Calling a Python file with arguments form within another Python file

Question

Suppose I need to run a python file, x.py, from within y.py, passing variable z from y.py to x.py.

How would I accomplish this? Would something like this work?

call(["python","x.py",z])

You should encapsulate the functionality of y in a function, then import that function in to x and call it appropriately. — Silas Ray
– Silas Ray, Commented Apr 15, 2013 at 17:47
That will work if call is subprocess.call and z is a string. You need to write some code and do some experimenting. — tdelaney
– tdelaney, Commented Apr 15, 2013 at 18:02

Silas Ray · Accepted Answer · 2013-04-15 17:53:32Z

You need to encapsulate your code correctly. You should have something like this:

y.py

def y_func(*args): # everything y does if __name__ == "__main__": import sys y_func(*sys.argv)

x.py

from y import y_func def x_func(*args): #do stuff y_result = y_func(*yargs) #finish x stuff if __name__ == "__main__": import sys x_func(*sys.argv)

Çağdaş Salur · Accepted Answer · 2013-04-15 17:46:35Z

0

Try this

# y.py import os z='argument' os.system('Python x.py '+z)

answered Apr 15, 2013 at 17:46

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Comments

0x90 · Accepted Answer · 2013-04-15 18:13:29Z

Using subprocess is better than os.system

import subprocess subprocess.check_output(["ls", "-ltr"])

In your case let's say you have a.py:

import sys print sys.argv[1:]

and in the prompt:

>>> import subprocess >>> subprocess.check_output(["python", 'a.py', 'aaa bbb ccc']) "['aaa bbb ccc']\n" >>>