How to start a background process in Python?

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 subprocesssubprocess.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 subprocessls_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.

Note: This answer is less current than it was when posted in 2009. Using the subprocess module shown in other answers is now recommended in the docs

(Note that the subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using these functions.)

If you want your process to start in the background you can either use system() and call it in the same way your shell script did, or you can spawn it:

import osos.spawnl(os.P_DETACH, 'some_long_running_command')

(or, alternatively, you may try the less portable os.P_NOWAIT flag).

See the documentation here.

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 osos.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.