Specifying default filenames with argparse, but not opening them on --help?

Question

Let\'s say I have a script that does some work on a file. It takes this file\'s name on the command line, but if it\'s not provided, it defaults to a known filename (conte

Answer 1

You could subclass argparse.FileType:

import argparse
import warnings

class ForgivingFileType(argparse.FileType):
    def __call__(self, string):
        try:
            super(ForgivingFileType,self).__call__(string)
        except IOError as err:
            warnings.warn(err)

parser = argparse.ArgumentParser(description='my illustrative example')
parser.add_argument('--content', metavar='file', 
                     default='content.txt', type=ForgivingFileType('r'),
                     help='file to process (defaults to content.txt)')
args = parser.parse_args()

This works without having to touch private methods like ArgumentParser._parse_known_args.