Display help message with Python argparse when script is called without any arguments

The cleanest solution will be to manually pass default argument if none were given on the command line:

parser.parse_args(args=None if sys.argv[1:] else ['--help'])

Complete example:

import argparse, sys

parser = argparse.ArgumentParser()
parser.add_argument('--host', default='localhost', help='Host to connect to')
# parse arguments
args = parser.parse_args(args=None if sys.argv[1:] else ['--help'])

# use your args
print("connecting to {}".format(args.host))

This will print complete help (not short usage) if called w/o arguments.

There are a pair of one-liners with sys.argv[1:] (a very common Python's idiom to refer the command line arguments, being sys.argv[0] the script's name) that can do the job.

The first one is self-explanatory, clean and pythonic:

args = parser.parse_args(None if sys.argv[1:] else ['-h'])

The second one is a little hackier. Combining the previously evaluated fact that an empty list is False with the True == 1 and False == 0 equivalences you get this:

args = parser.parse_args([None, ['-h']][not sys.argv[1:]])

Maybe too many brackets, but pretty clear if a previous argument selection was made.

_, *av = sys.argv
args = parser.parse_args([None, ['-h']][not av])

Most of the answers here required another module, such as sys, to be imported or were using optional arguments. I wanted to discover an answer that used only argparse, worked with required arguments, and if possible worked without catching exceptions. I ended up with the following:

import argparse

if __name__ == '__main__':

    arg_parser = argparse.ArgumentParser(add_help=False)
    arg_parser.add_argument('input_file', type=str, help='The path to the input file.')
    arg_parser.add_argument('output_file', type=str, help='The path to the output file.')
    arg_parser.add_argument('-h','--help', action='store_true', help='show this help message and exit')
    arg_parser.usage = arg_parser.format_help()
    args = arg_parser.parse_args()

The main idea was to use the format_help function in order to provide the help string to the usage statement. Setting add_help to False in the call to ArgumentParser() prevents the help statement from printing twice in certain circumstances. However, I had to create an argument for the optional help argument that mimicked the typical help message once it was set to False in order to display the optional help argument in the help message. The action is set to store_true in the help argument to prevent the help message from filling in a value like HELP for the parameter when it prints the help message.

So for a really simple answer. Most of the time with argparse you are checking to see if parameters are set anyway, to call a function that does something.

If no parameters, just else out at the end and print the help. Simple and works.

import argparse
import sys
parser = argparse.ArgumentParser()

group = parser.add_mutually_exclusive_group()
group.add_argument("--holidays", action='store_true')
group.add_argument("--people", action='store_true')

args=parser.parse_args()
if args.holidays:
    get_holidays()
elif args.people:
    get_people()
else:
    parser.print_help(sys.stderr)

I like to keep things as simple as possible, this works great:

#!/usr/bin/env python3
Description = """Tool description"""
Epilog  = """toolname.py -a aflag -b bflag  with these combined it does blah"""
arg_parser = argparse.ArgumentParser(
    formatter_class=argparse.RawDescriptionHelpFormatter,
    description=Description, 
    epilog=Epilog,
)
    try:
        if len(sys.argv) == 1:
            arg_parser.print_help()
    except Exception as e:
        print(e)

This is how I start all my tools as its always good to include some examples

When call add_subparsers method save the first positional argument to dest= and check value after argparse has been initialized, like this:

subparsers = parser.add_subparsers(dest='command')

And just check this this variable:

if not args.command:
    parser.print_help()
    parser.exit(1)  # If exit() - exit code will be zero (no error)

Full example:

#!/usr/bin/env python

""" doc """

import argparse
import sys

parser = argparse.ArgumentParser(description=__doc__)
subparsers = parser.add_subparsers(dest='command',
                                   help='List of commands')

list_parser = subparsers.add_parser('list',
                                    help='List contents')
list_parser.add_argument('dir', action='store',
                         help='Directory to list')

create_parser = subparsers.add_parser('create',
                                      help='Create a directory')
create_parser.add_argument('dirname', action='store',
                           help='New directory to create')
create_parser.add_argument('--read-only', default=False, action='store_true',
                           help='Set permissions to prevent writing to the directory')

args = parser.parse_args()

if not args.command:
    parser.print_help()
    parser.exit(1)

print(vars(args))  # For debug

This approach is a lot more elegant than most others. Instead of overriding error(), you can control the behaviour a lot more precisely by wrapping the parse_args() method:

import sys
import argparse


HelpFlags = ('help', '--help', '-h', '/h', '?', '/?', )


class ArgParser (argparse.ArgumentParser):
    
    def __init__(self, *args, **kws):
        super().__init__(*args, **kws)
    
    def parse_args(self, args=None, namespace=None):
        
        if args is None:
            args = sys.argv[1:]
        
        if len(args) < 1 or (args[0].lower() in HelpFlags):
            self.print_help(sys.stderr)
            sys.exit()
        
        return super().parse_args(args, namespace)

If your command is something where a user needs to choose some action, then use a mutually exclusive group with required=True.

This is kind of an extension to the answer given by pd321.

import argparse

parser = argparse.ArgumentParser()
group = parser.add_mutually_exclusive_group(required=True)
group.add_argument("--batch", action='store', type=int,  metavar='pay_id')
group.add_argument("--list", action='store_true')
group.add_argument("--all", action='store_true', help='check all payments')

args=parser.parse_args()

if args.batch:
    print('batch {}'.format(args.batch))

if args.list:
    print('list')

if args.all:
    print('all')

Output:

$ python3 a_test.py
usage: a_test.py [-h] (--batch pay_id | --list | --all)
a_test.py: error: one of the arguments --batch --list --all is required

This only give the basic help. And some of the other answers will give you the full help. But at least your users know they can do -h

This isn't good (also, because intercepts all errors), but:

def _error(parser):
    def wrapper(interceptor):
        parser.print_help()

        sys.exit(-1)

    return wrapper

def _args_get(args=sys.argv[1:]):
    parser = argparser.ArgumentParser()

    parser.error = _error(parser)

    parser.add_argument(...)
    ...

Here is definition of the error function of the ArgumentParser class.

As you see, the following signature takes two arguments. However, functions outside the class know nothing about first argument self, because, roughly speaking, this argument is for the class.

def _error(self, message):
    self.print_help()

    sys.exit(-1)

def _args_get(args=sys.argv[1:]):
    parser = argparser.ArgumentParser()

    parser.error = _error
    ...

will output:

"AttributeError: 'str' object has no attribute 'print_help'"

You can pass parser (self) in _error function, by calling it:

def _error(self, message):
    self.print_help()

    sys.exit(-1)

def _args_get(args=sys.argv[1:]):
    parser = argparser.ArgumentParser()

    parser.error = _error(parser)
    ...

But if you don't want exit the program right now, return it:

def _error(parser):
    def wrapper():
        parser.print_help()

        sys.exit(-1)

    return wrapper

Nonetheless, parser doesn't know that it has been modified. Thus, when an error occurs, it will print the cause of it (by the way, it's a localized translation). So intercept it:

def _error(parser):
    def wrapper(interceptor):
        parser.print_help()

        sys.exit(-1)

    return wrapper

Now, when an error occurs, parser will print the cause of it, and you'll intercept it, look at it, and... throw out.