running a .py file from another Python script using os

Question

Can someone give me a tip on how I can use os to run a different .py file from my python script? This code below works, but only because I specify the complete file path.

How can I modify the code to incorporate running plots.py from the same directory as my main script app.py ? Im using Windows at the moment but hoping it can work on any operating system. Thanks

import os

os.system('py C:/Users/benb/Desktop/flaskEconServer/plots.py')

Answer 1

You can execute an arbitrary Python script as a separate process using the subprocess.run() function something like this:

import os
import subprocess
import sys

#py_filepath = 'C:/Users/benb/Desktop/flaskEconServer/plots.py'
py_filepath = 'plots_test.py'

args = '"%s" "%s" "%s"' % (sys.executable,                  # command
                           py_filepath,                     # argv[0]
                           os.path.basename(py_filepath))   # argv[1]

proc = subprocess.run(args)
print('returncode:', proc.returncode)

If you would like to communicate with the process while it's running, that can also be done, plus there are other subprocess functions, including the lower-level but very general subprocess.Popen class that support doing those kind of things.

Answer 2

Python has built-in support for executing other scripts, without the need for the os module.

Try:

from . import plots

If you want to execute it in an independent python process, look into the multiprocessing or subprocess modules.

Answer 3

You can get the directory of the app.py file by using the following call in app.py

dir_path = os.path.dirname(os.path.realpath(__file__))

then join the file name you want

file_path = os.path.join(dir_path,'plot.py')

Finally your system call

os.system(f'py {file_path}') # if you're on 3.6 and above.
os.system('py %s' % file_path) # 3.5 and below

As others have said sub-processes and multi-threading may be better, but for your specific question this is what you want.