Argparse. Funny issue with variable named "args"

Question

I noticed one of my scripts was not running because an argparse parser was not able to parse_args().

if __name__ == "__main__":
    import argparse

    parser = argparse.ArgumentParser(description="Save a plot to file")
    parser.add_argument('input_directory', metavar='i', type=str, default='.',
                        help='The input directory')

    parser.add_argument('output_file', metavar='o', type=str,
                        help='The output filename')

    parser.add_argument('--fix', type=str, default=None,
                        help='If FIX, the txt containing fix classification results')

    args = parser.parse_args()

    import ipdb; ipdb.set_trace()

Strangely enough, I noticed that the problem has to do with the variable called args. I cannot retrieve any information from that variable, since it always returns empty. If I change the name of the args variable to anything else, then the script works fine.

I'm asking this question because I could not find any information telling me that args is a keyword, or anything else.

I invoked my script as:

python3 plotter.py --fix $(pwd)/fix.txt $(pwd) $(pwd)/plot.png

Here's a screenshot of the funny behaviour:

Note how any statement containing the word args does not return.

Any ideas of what may be happening here?

Does a plain `print(args)` work? Maybe `ipdb` is using `args` for its own purpose? — hpaulj, Jul 12 '17 at 15:42

score 2 · Accepted Answer · answered Nov 30 '21 at 11:31

2

You can refer to the args variable name by prepending it with an exclamation mark:

ipdb> !args

args is actually a command used by ipdb, see here:

Explanation:

a(rgs)
Print the argument list of the current function.

answered Nov 30 '21 at 11:31

ponadto

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Arpit Solanki · Answer 2 · 2017-07-12T13:41:25.817

1

Try using vars()

args = vars(parser.parse_args())

then retrieve using

some_val = args['some']

Also refer this stackoverflow post

edited Jul 12 '17 at 13:41

answered Jul 12 '17 at 13:35

Arpit Solanki

9,567
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score 1 · Answer 3 · answered Jul 13 '17 at 01:08

ipdb/__main__.py sets args in its main:

def main():
    ...
    opts, args = getopt.getopt(sys.argv[1:], 'hc:', ['--help', '--command='])

I don't know what's visible when run interactively, but I suspect its own use of args is interfering with what you see with

args = parser.parse_args(...)

When testing argparse, I routinely add a

print(args)

statement. Or I test it in an ipython session, where I can look at the result of parse_args without further work.

Argparse. Funny issue with variable named "args"

3 Answers3