How do I start a subprocess in python and not wait for it to return

Question

I'm building a site in django that interfaces with a large program written in R, and I would like to have a button on the site that runs the R program. I have that working, using subprocess.call(), but, as expected, the server does not continue rendering the view until subprocess.call() returns. As this program could take several hours to run, that's not really an option.
Is there any way to run the R program and and keep executing the python code? I've searched around, and looked into subprocess.Popen(), but I couldn't get that to work.
Here's the generic code I'm using in the view:

if 'button' in request.POST:
    subprocess.call('R CMD BATCH /path/to/script.R', shell=True)
    return HttpResponseRedirect('')

Hopefully I've just overlooked something simple.
Thank you.

Answer 1

subprocess.Popen(['R', 'CMD', 'BATCH', '/path/to/script.R'])

The process will be started asynchronously.

Example:

$ cat 1.py
import time
import subprocess

print time.time()
subprocess.Popen(['sleep', '1000'])
print time.time()

$ python 1.py
1340698384.08
1340698384.08

You must note that the child process will run even after the main process stops.