Python command line argument assign to a variable

Question

I have to either store the command line argument in a variable or assign a default value to it.

What i am trying is the below

import sys
Var=sys.argv[1] or "somevalue"

I am getting the error out of index if i don't specify any argument. How to solve this?

Maybe use a more structured approach, see http://stackoverflow.com/questions/1009860/command-line-arguments-in-python?rq=1 — Barmar, Feb 11 '14 at 07:59
@Barmar: In principle yes, but that's overkill for this scenario, don't you think? — Tim Pietzcker, Feb 11 '14 at 08:01

score 8 · Accepted Answer · answered Feb 11 '14 at 07:52

8

Var=sys.argv[1] if len(sys.argv) > 1 else "somevalue"

answered Feb 11 '14 at 07:52

Michael Schlottke-Lakemper

8,763
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Nice one-liner, I only began using this 'perl-like' method a few weeks ago. – bjd2385 Aug 15 '16 at 04:56

score 5 · Answer 2 · answered Feb 11 '14 at 08:22

The builtin argparse module is intended for exactly these sorts of tasks:

import argparse

# Set up argument parser
ap = argparse.ArgumentParser()

# Single positional argument, nargs makes it optional
ap.add_argument("thingy", nargs='?', default="blah")

# Do parsing
a = ap.parse_args()

# Use argument
print a.thingy

Or, if you are stuck with Python 2.6 or earlier, and don't wish to add a requirement on the backported argparse module, you can do similar things manually like so:

import optparse

opter = optparse.OptionParser()
# opter.add_option("-v", "--verbose") etc
opts, args = opter.parse_args()

if len(args) == 0:
    var = "somevalue"
elif len(args) == 1:
    var = args[0]
else:
    opter.error("Only one argument expected, got %d" % len(args))

print var

why not directly `opter.add_option('--var', default="somevalue")`? See my answer. — wolfrevo, Feb 11 '14 at 08:31

score 3 · Answer 3 · answered Feb 11 '14 at 07:52

3

Good question.

I think the best solution would be to do

try:
    var = sys.argv[1]
except IndexError:
    var = "somevalue"

answered Feb 11 '14 at 07:52

Tim Pietzcker

328,213
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503
561

wolfrevo · Answer 4 · 2014-02-11T08:20:50.533

Try the following with a command-line-processing template:

def process_command_line(argv):
    ...
    # add your option here
    parser.add_option('--var',
        default="somevalue",
        help="your help text")

def main(argv=None):
    settings, args = process_command_line(argv)
    ...
    print settings, args # <- print your settings and args

Running ./your_script.py with the template below and your modifications above prints {'var': 'somevalue'} []

For an example of a command-line-processing template see an example in Code Like a Pythonista: Idiomatic Python (http://python.net/~goodger/projects/pycon/2007/idiomatic/handout.html#command-line-processing):

#!/usr/bin/env python

"""
Module docstring.
"""

import sys
import optparse

def process_command_line(argv):
    """
    Return a 2-tuple: (settings object, args list).
    `argv` is a list of arguments, or `None` for ``sys.argv[1:]``.
    """
    if argv is None:
        argv = sys.argv[1:]

    # initialize the parser object:
    parser = optparse.OptionParser(
        formatter=optparse.TitledHelpFormatter(width=78),
        add_help_option=None)

    # define options here:
    parser.add_option(      # customized description; put --help last
        '-h', '--help', action='help',
        help='Show this help message and exit.')

    settings, args = parser.parse_args(argv)

    # check number of arguments, verify values, etc.:
    if args:
        parser.error('program takes no command-line arguments; '
                     '"%s" ignored.' % (args,))

    # further process settings & args if necessary

    return settings, args

def main(argv=None):
    settings, args = process_command_line(argv)
    # application code here, like:
    # run(settings, args)
    return 0        # success

if __name__ == '__main__':
    status = main()
    sys.exit(status)

Python command line argument assign to a variable

4 Answers4