One script to run multiple scripts

Question

I am trying to simplify a workflow that requires several individual scripts to be run. So far, I have been able to write a script that runs the other scripts but I have one issue that I can't seem to resolve. Each of the sub-scripts requires a file path and one argument within the path needs to be changed depending on who runs the scripts. Currently, I have to open each sub-script and manually changing this argument.

Is it possible to set this argument to a variable in the parent script, which can then be passed to the subscripts? Thus, it will only need to be set once and will no longer require it to be updated in each sub-script.

So far I have.....

import os

def driver(path: str):
    path_base = path
    path_use = os.path.join(path_base, 'docs', 'analysis', 'forecast')
    
    file_cash = os.path.join(path_use, 'cash.py')
    file_cap = os.path.join(path_use, 'cap.py')

    exec(open(file_cash).read())
    exec(open(file_cap).read())

    return

if __name__ == '__main__':
   driver(path=r'c:\users\[username]')

I would like to set path=r'c:\users\[username]' and then pass that to cash.py and cap.py.

Answer 1

Instead of trying to replicate the behaviour of the import statement, you should directly import these subscripts and pass the values you need them to use as function / method aruments. To import a script from a specific path, you can use importlib.import(), like this:

main.py

import os

def driver(path: str):
    path_use = os.path.join(path, 'docs', 'analysis', 'forecast')
    
    file_cash = os.path.join(path_use, 'cash.py')
    file_cap = os.path.join(path_use, 'cap.py')

    importlib.import(file_cash)
    importlib.import(file_cap)
    
    cash.cash("some_arg")
    cap.cap("some_other_arg")

if __name__ == '__main__':
   driver(path=r'c:\users\[username]')