CODEWITHSUNDEEP
pythonparsingargumentsargs
user1601118
user1601118

Reputation: 407

Python Command Args

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. :)

Upvotes: 24

Views: 23071

Answers (6)

funerr
funerr

Reputation: 8156

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.

Upvotes: 0

c24b
c24b

Reputation: 5552

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

Upvotes: 2

Mateusz Poreba
Mateusz Poreba

Reputation: 79

You should simply do something like this:

x = args.x

Upvotes: 2

hpaulj
hpaulj

Reputation: 231385

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.

Upvotes: 14

tdelaney
tdelaney

Reputation: 77347

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

Upvotes: 40

Claudiu
Claudiu

Reputation: 229361

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

Upvotes: -2

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