How to make Shell script style arguments passing with argparse in Python

Question

I have a python parsing arguments like below:

Code:

if __name__ == "__main__":
    parser = argparse.ArgumentParser(description='Args test')
    parser.add_argument('-myarg1', '--myarg1', type=str, dest='myarg1', required=True, help='Help yourself')
    parser.add_argument('-myarg2', '--myarg2', type=str, dest='myarg2', default=' ', help='Help yourself')

    args = parser.parse_args()
    print(args.myarg1)
    print(args.myarg2)

Above works if I call the script like below:

python myargs.py -myarg1 something -myarg2 somethingelse

But it does not work if I call it like below:

python myargs.py -myarg1 something -myarg2

And throws the below error obviously because it expects caller to pass value for the second argument.

usage: myargs.py [-h] -myarg1 MYARG1 [-myarg2 MYARG2]
myargs.py: error: argument -myarg2/--myarg2: expected one argument

Quesion:
I understand the reason for python complaining about it above. But, I want the user of my python script to be able to call the second argument with just saying -myarg2 or --myarg2 without specifying the type. Just like shell script style. Is it possible to do it with argparse?

I am using python 2.7 and above.

I updated the value type. Its str. Both arguments are string type — AdeleGoldberg
– AdeleGoldberg, Commented Nov 17, 2020 at 8:15

S. Ferard · Accepted Answer · 2020-11-17 08:46:44Z

2

It is possible. You can use the action="store_true" attribute to turn an argument into a Boolean (flag).

parser.add_argument('-myarg2', '--myarg2', dest='myarg2', action="store_true", help='Help yourself')
args = parser.parse_args()
print(args.myarg2) # True if "python myargs.py -myarg2", False if "python myargs.py"

Edit

If you want the user to be able to pass an optional argument to the myarg2 flag, you need to use the nargs='?' attribute. You also need to define a default attribute which will be called if the flag isn't used, and a const attribute which will be called if the flag is used but without arguments.

parser.add_argument('-myarg2', '--myarg2', dest='myarg2', nargs='?', const="no value", default='no flag', help='Help yourself')

edited Nov 17, 2020 at 8:46

answered Nov 17, 2020 at 8:23

S. Ferard

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2 Comments

Tom Ron Over a year ago

But if she wants to pass a str argument this will result raise an error

S. Ferard Over a year ago

Indeed, I've modified my answer to take care of this case

Tom Ron · Accepted Answer · 2020-11-17 08:19:57Z

1

The problem is that you call the argument flag without a value. If you want the value to be an empty string do -

python myargs.py -myarg1 something -myarg2 ' '

answered Nov 17, 2020 at 8:19

Tom Ron

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1 Comment

AdeleGoldberg Over a year ago

I know that I can pass ' ' or ''. It works but I wanted to avoid it

Lior Cohen · Accepted Answer · 2020-11-17 08:24:42Z

1

if you want to use --myarg and interpret it as true (otherwise it is false) you need to use

parser.add_argument("-myarg2", "--myarg2", action="store_true")

answered Nov 17, 2020 at 8:24

Lior Cohen

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5 Comments

Tom Ron Over a year ago

But if she wants to pass a str argument this will result raise an error

AdeleGoldberg Over a year ago

@Lior Cohen good point. Strange that python does not give a way to pass string or not pass a string. But if this is the constraint in python, then I would have to live with it.

AdeleGoldberg Over a year ago

Ok. Just one question. action='store_false' makes it false?

AdeleGoldberg Over a year ago

How to check if user has passed an input or not?

S. Ferard Over a year ago

@AdeleGoldberg Indeed, action="store_false"will make it false, and action="store_const" will make it the value of the const attribute. You can learn more here

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