Python: Run all functions in all python scripts in a folder dynamically

Question

main_folder/
  __init__.py
  sub_folder1/
    __init__.py
    run_all_functions.py
  sub_folder2/
    __init__.py
    script1.py
    script2.py
    script3.py

script1.py

def something():
 ....
def something2():
 ....

script2.py

def example():
 ....
def example2():
 ....

script3.py

def example3():
 ....

Would like to check if there are ways that i can run all the functions in different scripts in folder2 and consolidate them in run_all_functions.py in folder1 dynamically. I might add more scripts with functions in folder 2 in the near future.

If the files define a module you could use `inspect` to load all the methods declared for each module. Then you only have to repeat the process through all the files. This might be useful: https://stackoverflow.com/questions/139180/how-to-list-all-functions-in-a-python-module — Cristian Ramon-Cortes, Aug 31 '21 at 07:52
You could use exec, ie. open the .py files one by one and run exec on the text, like: `with open(filename, 'r') as file: exec(file.read())` — Vinzent, Aug 31 '21 at 07:54

score 3 · Answer 1 · answered Aug 31 '21 at 08:23

Provided that you add a run_all() entry-point function to each .py file in a directory with your scripts, you could load all these file and call their respective run_all() functions.

Here is a little demonstration, adjust DIR.

import importlib.util
import pathlib

DIR = '/tmp/test'

for pyfile in pathlib.Path(DIR).glob('*.py'): # or perhaps 'script*.py'
    spec = importlib.util.spec_from_file_location(f"{__name__}.imported_{pyfile.stem}" , pyfile)
    module = importlib.util.module_from_spec(spec)
    spec.loader.exec_module(module)
    module.run_all()

Without a per module run_all(), you must use some kind of introspection to find all functions to be run.

Python: Run all functions in all python scripts in a folder dynamically

1 Answers1