How to read integer command line arguments in Python?

Question

I have been trying the following, as suggested by many post, but I am still getting

ValueError: invalid literal for int() with base 10: 'gap_interval.py'. What am I missing?

import sys 

if __name__ == '__main__':
    n = 4
    a = []
    for i in sys.argv:
        print int(i.strip())

gap_interval.py is not a string. It is also part of sys.argv which it's telling you. — metatoaster
– metatoaster, Commented Apr 9, 2014 at 23:08

sshashank124 · Accepted Answer · 2014-04-09 23:46:33Z

4

Python sys.argv always contains the filename as the first argument:

>>> python test.py 1 2 3
>>> sys.argv
['test.py', '1', '2', '3']

If you want to convert it to int, you can do:

nums = map(int, sys.argv[1:])

>>> nums
[1,2,3]

answered Apr 9, 2014 at 23:46

sshashank124

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Comments

Hai Vu · Accepted Answer · 2014-04-09 23:42:02Z

3

You need to skip sys.argv[0], which is the script's name:

for i in sys.argv[1:]:
    # ...

answered Apr 9, 2014 at 23:42

Hai Vu

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