Python argparser repeat subparse

Question

I\'m using pythons(2.7.2) argparse (1.1) to parse command line and what I want is to create subparser and make it possible to enter subparser commands multiple times. Like t

Answer 1

To the main parser, the subparsers argument is a positional that takes choices. But it also allocates ALL of the remaining argument strings to the subparser.

I expect that your string is parsed as follows:

./script.py version 1 --file 1 2 3 version 3 --file 4 5 6

version is accepted as a subparser name. 1 is accepted as value to positional argument n. (of the subparser). --file is accepted as a optional argument (by the subparser). The values from the second invocation overwrite the values from the first. I'm guessing --file has nargs='*'. If so, the first one writes ['1','2','3','version','3'] to the namespace, while the second overwrites it with ['4','5','6']. If nargs=3, I would expect the subparser to choke on the second version, which it would see as an unknown positional.

So the basic point is - once the 'version' subparser gets the argument list, it does not let go until it has parsed everything it can. In this case it parses both --file occurrences. Anything it can't handle comes back to the main parser as 'UNKNOWNS', which normally raises an error.

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If you want values from repeated optionals, use an append action

parser.add_argument('--foo',action='append', nargs=3)

import argparse
parser = argparse.ArgumentParser()
sp = parser.add_subparsers(dest='version')
spp = sp.add_parser('version')
spp.add_argument('n',nargs='*',type=int)
spp.add_argument('--file',nargs=3,action='append')
str = 'version 1 --file 1 2 3 version 3 --file 4 5 6'
print(parser.parse_known_args(str.split()))

produces

(Namespace(file=[['1', '2', '3'], ['4', '5', '6']], n=[1], version='version'), ['version', '3'])

Still only one call to version subparser, but all the data is present.

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A different approach would be to nest subparsers

parser = argparse.ArgumentParser()
sp = parser.add_subparsers(dest='sub')
spp = sp.add_parser('version')
spp.add_argument('n',nargs=1,type=int)
spp.add_argument('--file',nargs=3)

sp = spp.add_subparsers(dest='sub1')
spp = sp.add_parser('version')
spp.add_argument('n1',nargs=1,type=int)
spp.add_argument('--file',dest='file1',nargs=3)

str = 'version 1 --file 1 2 3 version 3 --file 4 5 6'
print(parser.parse_args(str.split()))

Note that I have to change the 'dest' to avoid over writing values. This produces

Namespace(file=['1', '2', '3'], file1=['4', '5', '6'], n=[1], n1=[3], sub='version', sub1='version')