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: stackoverflow.com/questions/139180/… — Cristian Ramon-Cortes
– Cristian Ramon-Cortes, Commented Aug 31, 2021 at 7:52
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()) — user9413641
– user9413641, Commented Aug 31, 2021 at 7:54

VPfB · Accepted Answer · 2021-08-31 08:23:55Z

3

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.

answered Aug 31, 2021 at 8:23

VPfB

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