Can argparse in python 2.7 be told to require a minimum of TWO arguments?

Question

My application is a specialized file comparison utility and obviously it does not make sense to compare only one file, so nargs='+' is not quite appropriate.

nargs=N only excepts a maximum of N arguments, but I need to accept an infinite number of arguments as long as there are at least two of them.

Answer 1

Short answer is you can't do that because nargs doesn't support something like '2+'.

Long answer is you can workaround that using something like this:

parser = argparse.ArgumentParser(usage='%(prog)s [-h] file file [file ...]')
parser.add_argument('file1', nargs=1, metavar='file')
parser.add_argument('file2', nargs='+', metavar='file', help=argparse.SUPPRESS)
namespace = parser.parse_args()
namespace.file = namespace.file1 + namespace.file2

The tricks that you need are:

• Use usage to provide you own usage string to the parser
• Use metavar to display an argument with a different name in the help string
• Use SUPPRESS to avoid displaying help for one of the variables
• Merge two different variables just adding a new attribute to the Namespace object that the parser returns

The example above produces the following help string:

usage: test.py [-h] file file [file ...]

positional arguments:
  file

optional arguments:
  -h, --help  show this help message and exit

and will still fail when less than two arguments are passed:

$ python test.py arg
usage: test.py [-h] file file [file ...]
test.py: error: too few arguments

Answer 2

Couldn't you do something like this:

import argparse

parser = argparse.ArgumentParser(description = "Compare files")
parser.add_argument('first', help="the first file")
parser.add_argument('other', nargs='+', help="the other files")

args = parser.parse_args()
print args

When I run this with -h I get:

usage: script.py [-h] first other [other ...]

Compare files

positional arguments:
  first       the first file
  other       the other files

optional arguments:
  -h, --help  show this help message and exit

When I run it with only one argument, it won't work:

usage: script.py [-h] first other [other ...]
script.py: error: too few arguments

But two or more arguments is fine. With three arguments it prints:

Namespace(first='one', other=['two', 'three'])