Python argparse: default value or specified value
The difference between:
parser.add_argument("--debug", help="Debug", nargs='?', type=int, const=1, default=7)
and
parser.add_argument("--debug", help="Debug", nargs='?', type=int, const=1)
is thus:
myscript.py
=> debug is 7 (from default) in the first case and "None" in the second
myscript.py --debug
=> debug is 1 in each case
myscript.py --debug 2
=> debug is 2 in each case
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('--example', nargs='?', const=1, type=int)
args = parser.parse_args()
print(args)
% test.py
Namespace(example=None)
% test.py --example
Namespace(example=1)
% test.py --example 2
Namespace(example=2)
nargs='?'
means 0-or-1 argumentsconst=1
sets the default when there are 0 argumentstype=int
converts the argument to int
If you want test.py
to set example
to 1 even if no --example
is specified, then include default=1
. That is, with
parser.add_argument('--example', nargs='?', const=1, type=int, default=1)
then
% test.py
Namespace(example=1)