Python Argparser显示instancemethod的解析器

Question

Still plugging away at Python so I will apologize if this comment seems a little too juvenile but I am still having issues figuring this out:

I have a subparser 'foo' and a parser 'test', I will include an abbreviated code snippet below:

my_parser = test_subparsers.add_parser('foo')

my_parser.set_defaults( func =myfunction, which ='Foo')

When I print out the output from:

parser.parse_known_args()

I get something that looks like this:

(Namespace(config = 'config_file_path', func = 'function myfunction at 0x123456>', which = 'Foo') [] )

The type for parser.parse_known_args() is determined to be an 'instancemethod'.

When I try to convert the string to a dictionary object (using 'vars'), I get an empty dictionary, which I wasn't really surprised by, thanks to you informative people on stack overflow!

I am trying to retrieve the "which" item in particular (Foo), does anyone know how to do this??

Answer 1

parse_known_args returns a tuple , the namespace and a list of 'extras' (here empty)

In [10]: parser.parse_known_args(['foo'])
Out[10]: (Namespace(func=<function myfunction at 0x9da864c>, which='Foo'), [])

So add the [0] to access the namespace

In [11]: parser.parse_known_args(['foo'])[0]
Out[11]: Namespace(func=<function myfunction at 0x9da864c>, which='Foo')

The usual way of accessing attributes of the namespace

In [12]: parser.parse_known_args(['foo'])[0].func
Out[12]: <function __main__.myfunction>

In [13]: parser.parse_known_args(['foo'])[0].which
Out[13]: 'Foo'

As a dictionary:

In [14]: vars(parser.parse_known_args(['foo'])[0])
Out[14]: {'func': <function __main__.myfunction>, 'which': 'Foo'}