How to start a background process in Python?

You probably want the answer to “How to call an external command in Python”.

The simplest approach is to use the os.system function, e.g.:

import os
os.system("some_command &")

Basically, whatever you pass to the system function will be executed the same as if you’d passed it to the shell in a script.

You probably want to start investigating the os module for forking different threads (by opening an interactive session and issuing help(os)). The relevant functions are fork and any of the exec ones. To give you an idea on how to start, put something like this in a function that performs the fork (the function needs to take a list or tuple ‘args’ as an argument that contains the program’s name and its parameters; you may also want to define stdin, out and err for the new thread):

try:
    pid = os.fork()
except OSError, e:
    ## some debug output
    sys.exit(1)
if pid == 0:
    ## eventually use os.putenv(..) to set environment variables
    ## os.execv strips of args[0] for the arguments
    os.execv(args[0], args)

While jkp‘s solution works, the newer way of doing things (and the way the documentation recommends) is to use the subprocess module. For simple commands its equivalent, but it offers more options if you want to do something complicated.

Example for your case:

import subprocess
subprocess.Popen(["rm","-r","some.file"])

This will run rm -r some.file in the background. Note that calling .communicate() on the object returned from Popen will block until it completes, so don’t do that if you want it to run in the background:

import subprocess
ls_output=subprocess.Popen(["sleep", "30"])
ls_output.communicate()  # Will block for 30 seconds

See the documentation here.

Also, a point of clarification: “Background” as you use it here is purely a shell concept; technically, what you mean is that you want to spawn a process without blocking while you wait for it to complete. However, I’ve used “background” here to refer to shell-background-like behavior.

I found this here:

On windows (win xp), the parent process will not finish until the longtask.py has finished its work. It is not what you want in CGI-script. The problem is not specific to Python, in PHP community the problems are the same.

The solution is to pass DETACHED_PROCESS Process Creation Flag to the underlying CreateProcess function in win API. If you happen to have installed pywin32 you can import the flag from the win32process module, otherwise you should define it yourself:

DETACHED_PROCESS = 0x00000008

pid = subprocess.Popen([sys.executable, "longtask.py"],
                       creationflags=DETACHED_PROCESS).pid

Use subprocess.Popen() with the close_fds=True parameter, which will allow the spawned subprocess to be detached from the Python process itself and continue running even after Python exits.

https://gist.github.com/yinjimmy/d6ad0742d03d54518e9f

import os, time, sys, subprocess

if len(sys.argv) == 2:
    time.sleep(5)
    print 'track end'
    if sys.platform == 'darwin':
        subprocess.Popen(['say', 'hello'])
else:
    print 'main begin'
    subprocess.Popen(['python', os.path.realpath(__file__), '0'], close_fds=True)
    print 'main end'

Both capture output and run on background with threading

As mentioned on this answer, if you capture the output with stdout= and then try to read(), then the process blocks.

However, there are cases where you need this. For example, I wanted to launch two processes that talk over a port between them, and save their stdout to a log file and stdout.

The threading module allows us to do that.

First, have a look at how to do the output redirection part alone in this question: Python Popen: Write to stdout AND log file simultaneously

Then:

main.py

#!/usr/bin/env python3

import os
import subprocess
import sys
import threading

def output_reader(proc, file):
    while True:
        byte = proc.stdout.read(1)
        if byte:
            sys.stdout.buffer.write(byte)
            sys.stdout.flush()
            file.buffer.write(byte)
        else:
            break

with subprocess.Popen(['./sleep.py', '0'], stdout=subprocess.PIPE, stderr=subprocess.PIPE) as proc1, 
     subprocess.Popen(['./sleep.py', '10'], stdout=subprocess.PIPE, stderr=subprocess.PIPE) as proc2, 
     open('log1.log', 'w') as file1, 
     open('log2.log', 'w') as file2:
    t1 = threading.Thread(target=output_reader, args=(proc1, file1))
    t2 = threading.Thread(target=output_reader, args=(proc2, file2))
    t1.start()
    t2.start()
    t1.join()
    t2.join()

sleep.py

#!/usr/bin/env python3

import sys
import time

for i in range(4):
    print(i + int(sys.argv[1]))
    sys.stdout.flush()
    time.sleep(0.5)

After running:

./main.py

stdout get updated every 0.5 seconds for every two lines to contain:

0
10
1
11
2
12
3
13

and each log file contains the respective log for a given process.

Inspired by: https://eli.thegreenplace.net/2017/interacting-with-a-long-running-child-process-in-python/

Tested on Ubuntu 18.04, Python 3.6.7.

You can use

import os
pid = os.fork()
if pid == 0:
    Continue to other code ...

This will make the python process run in background.

I haven’t tried this yet but using .pyw files instead of .py files should help. pyw files dosen’t have a console so in theory it should not appear and work like a background process.