Python 实用宝典

Question 1

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.

Question 2

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)

Question 3

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.

Question 4

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

Python 实用宝典

Python argparse：默认值或指定值

问题：Python argparse：默认值或指定值

回答 0

回答 1

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