A slurm sbatch wrapper that automatically opens the output file

sbatlog.py

#!/bin/python3

import os
import sys
import subprocess
import re


def run():
    arguments = sys.argv[1:]
    process = subprocess.run(['sbatch', *arguments],
                             stdout=subprocess.PIPE,
                             universal_newlines=True)
    jobinfo = process.stdout.strip()
    jobid = jobinfo.split()[-1]
    print(jobinfo)

    process = subprocess.run(['scontrol', 'show', 'job', jobid],
                             stdout=subprocess.PIPE,
                             universal_newlines=True)

    # splits consecutive spaces, spaces, =, \n
    scontrol = re.split(' +| |=|\n', process.stdout.strip())
    index = scontrol.index("StdOut")
    fileout = scontrol[index + 1].strip()

    print(f'Waiting for {fileout}... ', end='')
    while not os.path.exists(fileout):
        pass
    print('Done!\n'
          'Output:\n')

    os.system(f'tail -f {fileout}')


def usage():
    print('sbatch wrapper that automatically opens the output file\n\n'
          'Usage:\n'
          '$ sbatlog <script.sh> <--any-additional-slurm-commands>')


if __name__ == '__main__':
    if len(sys.argv) > 1:
        run()
    else:
        usage()