list.txt file are relative to the path of input_file_list.txt file.

So when python script shall executed the current working directory (use os.getcwd() to get the path)

As my relative path is relative to input_file_list.txt, that is “D:Copyofconc”, i have to change the current working directory to “D:Copyofconc”.

So i have to use os.chdir(‘D:Copyofconc’), so the current working directory shall be “D:Copyofconc”.

Now to get the files input1.dic and input2.dic, i will read the lines “..concinput1.dic” then shall use the command

input1_path= os.path.abspath(‘..concinput1.dic’) (to change relative path to absolute path. Here as current working directory is “D:Copyofconc”, the file “.concinput1.dic” shall be accessed relative to “D:Copyofconc”)

so input1_path shall be “D:concinput1.dic”

I’m not sure if this applies to some of the older versions, but I believe Python 3.3 has native relative path support.

For example the following code should create a text file in the same folder as the python script:

open("text_file_name.txt", "w+t")

(note that there shouldn’t be a forward or backslash at the beginning if it’s a relative path)

An alternative which works for me:

this_dir = os.path.dirname(__file__) 
filename = os.path.realpath("{0}/relative/file.path".format(this_dir))

Consider my code:

import os


def readFile(filename):
    filehandle = open(filename)
    print filehandle.read()
    filehandle.close()



fileDir = os.path.dirname(os.path.realpath('__file__'))
print fileDir

#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)

#For accessing the file in a folder contained in the current folder
filename = os.path.join(fileDir, 'Folder1.1/same.txt')
readFile(filename)

#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)

#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)

Instead of using

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

as in the accepted answer, it would be more robust to use:

import inspect
import os
dirname = os.path.dirname(os.path.abspath(inspect.stack()[0][1]))
filename = os.path.join(dirname, 'relative/path/to/file/you/want')

because using __file__ will return the file from which the module was loaded, if it was loaded from a file, so if the file with the script is called from elsewhere, the directory returned will not be correct.

These answers give more detail: https://stackoverflow.com/a/31867043/5542253 and https://stackoverflow.com/a/50502/5542253

This code will return the absolute path to the main script.

import os
def whereAmI():
    return os.path.dirname(os.path.realpath(__import__("__main__").__file__))

This will work even in a module.

As mentioned in the accepted answer

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

I just want to add that

the latter string can’t begin with the backslash , infact no string
should include a backslash

It should be something like

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

The accepted answer can be misleading in some cases , please refer to this link for details

What worked for me is using sys.path.insert. Then I specified the directory I needed to go. For example I just needed to go up one directory.

import sys
sys.path.insert(0, '../')

It’s 2018 now, and Python has already evolved 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 its relative path to the mod_path.

Now let’s solve this with 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)

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.

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')

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()

A simple solution would be

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

From C:UsersxyzmyFolder to C:Usersxyztestdata :

import os
working_dir = os.path.abspath(os.path.dirname(os.path.dirname(__file__)))
# C:UsersxyzmyFolder
print(working_dir)
updated_working_dir = os.path.join(os.path.realpath(working_dir + '/../'), 'testdata')
# C:Usersxyztestdata
print(updated_working_dir)

Output

C:UsersxyzmyFolder
C:Usersxyztestdata

Example

Here’s an example, tested in Python ‘3.9.5`:

your current directory: 'c:project1code'
and you want to access the following folder: 'c:project1datasettrain'.
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

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 WorkshopTemplates, 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)]