CODEWITHSUNDEEP
pythonsys
JohnDoe122
JohnDoe122

Reputation: 688

Passing values from command line python

I have a python function that requires several parameters. Something like:

def func(par1=1, par2=2, par3=3):
...

I want to make this callable from command line such as: function.py par1=1 par2=2 par3=3.

I was thinking of doing it something like this:

import sys

if sys.argv[0][0:4]=="par1":
   par1=int(sys.argv[0][5:])
else:
   par1=1

but it doesn't look very nice, as the user might pass just some parametrs, or they might pass it as par1 =1 or par1= 1 or par1 = 1 so I would have to hard code with if-else all these possibilities. Is there a nicer, more professional solution? Thank you!

Upvotes: 0

Views: 66

Answers (2)

FiddleStix
FiddleStix

Reputation: 3731

As mentioned in Amadan's comment, you probably want function.py --par1=1 --par2=2 --par3=3 and not function.py par1=1 par2=2 par3=3

For the former, you can use

function.py

import argparse

parser = argparse.ArgumentParser()
parser.add_argument("--par1", default=11)
parser.add_argument("--par2", default=22)
parser.add_argument("--par3", default=33)
parser.add_argument("--par4", default=44)
opts = parser.parse_args()

print(opts.par1, opts.par2, opts.par3, opts.par4)

$ python function.py  # default values
11 22 33 44

$ python function.py --par1 99  # spaces are OK
99 22 33 44

$ python function.py --par1=99  # = are OK
99 22 33 44

Upvotes: 0

ted
ted

Reputation: 14744

Use argparse from the standard library

import argparse

parser = argparse.ArgumentParser()
parser.add_argument("--par1", type=int, default=1, help="This is the description")
opts = parser.parse_args()

print(opts.part1)

checkout the docs for more details: https://docs.python.org/3/library/argparse.html

Upvotes: 3

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