How to pass a variable into argparse in python

Question

I have a python script func.py that contains argparse and takes a file

parser = argparse.ArgumentParser(description='test')
parser.add_argument('-i', dest='ifile', metavar="FILE", help='input file')
args = parser.parse_args()

This ifile is basically just a name of that file in a directory

filename = 'PATH TO FILE' + args.ifile

If I have a variable that contains the name of the file how can I pass the variable to the argparse so that it would take the string value of the variable? Something like this :

new_file = 'text_file.txt'
func.py -i new_file
filename = 'PATH TO FILE' + 'text_file.txt'

What I am getting now is this:

new_file = "text_file.txt'
python func.py -I new_file
filename = 'PATH TO FILE' + 'new_file'

And then obviously an error message that it cannot find the file.

Can you post a more detailed example? I cant even figure out where are those lines coming from — JoshuaCS
– JoshuaCS, Commented Dec 14, 2018 at 0:37
For what reason do you want the command line argument to refer to a variable name? I suspect that you are going the wrong way to solve whatever problem you are trying to solve. — user10605163
– user10605163, Commented Dec 14, 2018 at 0:37
@user10605163 I need to run func.py inside jupyter notebook (! func.py -i new_file) Unfortunately there is no other way I can run this script — Sabrina
– Sabrina, Commented Dec 14, 2018 at 0:41
Please add Jupyter to the question. Normally this isn't a problem in python. — kokeen
– kokeen, Commented Dec 14, 2018 at 0:43

l'L'l · Accepted Answer · 2018-12-14 00:42:44Z

2

You should simply be able to do:

$ new_file="text_file.txt"
$ ./func.py -I $new_file

Your args variable in Python should then contain ifile='text_file.txt':

text_file.txt

answered Dec 14, 2018 at 0:42

l'L'l

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