Relative paths break when executing Python script from Windows batch?

Question

My Python script works perfectly if I execute it directly from the directory it's located in. However if I back out of that directory and try to execute it from somewhere else (without changing any code or file locations), all the relative paths break and I get a FileNotFoundError.

The script is located at ./scripts/bin/my_script.py. There is a directory called ./scripts/bin/data/. Like I said, it works absolutely perfectly as long as I execute it from the same directory... so I'm very confused.

Successful Execution (in ./scripts/bin/): python my_script.py

Failed Execution (in ./scripts/): Both python bin/my_script.py and python ./bin/my_script.py

Failure Message:

Traceback (most recent call last):
  File "./bin/my_script.py", line 87, in <module>
    run()
  File "./bin/my_script.py", line 61, in run
    load_data()
  File "C:\Users\XXXX\Desktop\scripts\bin\tables.py", line 12, in load_data
DATA = read_file("data/my_data.txt")
  File "C:\Users\XXXX\Desktop\scripts\bin\fileutil.py", line 5, in read_file
    with open(filename, "r") as file:
FileNotFoundError: [Errno 2] No such file or directory: 'data/my_data.txt'

Relevant Python Code:

def read_file(filename):
    with open(filename, "r") as file:
        lines = [line.strip() for line in file]
        return [line for line in lines if len(line) == 0 or line[0] != "#"]

def load_data():
    global DATA
    DATA = read_file("data/my_data.txt")

Answer 1

Yes, that is logical. The files are relative to your working directory. You change that by running the script from a different directory. What you could do is take the directory of the script you are running at run time and build from that.

import os

def read_file(filename):
    #get the directory of the current running script. "__file__" is its full path
    path, fl = os.path.split(os.path.realpath(__file__))
    #use path to create the fully classified path to your data
    full_path = os.path.join(path, filename)
    with open(full_path, "r") as file:
       #etc

Answer 2

Your resource files are relative to your script. This is OK, but you need to use

os.path.realpath(__file__)

or

os.path.dirname(sys.argv[0])

to obtain the directory where the script is located. Then use os.path.join() or other function to generate the paths to the resource files.