Relative paths in Python

Question

I\'m building a simple helper script for work that will copy a couple of template files in our code base to the current directory. I don\'t, however, have the absolute path

Answer 1

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')

Answer 2

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:\conc\input1.dic

D:\conc\input2.dic

D:\Copyioconc\input_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:

..\conc\input1.dic

..\conc\input2.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 "..\conc\input1.dic" then shall use the command

input1_path= os.path.abspath('..\conc\input1.dic') (to change relative path to absolute path. Here as current working directory is "D:\Copyofconc", the file ".\conc\input1.dic" shall be accessed relative to "D:\Copyofconc")

so input1_path shall be "D:\conc\input1.dic"

Answer 3

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.

Answer 4

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.

Answer 5

From what suggest others and from pathlib documentation, a simple and clear solution is the following (suppose the file we need to refer to: Test/data/users.csv:

# This 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())

Now this looks a bit odd to me, because if you move some_script.py around, the path to the root of our project may change (we would need to modify parents[4]). On the other hand I found a solution that I prefer based on the same idea.

Suppose we have the following directory structure:

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

The paths.py file will be responsible for storing the root location of our projet:

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()

Answer 6

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.