Python argparse: default value or specified value

Question

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.

unutbu · Accepted Answer · 2013-03-08 18:52:43Z

458

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)

answered Mar 8, 2013 at 18:52

unutbu

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3 Comments

0andriy Over a year ago

How to do this with strings? I have a dilemma with differentiation of "" (empty string as default) and "" (empty string as entered by user). In the code for now I'm using the default and since I need to do some ops, I have something like this self.foo = (args.bar or some_else_source).upper(). It will break on None object AFAIUC.

dsz Over a year ago

BE CAREFUL; if you try to have positional args afterwards, the first position arg will become the value for --example (if it doesn't have an explicit one). Just spent 10 minutes debugging that!

Frederick Nord Over a year ago

is there a way to make it accept an optional int and still distinguish a positional string argument?

Murray · Accepted Answer · 2020-02-04 01:56:11Z

112

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

edited Feb 4, 2020 at 1:56

user3657941

answered Feb 4, 2020 at 1:50

Murray

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4 Comments

TurnipEntropy Over a year ago

Excellent way to describe this difference

urie Over a year ago

There is a contradiction between your answer and the previous answer... According to previous answer: myscript.py --debug should bring 7

Murray Over a year ago

You’re welcome to test the code for yourself. If I’ve made a mistake it’s helpful for us all. However, I believe I have correctly described the results of my testing. Test.py gives 7 because no —debug argument was provided so the default value is used. Test.py —debug gives 1, the value set by const.

AwfulPersimmon Over a year ago

There may be issues with running code, but this answer does correctly demonstrate how to assign default values with argparse, which is the crux of the question.

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