CODEWITHSUNDEEP
pythonarrayssorting
Cocotton
Cocotton

Reputation: 912

Python sort string array with subtring

I have an array which contains the ouput of the "ps aux" command. My goal is to sort the array with the command name column but I have no idea how to do this and I can't manage to find an answer.

Here's my code so far

#!/usr/bin/python
import subprocess

ps = subprocess.Popen(['ps', 'aux'], stdout=subprocess.PIPE).communicate()[0]
processes = ps.split('\n')

nfields = len(processes[0].split()) - 1
for row in processes[1:]:
#    print row.split(None, nfields) //This is used to split all the value in the string
     print row

The output of this code snipet is something like

...
root        11  0.0  0.0      0     0 ?        S<    2012   0:00 [kworker/1:0H]
root        12  0.0  0.0      0     0 ?        S     2012   0:00 [ksoftirqd/1]
root        13  0.0  0.0      0     0 ?        S     2012   0:00 [migration/2]

...

So my goal would have a similar output but sorted on the last column so in the end it would looks like this

...
root        13  0.0  0.0      0     0 ?        S     2012   0:00 [migration/2]
root        12  0.0  0.0      0     0 ?        S     2012   0:00 [ksoftirqd/1]
root        11  0.0  0.0      0     0 ?        S<    2012   0:00 [kworker/1:0H]
...

Anyone of you have any clues on how to do this?

Upvotes: 1

Views: 215

Answers (2)

hughdbrown
hughdbrown

Reputation: 49013

Something like this:

#!/usr/bin/env python
import subprocess
from operator import itemgetter

ps = subprocess.Popen(['ps', 'aux'], stdout=subprocess.PIPE).communicate()[0]
processes = [p for p in ps.split('\n') if p]
split_processes = [p.split() for p in processes]

And then print out your results like this:

for row in sorted(split_processes[1:], key=itemgetter(10)):
    print " ".join(row)

or like this (if you want only the process name and arguments):

for row in sorted(split_processes[1:], key=itemgetter(10)):
    print " ".join(row[10:])

Upvotes: 2

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798746

sorted(..., key=lambda x: x.split()[10])

Upvotes: 2

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