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

This answer comes from Steven Bethard on Google groups. I'm reposting it here to make it easier for people without a Google account to access.

You can override the default behavior of the error method:

import argparse
import sys

class MyParser(argparse.ArgumentParser):
    def error(self, message):
        sys.stderr.write('error: %s\n' % message)
        self.print_help()
        sys.exit(2)

parser = MyParser()
parser.add_argument('foo', nargs='+')
args = parser.parse_args()

---

Note that the above solution will print the help message whenever the error method is triggered. For example, test.py --blah will print the help message too if --blah isn't a valid option.

If you want to print the help message only if no arguments are supplied on the command line, then perhaps this is still the easiest way:

import argparse
import sys

parser=argparse.ArgumentParser()
parser.add_argument('foo', nargs='+')
if len(sys.argv)==1:
    parser.print_help(sys.stderr)
    sys.exit(1)
args=parser.parse_args()

---

Note that parser.print_help() prints to stdout by default. As init_js suggests, use parser.print_help(sys.stderr) to print to stderr.

Instead of writing a class, a try/except can be used instead

try:
    options = parser.parse_args()
except:
    parser.print_help()
    sys.exit(0)

The upside is that the workflow is clearer and you don't need a stub class. The downside is that the first 'usage' line is printed twice.

This will need at least one mandatory argument. With no mandatory arguments, providing zero args on the commandline is valid.

With argparse you could do:

parser.argparse.ArgumentParser()
#parser.add_args here

#sys.argv includes a list of elements starting with the program
if len(sys.argv) < 2:
    parser.print_usage()
    sys.exit(1)