python argparse可选参数

Question

I am evaluating argparse for my python program and I am not sure if I can accomplish following:

Usage:
prog --file <filename> --function <function name> --mode <mode name>
prog --file <filename> --function <function name>  {in this case default mode is applied on specified function in a specified file}
prog --file <filename> --mode <mode name> {in this case specified mode is applied on a specified file}

In short I want to make entire --function and it's follower optional. Currently I have passed nargs = "?" in that but that confuses argparse and when I pass --mode it takes it as --function.

parser = argparse.ArgumentParser(description = "program")
parser.add_argument("file", nargs=2)
parser.add_argument("function", nargs='?')
parser.add_argument("mode", nargs='?', default='read')

Is there anyway I can accomplish this using argparse? Or I would have to write my own parser using sys.args ???

Thanks.

Answer 1

I found this while researching a different argparse question. Maybe my answer will help someone else. If I understand the question, you need to add "--" to the optional arguments. Declarations would look like this:

parser = argparse.ArgumentParser(description = "program")
parser.add_argument("file")
parser.add_argument("--function")
parser.add_argument("--mode", default='read')

When you do this, 'file' is required, and 'function' and 'mode' are optional.
Add the following lines to your code snippet, and you will get output shown.

args = parser.parse_args('my_file --function func_1'.split())
print (args)
Namespace(file='my_file', function='func_1', mode='read')

args = parser.parse_args('other_file --mode write'.split())
print (args)
Namespace(file='other_file', function=None, mode='write')

You can access individual values with args.file, args.function, and args.mode .
What you do with them is up to you. :-)