Change working directory when calling another script in Python

Question

I have three files, base.py, read_a_file.py, and file_to_be_read.txt. read_a_file.py contains the function read_file which opens file_to_be_read.txt to be read. base.py imports read_a_file.py and calls the read_file function. I also have the following file structure:

-directory 
   -base.py
-read_a_file.py
-file_to_be_read.txt

The problem is that when I run base.py, I receive a FileNotFound error since the working directory is located at root/directory/ rather than at the root. I tried sys.path.append(root) but that didn't seem to work. Any help would be appreciated!

Here's the full traceback:

Error
Traceback (most recent call last):
  File "C:\Users\agctute\PycharmProjects\sapaigenealg\myAI\tests\aitests.py", line 26, in test_AI
    self.ai = AI(player=Player())
  File "C:\Users\agctute\PycharmProjects\sapaigenealg\myAI\playerai.py", line 24, in __init__
    self.tier_list = read_tier_list()
  File "C:\Users\agctute\PycharmProjects\sapaigenealg\myAI\stat_comparison.py", line 47, in read_tier_list
    with open('stat_tier_list.txt', 'r') as f:
FileNotFoundError: [Errno 2] No such file or directory: 'stat_tier_list.txt'

please include the full traceback from the error. and it sounds like you need to change your code to work on paths relative to the module, so if you don't know how to do that you should include your actual code to. or at least the parts that deal with paths and imports — Alexander
– Alexander, Commented Aug 11, 2022 at 4:57
I could just change the relative path to the file, but I was wondering if there was any other way to do it. — agctute
– agctute, Commented Aug 11, 2022 at 5:16
You need to change it so that it is relative to the python module that is opening it. That way no matter what directory is the CWD it will always find the right path. — Alexander
– Alexander, Commented Aug 11, 2022 at 5:25
sys.path only helps with importing python modules, not with regular files. — John Gordon
– John Gordon, Commented Aug 12, 2022 at 0:43

agctute · Accepted Answer · 2022-08-12 07:16:46Z

3

You identified the problem correctly but you are searching for the wrong answer. In most cases it is not a good idea to fiddle around with the working directory. If this is done at multiple occasions, you might end up in a place you did not expect.

You run into your problem because your script failed to find a file. This is because you hard-coded the path to it. Something like

with open('stat_tier_list.txt', 'r') as f:

can hardly be good idea. Use

path_to_file = 'stat_tier_list.txt'
with open(path_to_file , 'r') as f:

instead. This way you can make the path an argument of your function

# read_a_file.py
def read_file(path_to_file):
    with open(path_to_file, 'r') as f:

Now, go back to base.py and call the function properly:

# base.py
from pathlib import Path
from read_a_file import read_file # You may have to reorganise your folder structure to make this work

PATH_ROOT = Path('C:\Users\agctute\PycharmProjects\sapaigenealg')

path_to_file = PATH_ROOT.joinpath('file_to_be_read.txt')

read_a_file(path_to_file)

I suppose, that is what you really want. A function read_a_file should always accept an argument which specifies what file has to be read.

edited Aug 12, 2022 at 7:16

agctute

1511 silver badge9 bronze badges

answered Aug 11, 2022 at 7:29

Durtal

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1 Comment

agctute Over a year ago

In hindsight I definitely should have done that huh. Thanks!

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