python - How to create an argument that is optional? -

instead of user having use script.py --file c:/stuff/file.txt there way let user optionally use --file? instead, script.py c:/stuff/file.txt parser still know user referring --file argument (because it's implied).

try this

import argparse  class donotreplaceaction(argparse.action):     def __call__(self, parser, namespace, values, option_string=none):         if not getattr(namespace, self.dest):             setattr(namespace, self.dest, values)  parser = argparse.argumentparser(description="this example.") parser.add_argument('file', nargs='?', default='', help='specifies file.', action=donotreplaceaction) parser.add_argument('--file', help='specifies file.')  args = parser.parse_args() # check file argument if not args.file:     raise exception('missing "file" argument') 

look @ message. arguments optional

usage: test.py [-h] [--file file] [file]  example.  positional arguments:   file         specifies file.  optional arguments:   -h, --help   show message , exit   --file file  specifies file. 

one thing notice positional file override optional --file , set args.file default ''. overcome used custom action positional file. forbids overriding set properties.

the other thing notice rather raising exception specify default value.