Using a Python subprocess call to invoke a Python script

Question

I have a Python script that needs to invoke another Python script in the same directory. I did this:

from subprocess import call
call('somescript.py')

I get the following error:

call('somescript.py')
File "/usr/lib/python2.6/subprocess.py", line 480, in call
return Popen(*popenargs, **kwargs).wait()
File "/usr/lib/python2.6/subprocess.py", line 633, in __init__
errread, errwrite)
File "/usr/lib/python2.6/subprocess.py", line 1139, in _execute_child

raise child_exception
OSError: [Errno 2] No such file or directory

I have the script somescript.py in the same folder though. Am I missing something here?

Answer 1

If 'somescript.py' isn't something you could normally execute directly from the command line (I.e., $: somescript.py works), then you can't call it directly using call.

Remember that the way Popen works is that the first argument is the program that it executes, and the rest are the arguments passed to that program. In this case, the program is actually python, not your script. So the following will work as you expect:

subprocess.call(['python', 'somescript.py', somescript_arg1, somescript_val1,...])

This correctly calls the Python interpreter and tells it to execute your script with the given arguments.

Note that this is different from the above suggestion:

subprocess.call(['python somescript.py'])

That will try to execute the program called python somscript.py, which clearly doesn't exist.

call('python somescript.py', shell=True)

Will also work, but using strings as input to call is not cross platform, is dangerous if you aren't the one building the string, and should generally be avoided if at all possible.

Answer 2

Windows? Unix?

Unix will need a shebang and exec attribute to work:

#!/usr/bin/env python

as the first line of script and:

chmod u+x script.py

at command-line or

call('python script.py'.split())

as mentioned previously.

Windows should work if you add the shell=True parameter to the "call" call.

Answer 3

Check out this.

from subprocess import call 
with open('directory_of_logfile/logfile.txt', 'w') as f:
   call(['python', 'directory_of_called_python_file/called_python_file.py'], stdout=f)

Answer 4

import subprocess
command = 'home/project/python_files/run_file.py {} {} {}'.format(
        arg1, arg2, arg3) # if you want to pass any arguments
p = subprocess.Popen(
        [command],
        shell=True,
        stdin=None,
        stdout=subprocess.PIPE,
        stderr=subprocess.PIPE,
        close_fds=True)
out, err = p.communicate()

Answer 5

subprocess.call expects the same arguments as subprocess.Popen - that is a list of strings (the argv in C) rather than a single string.

It's quite possible that your child process attempted to run "s" with the parameters "o", "m", "e", ...

Answer 6

If you're on Linux/Unix you could avoid call() altogether and not execute an entirely new instance of the Python executable and its environment.

import os

cpid = os.fork()
if not cpid:
    import somescript
    os._exit(0)

os.waitpid(cpid, 0)

For what it's worth.

Answer 7

What's wrong with

import sys
from os.path import dirname, abspath

local_dir = abspath(dirname(__file__))
sys.path.append(local_dir)

import somescript

or better still wrap the functionality in a function, e.g. baz, then do this.

import sys
from os.path import dirname, abspath

local_dir = abspath(dirname(__file__))
sys.path.append(local_dir)

import somescript
somescript.baz()

There seem to be a lot of scripts starting python processes or forking, is that a requirement?

Answer 8

def main(argv):
    host = argv[0]
    type = argv[1]
    val = argv[2]

    ping = subprocess.Popen(['python ftp.py %s %s %s'%(host,type,val)],stdout = subprocess.PIPE,stderr = subprocess.PIPE,shell=True)
    out = ping.communicate()[0]
    output = str(out)
    print output

Answer 9

The subprocess call is a very literal-minded system call. it can be used for any generic process...hence does not know what to do with a Python script automatically.

Try

call ('python somescript.py')

If that doesn't work, you might want to try an absolute path, and/or check permissions on your Python script...the typical fun stuff.

Answer 10

First, check if somescript.py is executable and starts with something along the lines of #!/usr/bin/python. If this is done, then you can use subprocess.call('./somescript.py').

Or as another answer points out, you could do subprocess.call(['python', 'somescript.py']).