Python argparse command line flags without arguments

Question

How do I add an optional flag to my command line args?

eg. so I can write

python myprog.py 

or

python myprog.py -w

I tried

parser.add_argument('-w')

But I just get an error message saying

Usage [-w W]
error: argument -w: expected one argument

which I take it means that it wants an argument value for the -w option. What's the way of just accepting a flag?

I'm finding http://docs.python.org/library/argparse.html rather opaque on this question.

Answer 1

As you have it, the argument w is expecting a value after -w on the command line. If you are just looking to flip a switch by setting a variable True or False, have a look at http://docs.python.org/dev/library/argparse.html#action (specifically store_true and store_false)

import argparse

parser = argparse.ArgumentParser()
parser.add_argument('-w', action='store_true')

Edit: As Sven points out, a default value in this case is superfluous.

Answer 2

Adding a quick snippet to have it ready to execute:

Source: myparser.py

import argparse
parser = argparse.ArgumentParser(description="Flip a switch by setting a flag")
parser.add_argument('-w', action='store_true')

args = parser.parse_args()
print args.w

Usage:

python myparser.py -w
>> True

Answer 3

Here's a quick way to do it, won't require anything besides sys.. though functionality is limited:

flag = "--flag" in sys.argv[1:]

[1:] is in case if the full file name is --flag