Suppress output from subprocess.Popen

Question

How do you stop the output from subprocess.Popen from being output? Printing can sometimes be slow if there is a great deal of it.

Almost duplicate of [python - How to hide output of subprocess - Stack Overflow](https://stackoverflow.com/questions/11269575/how-to-hide-output-of-subprocess) (Popen versus call) — user202729, Dec 05 '21 at 16:49

score 30 · Answer 1 · answered Jul 16 '13 at 14:50

30

In Python 3.3+ you could use subprocess.DEVNULL, to suppress the output:

from subprocess import DEVNULL, STDOUT, check_call

check_call([cmd, arg1, arg2], stdout=DEVNULL, stderr=STDOUT)

Remove stderr=STDOUT if you don't want to suppress stderr also.

answered Jul 16 '13 at 14:50

jfs

This also worked for me (Python 3.6, Ubuntu Linux OS): subprocess.Popen(cmd, shell=False, stdout=DEVNULL) – Jamie Nicholl-Shelley May 07 '21 at 13:31

score 30 · Accepted Answer · answered Aug 16 '11 at 17:56

30

If you want to totally throw it away:

import subprocess
import os
with open(os.devnull, 'w') as fp:
    cmd = subprocess.Popen(("[command]",), stdout=fp)

If you are using Python 2.5, you will need from __future__ import with_statement, or just don't use with.

answered Aug 16 '11 at 17:56

Brent Newey

1

This no longer works. Please update. TypeError: popen() got an unexpected keyword argument 'stdout' – James Dalgleish Jun 14 '22 at 21:35

score 0 · Answer 3 · answered May 07 '21 at 13:32

0

This also worked for me (Python 3.6, Ubuntu Linux OS):

subprocess.Popen(cmd, shell=False, stdout=DEVNULL)

-This assumes you want non blocking call and no junk in the console from the cmd.

answered May 07 '21 at 13:32

3 Answers3