你如何让 argparse 选择一个默认的子解析器？

Question

I have the following code in script.py :

import argparse

parser = argparse.ArgumentParser()
sp = parser.add_subparsers(dest='command')
sp.default = 'a'

a_parser = sp.add_parser('a')
b_parser = sp.add_parser('b')

a_parser.add_argument('--thing', default='thing')
b_parser.add_argument('--nothing', default='nothing')

args = parser.parse_args()
print(args)

I can call this three different ways:

$ python3 script.py
Namespace(command='a')
$ python3 script.py a
Namespace(command='a', thing='thing')
$ python3 script.py b
Namespace(command='b', nothing='nothing')

There's only one problem with this: what I want is that if I provide zero arguments on the command line is for a_parser to be the one that ends out parsing and doing stuff. Clearly it's not, the sp.default is just setting command='a' , not what I expect, which is to say, "Oh yeah, the user didn't provide any arguments on the command line, but I know that this should be processed by a_parser . Here's Namespace(command='a', thing='thing') !"

Is there a way that I can do this with argparse? I've looked for a few different options, but none of them really seem to provide what I'm after. I guess I could do some jiggery with making 3 distinct ArgumentParsers and then passing on the arguments to each of them, though that sounds a bit gross.

Any better options?

Answer 1

First a historical note - subparsers were not optional, and they still aren't in Python2. The fact that they are optional in Py3 is something of a bug that was introduced several years ago. There was a change in the test for required arguments, and subparsers (a kind of positional) fell through the cracks. If done right you should have had to explicitly set subparsers as not-required.

Subparsers don't behave like other non-required arguments, ones with nargs='?' or flagged without the required parameter.

In any case, your sp.default defines the value that will be put in the command dest, but it does not trigger the use of a_parser . That command='a' is never 'evaluated'.

The use of parse_known_args might allow you to devaluate the remaining strings with a_parser .

Without any arguments, we can do:

In [159]: args, extras = parser.parse_known_args([])
In [160]: args
Out[160]: Namespace(command='a')
In [161]: extras
Out[161]: []

Then conditionally run a_parser (if command is 'a' but no 'thing')

In [163]: a_parser.parse_args(extras,namespace=args)
Out[163]: Namespace(command='a', thing='thing')

But if I try to include a --thing value:

In [164]: args, extras = parser.parse_known_args('--thing ouch'.split())
usage: ipython3 [-h] {a,b} ...
ipython3: error: argument command: invalid choice: 'ouch' (choose from 'a', 'b')

It tries to parse the 'ouch' as subparser name. The main parser doesn't known anything about the --thing argument.

As I explained in the other argparse question today, the toplevel parser parses the inputs until it finds something that fits the 'subparsers' command (or in this example raises an error). Then it passes the parsing to the subparser. It does not resume parsing after.

Add top level argparse arguments after subparser args

How to Set a Default Subparser using Argparse Module with Python 2.7

My answer to this Py2 request might work for you. I first run a parse_known_args with a parser that doesn't have subparsers, and conditionally run a second parser that handles the subparsers.

In [165]: firstp = argparse.ArgumentParser()
In [166]: args, extras = firstp.parse_known_args('--thing ouch'.split())
In [167]: args
Out[167]: Namespace()

If extras doesn't have 'a' or 'b' call a_parser (alternatively just look at sys.argv[1:] directly):

In [168]: extras
Out[168]: ['--thing', 'ouch']
In [169]: a_parser.parse_args(extras)
Out[169]: Namespace(thing='ouch')

Or modify extras to include the missing subparser command:

In [170]: extras = ['a']+extras
In [171]: parser.parse_args(extras)
Out[171]: Namespace(command='a', thing='ouch')

In any case, optional subparsers is not well developed in argparse . It's the side effect of a change made a while back, rather than a well thought out feature.