Python argparse: default value or specified value

问题

I would like to have a optional argument that will default to a value if only the flag is present with no value specified, but store a user-specified value instead of the default if the user specifies a value. Is there already an action available for this?

An example:

python script.py --example
# args.example would equal a default value of 1
python script.py --example 2
# args.example would equal a default value of 2

I can create an action, but wanted to see if there was an existing way to do this.

回答1:

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 arguments
const=1 sets the default when there are 0 arguments
type=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)

回答2:

Actually, you only need to use the default argument to add_argument as in this test.py script:

import argparse

if __name__ == '__main__':

    parser = argparse.ArgumentParser()
    parser.add_argument('--example', default=1)
    args = parser.parse_args()
    print(args.example)

test.py --example
% 1
test.py --example 2
% 2

Details are here.

来源：https://stackoverflow.com/questions/15301147/python-argparse-default-value-or-specified-value

标签

python

default-value

argparse