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. :)

Search for `python command line arguments` and you'll definitely find useful links. For instance, the first link gives you the technique described by Claudiu. — keyser, Jul 29 '13 at 21:06

score 40 · Answer 1 · answered Jul 29 '13 at 21:21

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

score 14 · Answer 2 · answered Jul 30 '13 at 01:06

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.

Being honest, I'd -1 for the scary first part (imo, haha) and +1 for the `vars` trick that is very useful if one requires a dict. — jeromej, Dec 11 '14 at 12:50

score 2 · Answer 3 · answered Sep 27 '17 at 10:02

2

You should simply do something like this:

x = args.x

answered Sep 27 '17 at 10:02

Mateusz Poreba

79
1
3

score 2 · Answer 4 · answered Feb 02 '20 at 15:01

2

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

answered Feb 02 '20 at 15:01

c24b

5,278
6
27
35

1

`vars()` is better. https://stackoverflow.com/questions/16878315/what-is-the-right-way-to-treat-argparse-namespace-as-a-dictionary – cbmanica May 31 '21 at 18:45
I strongly disagree. – c24b Jun 09 '21 at 23:12
Why? (actual question, as I thought I was providing a useful answer) – cbmanica Jun 10 '21 at 16:38
Using global vars can have strong hidden side effects and lack of readabilty. Basically we don't exactly know what we manipulate. Using global variable is considered as a bad practise beside the fact that you can do it in a more readable and cleaner way... I don't pretend my answer is the better but vars() is hackish. – c24b Jun 10 '21 at 20:04

score 0 · Answer 5 · answered Apr 21 '23 at 16:53

It might also be that you have multiple word variable like (with a dash here): parser.add_argument('-d', '--extracted-dir', type=str,...

To access it you can use: args.extracted_dir

You probably tried doing args['extracted-dir'] and that's why you got the error.

score -2 · Accepted Answer · answered Jul 29 '13 at 21:05

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

Thanks! For whatever reason, the examples in the python documentation don't mention this at all. http://docs.python.org/2/library/argparse.html The text does in passing, but I'd expect something more readable / usable from this python page. — user1601118, Jul 29 '13 at 21:10

Python Command Args

6 Answers6

Linked

Related