How do I redirect stdout to a file when using subprocess.call in python?

Question

I'm calling a python script (B) from another python script (A).

Using subprocess.call, how do I redirect the stdout of B to a file that specify?

I'm using python 2.6.1.

Why are you using subprocess.call() to execute another Python script? Wouldn't it be better to import it and call the appropriate functions? — johnsyweb
– johnsyweb, Commented Feb 19, 2012 at 5:40
Script B is a wrapper script that is calling A to which I want to send a list of arguments that I'm iterating through. — Frez
– Frez, Commented Feb 19, 2012 at 5:49

icktoofay · Accepted Answer · 2012-02-19 05:30:59Z

50

Pass a file as the stdout parameter to subprocess.call:

with open('out-file.txt', 'w') as f:
    subprocess.call(['program'], stdout=f)

answered Feb 19, 2012 at 5:30

icktoofay

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1 Comment

sage Over a year ago

...and this reminded me that I can also put stderr=f.

tabssum · Accepted Answer · 2015-12-02 10:32:24Z

2

Alternate approach

import subprocess
p=subprocess.Popen('lsblk -l|tee a.txt',stdout=subprocess.PIPE,shell=True)
(output,err)=p.communicate()
p_status=p.wait()
print output

Above code will write output of command to file a.txt

answered Dec 2, 2015 at 10:32

tabssum

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2 Comments

jabozzo Over a year ago

This one is useful if for some strange reason you need the subprocess to be in control of the file, not python (my case)

Charles Duffy Over a year ago

...though that can, and should, still be done without shell=True; you can have two separate Popen objects, one for lsblk and a second one for tee, with the first's stdout connected to the second's stdin; the subprocess module docs show how to do so.

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