Python argparse default string value

Question

using argparse:

parser.add_argument("-o", "--output", help="Log to file")

I want to achieve the following behavior:

The user doesn't specify the -o flag - no logging should be done.
User specifies the -o with nothing - I should log to a default location, defined within my program.
User specifies -o and a string(path) - I should log there.

Does anyone know the best way to use add_argument to achieve that? I saw a similar example with int values, but in my case, it doesn't get my default value.

Do you understand why the example gets a `int` value? The role of the `type` parameter? — hpaulj, Jul 09 '18 at 15:35

score 19 · Accepted Answer · answered Jul 09 '18 at 09:33

You can use nargs='?' for this:

parser.add_argument('-o', '--output', 
                    nargs='?', default=None, const='my_default_location')

If not present, it will produce the default value, if present but without a value it'll use const, otherwise it'll use the supplied value.

Also read through the other examples in the docs, there's a sample for an optional output file which could be useful.

Python argparse default string value

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