CODEWITHSUNDEEP
pythonoperating-system
Mandar Deshkar
Mandar Deshkar

Reputation: 21

Avoiding Hard Coding to get Python file paths

I am working on the files with different locations. I want to avoid hard coding at every step. My files Involves, Input files(consisting of inputs), python scripts and output files. Now, I want to interact with each of the mentioned folder without hard-coding my scripts. Does python has any module for trimming down the path to root directory and accessing the subsequent folders.

Currently, I am using os.getcwd() for accessing my script's path. I want to use the output of this code for accessing other files in the different folder

import os
dirpath = os.getcwd()
root_dirpath, py_script = os.path.split(dirpath)

root_dirpath in the code gives me the path for my python script. I want to trim this path to user folder and then access the file from the other folder. Is there any module to do this with out actually hard coding.

Upvotes: 2

Views: 2648

Answers (1)

J_H
J_H

Reputation: 20450

There is current-working-directory, and then there is the directory your program is in. They are distinct concepts, and might have the same value or different.

This expression is pretty useful:

os.path.dirname(os.path.realpath(__file__))

Or more simply, just take dirname of __file__ if your $PYTHONPATH lacks troublesome entries like ..

Some projects will have a "top" directory that is one or more levels up from the location of your python source. You might find it useful to append '../../..' to __file__, or use os.path.join() to accomplish the same thing.

Here is one possibility for locating your favorite files:

from pathlib import Path

top = Path(__file__ + '../..').resolve()
arrow = top / 'assets/arrow.png'
print('source image:', arrow)
print('destination image:', top / 'output' / os.path.basename(arrow))

Caveat: sometimes library code runs directly from a .zip file or other unusual environment. In that case you might find __file__ is not quite as useful.

Upvotes: 2

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