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.
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.
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')