Python OSError: [Errno 2]

Question

I have the following code that is attempting to start each of the "commands" below in Linux. The module attempts to keep each of the 2 commands running if either should crash for whatever reason.

#!/usr/bin/env python
import subprocess

commands = [ ["screen -dmS RealmD top"], ["screen -DmS RealmD top -d 5"] ]
programs = [ subprocess.Popen(c) for c in commands ]
while True:
    for i in range(len(programs)):
        if programs[i].returncode is None:
            continue # still running
        else:
            # restart this one
            programs[i]= subprocess.Popen(commands[i])
        time.sleep(1.0)

Upon executing the code the following exception is thrown:

Traceback (most recent call last):
  File "./marp.py", line 82, in <module>
    programs = [ subprocess.Popen(c) for c in commands ]
  File "/usr/lib/python2.6/subprocess.py", line 595, in __init__
    errread, errwrite)
  File "/usr/lib/python2.6/subprocess.py", line 1092, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

I think I'm missing something obvious, can anyone see what's wrong with the code above?

Answer 1

commands = [ "screen -dmS RealmD top", "screen -DmS RealmD top -d 5" ]
programs = [ subprocess.Popen(c.split()) for c in commands ]

Answer 2

Just in case.. I also got stuck with this error and the issue was that my files were in DOS instead of UNIX so at :

 return subprocess.call(lst_exp)

where lst_exp is a list of args, one of them was "not found" because it was in DOS instead of UNIX but error thrown was the same :

File "/var/www/run_verifier.py", line 59, in main
return subprocess.call(lst_exp)
File "/usr/lib/python2.7/subprocess.py", line 522, in call
return Popen(*popenargs, **kwargs).wait()
File "/usr/lib/python2.7/subprocess.py", line 710, in __init__
errread, errwrite)
File "/usr/lib/python2.7/subprocess.py", line 1335, in _execute_child
raise child_exception
OSError: [Errno 2] No such file or directory

Answer 3

I got same error when i wrote like this :-

subprocess.Popen("ls" ,shell = False , stdout = subprocess.PIPE ,stderr = subprocess.PIPE)

And problem is solved when i made shell=True .It will work

subprocess.Popen("ls" ,shell = False , stdout = subprocess.PIPE ,stderr = subprocess.PIPE, shell=True)

Answer 4

Use ["screen", "-dmS", "RealmD", "top"] instead of ["screen -dmS RealmD top"].

Maybe also use the complete path to screen.

If the program still cannot be found you can also go through your shell with shell=True, but then you need to quote and escape your parameters etc. Make sure to read the information in the docs if you plan to do that.

Answer 5

The problem is that your command should be split. subprocces requires that the cmd is a list, not a string. It shouldn't be:

subprocess.call('''awk 'BEGIN {FS="\t";OFS="\n"} {a[$1]=a [$1] OFS $2 FS $3 FS $4} END
{for (i in a) {print i a[i]}}' 2_lcsorted.txt > 2_locus_2.txt''') 

That won't work. If you feed subprocess a string, it assumes that that is the path to the command you want to execute. The command needs to be a list. Check out http://www.gossamer-threads.com/lists/python/python/724330. Also, because you're using file redirection, you should use subprocess.call(cmd, shell=True). You can also use shlex.

Answer 6

Only guess is that it can't find screen. Try /usr/bin/screen or whatever which screen gives you.