Python Command Args

Question

I have been googling almost an hour and am just stuck.

for a script, stupidadder.py, that adds 2 to the command arg.

e.g. python stupidadder.py 4

prints 6

python stupidadder.py 12

prints 14

I have googled so far:

import argparse
parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('x', metavar='x', type=int, nargs='+',
                    help='input number')

...

args = parser.parse_args()
print args
x = args['x']  # fails here, not sure what to put
print x + 2

I can't find a straightforward answer to this anywhere. the documentation is so confusing. :( Can someone help? Please and thank you. :)

Answer 1

I'm not entirely sure what your goal is. But if that's literally all you have to do, you don't have to get very complicated:

import sys
print int(sys.argv[1]) + 2

Here is the same but with some nicer error checking:

import sys

if len(sys.argv) < 2:
    print "Usage: %s <integer>" % sys.argv[0]
    sys.exit(1)

try:
    x = int(sys.argv[1])
except ValueError:
    print "Usage: %s <integer>" % sys.argv[0]
    sys.exit(1)

print x + 2

Sample usage:

C:\Users\user>python blah.py
Usage: blah.py <integer>

C:\Users\user>python blah.py ffx
Usage: blah.py <integer>

C:\Users\user>python blah.py 17
19

Answer 2

Assuming that you are learning how to use the argparse module, you are very close. The parameter is an attribute of the returned args object and is referenced as x = args.x.

import argparse
parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('x', metavar='x', type=int, nargs='+',
                    help='input number')

...

args = parser.parse_args()
print args
#x = args['x']  # fails here, not sure what to put
x = args.x
print x + 2

Answer 3

A sample run in Ipython with your code, showing that args is a simple object, not a dictionary. In the argparse code the namespace is accessed with getattr and setattr

In [4]: args=parser.parse_args(['12','4','5'])
In [5]: args
Out[5]: Namespace(x=[12, 4, 5])
In [6]: args['x']
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-6-3867439e1f91> in <module>()
----> 1 args['x']
TypeError: 'Namespace' object is not subscriptable
In [7]: args.x
Out[7]: [12, 4, 5]
In [8]: getattr(args,'x')
Out[8]: [12, 4, 5]
In [9]: sum(getattr(args,'x'))
Out[9]: 21

vars() can be used to turn the namespace into a dictionary.

In [12]: vars(args)['x']
Out[12]: [12, 4, 5]

Review the Namespace section of the argparse documentation.

Answer 4

You should simply do something like this:

x = args.x

Answer 5

As you are manipulating directly args Namespace object as if it were a dictionnary, it raises a

TypeError: 'Namespace' object is not subscriptable

My quick, personnal and ugly workaround is to access to the internal dict using internal __dict__ type

user_args = args.__dict__

This is quite usefull if you need to iterate over the arguments and filter it