How can I get Python Argparse to list choices only once?

Question

I have code that looks like:

list_of_choices = ["foo", "bar", "baz"]
parser = argparse.ArgumentParser(description='some description')
parser.add_argument("-n","--name","-o","--othername",dest=name,
    choices=list_of_choices

and what I get for output looks like:

-n {foo,bar,baz}, --name {foo,bar,baz}, -o {foo,bar,baz}, 
--othername {foo,bar,baz}

What I would like is:

-n, --name, -o, --othername {foo,bar,baz}

For context, there are historical reasons why we need two names for the same option and the actual list of choices is 22 elements long, so it looks much worse than the above.

This problem is subtly different than Python argparse: Lots of choices results in ugly help output in that I am not working with two separate options and it is okay to have it all on the line as above.

score 20 · Answer 1 · edited Jan 22 '20 at 19:54

20

I think you might want multiple add_arguments() and only set choices on the one where you want the choices.

list_of_choices = ["foo", "bar", "baz"]
parser = argparse.ArgumentParser(description='some description')
parser.add_argument("-n")
parser.add_argument("--name")
parser.add_argument("-o")
parser.add_argument("--othername", dest='name',
    choices=list_of_choices)

edited Jan 22 '20 at 19:54

Michael

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answered Nov 20 '12 at 18:24

Thomas Schultz

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I don't think that quite works, in that if I pass -f foo, I later get an error saying name is None. – coyot Nov 20 '12 at 18:50
Are you trying to catch all arguments or just accept the ones you define? In this case you would have to do another add_argument() for -f. Also i think that dest is supposed to be a string dest='name'. http://docs.python.org/dev/library/argparse.html#dest – Thomas Schultz Nov 20 '12 at 19:18

coyot · Answer 2 · 2012-11-20T19:23:22.707

4

Thanks, @thomas-schultz. I wasn't aware of the sequential aspect of add_argument and your comment put me on the right track, combined with a comment from that other thread.

Basically, what I do now is put all four in a mutex group, suppress the output of the first three and then include them in the description of the group.

The output looks like:

group1
   use one of -n, --name, -o, --othername
-n {foo,bar,baz}

which is MUCH cleaner than the original.

edited Nov 20 '12 at 19:23

answered Nov 20 '12 at 19:15

coyot

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Can you post the code you have so I get what you're doing? :) – Thomas Schultz Nov 20 '12 at 19:46

score 2 · Answer 3 · edited Jan 22 '20 at 20:01

Here's the code I settled on after a bit more tweaking:

parser = argparse.ArgumentParser(description='some description', 
    epilog="At least one of -n, -o, --name, or --othername is required"
           " and they all do the same thing.") 
parser.add_argument('-d', '--dummy', dest='dummy',
    default=None, help='some other flag')
stuff = parser.add_mutually_exclusive_group(required=True)
stuff.add_argument('-n', dest='name', 
    action='store', choices=all_grids, help=argparse.SUPPRESS)
stuff.add_argument('-o', dest='name', 
    action='store', choices=all_grids, help=argparse.SUPPRESS)
stuff.add_argument('--name', dest='name', 
    action='store', choices=all_grids, help=argparse.SUPPRESS)
stuff.add_argument('--othername', dest='name', 
    action='store', choices=all_grids, help='')
args = parser.parse_args()

output with -h is usage, then options list, then this:

--othername {foo,bar,baz}

At least one of -n, -o, --name, or --othername is required and they all do the same thing.

Why not edit this into your other answer? It's part of the same answer, after all. — Michael, Dec 21 '16 at 18:50
Also, `action='store'` [is the default](https://docs.python.org/dev/library/argparse.html#action) and can be omitted. — Michael, Jan 22 '20 at 20:04

How can I get Python Argparse to list choices only once?

3 Answers3