Find top-level directory from subdirectory on Linux in Python

Question

I am creating various HTML file parts, image thumbnails etc. from within a CodeIgniter application tree using cron-scheduled Python 2.7 programs on Linux. The actual Python programs exist under the CodeIgniter tree in a subdirectory one level below the application directory as follows.

codeigniter/web-root 
 |
 application
 | | 
 | scripts
 | | |
 | | my-program.py
 | | 
 | database
 | |
 | database.sqlite 
 images

I want to determine the codeigniter/web-root directory from within my-program.py using methods from the os.path module. However, the absolute path to the codeigniter/web-root is different on the development and production environments so I prefer not to hardwire this path information into the Python program itself.

The current script(s) use the following construct to determine the absolute path of "codeigniter/web-root" which is two directory levels above the script itself in both environments.

#!/bin/env python2.7
import os.path
ci_root = os.path.dirname(os.path.dirname(os.path.dirname(os.path.abspath(__file__))))

Is there a cleaner way to determine the top level(ci_root) directory without using multiple os.path.dirname calls?

Answer 1

import os.path
print os.path.abspath(__file__+'/../../..') # the directory three levels up

I was pleasantly surprised that

• abspath() managed to parse correctly, without using os.path.join()
• We didn't have to strip out the filename before we build the path

For cross platform compatibility if abspath's parsing does not work, we could use something less readable, e.g.

• os.path.sep instead of '/'
• os.path.join()

Answer 2

Find the real path based on the relative path of "up three directories," and extract the last part of that real path.

[Edit: In response to your comment I have revised this. In my opinion it is getting complex enough that your chain of dirname's is better. At least it is easy to understand what it happening.]

from os import path as p
_, ci_root = p.split(p.abspath(p.join(__file__, '../../..')))

• Ochi: shtako... you were faster :) +1
• T.P.: That actually causes it too jump up one too many directory levels. Is there a means to perform this always based off of the file attribute so that starting directory does not impact the execution of the program. For instance, if I wanted to execute this from my $HOME directory in addition to the cronjob, the programs will break because a relative path is specified.