How does one make argparse print usage when no options are provided on the command line?

Thu Dec 6 00:06:29 EST 2012

On 12/5/2012 7:48 PM, rh wrote:
> On Wed, 5 Dec 2012 18:42:37 +0100
> Bruno Dupuis <python.ml.bruno.dupuis at lisael.org> wrote:
>>> On Wed, Dec 05, 2012 at 08:48:30AM -0800, rh wrote:
>>> I have argparse working with one exception. I wanted the program to
>>> print out usage when no command line options are given. But I only
>>> came across other examples where people didn't use argparse but
>>> instead printed out a separate usage statement. So they used
>>> argparse for everything but the case where no command line args are
>>> given.
>>>>>>> this is quite raw, but i'd add
>>>> import sys
>> if len(sys.argv) == 1:
>> sys.argv.append('-h')
>> This works too. I guess I like the print_usage() method better.
>> Being new to python I have noticed that I had copied a bit of code that did
>> if len(sys.argv[1:]) == 0:

This needlessly creates and tosses a new object.
>> You did this:
> if len(sys.argv) == 1:

This does not.
> The other reply did this:
> if len(sys.argv) <= 1:

This allows for the possibility that len(sys.argv) == 0. However, that 
can (according to the doc) only happen when starting the interpreter 
interactively without a script. Since that does not apply to code within 
a .py file, I prefer == 1.
"argv[0] is the script name (it is operating system dependent whether 
this is a full pathname or not). If the command was executed using the 
-c command line option to the interpreter, argv[0] is set to the string 
'-c'. If no script name was passed to the Python interpreter, argv[0] is 
the empty string."
-- 
Terry Jan Reedy