Python: how to add '-help' to argparse help command list?

Question

Is there a way to include '-help' command to argparse help list?

I wish to have something like this on output, if i am typing '-help'.

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

Thanks

Answer 1

As @Akaisteph7 suggested:

import argparse

parser = argparse.ArgumentParser()
parser.add_argument('-help', action="help", help="second help :)")
parser.add_argument('-f', '--foo')

parser.print_help()

0945:~/mypy$ python3 stack57058526.py 
usage: stack57058526.py [-h] [-help] [-f FOO]

optional arguments:
  -h, --help         show this help message and exit
  -help              second help :)
  -f FOO, --foo FOO

Changing to:

parser = argparse.ArgumentParser(add_help=False)
parser.add_argument('-h','--help','-help', action="help", help="replacement help")

0946:~/mypy$ python3 stack57058526.py 
usage: stack57058526.py [-h] [-f FOO]

optional arguments:
  -h, --help, -help  replacement help
  -f FOO, --foo FOO

Adding the '-help' flag to the default help requires modifying a couple of 'private' attributes:

parser = argparse.ArgumentParser()
parser._actions[0].option_strings += ['-help']                                                          
parser._option_string_actions['-help'] = parser._option_string_actions['-h']  

0947:~/mypy$ python3 stack57058526.py 
usage: stack57058526.py [-h] [-f FOO]

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

If you want to build this change into your local version of argparse, you could modify this block of code in the ArgumentParser.__init__ method:

    if self.add_help:
        self.add_argument(
            default_prefix+'h', default_prefix*2+'help',
            action='help', default=SUPPRESS,
            help=_('show this help message and exit'))

Whether you change a local copy of argparse.py, or subclass ArgumentParser is up to you.