Python: Using run time inputs to change key values in a function

Question

import sys

def fun( a=7, b=1 ):
    print a, b

fun()
fun( a=5 )
fun( sys.argv[1:] )

The first fun() prints out '7,1', the second prints out '5,1', but the third prints out '['a=8', 'b=6'] 1]'. I would like to be able to call my program with

python my_program.py a=5 b=6

to change the value of the printed a and b. But this doesn't work since sys.argv[1] is a list of strings.

Is there any way to convert the the string list to a form the function can understand?

Answer 1

Use ** for kwarg unpacking:

d = {}
for a in sys.argv[1:]:
    k, v = a.split('=')
    d[k] = int(v)
func(**d)

Another way, using the csv module:

import csv
func(**{k: int(v) for k, v in csv.reader(sys.argv[1:], delimiter='=')})

As @MartijnPieters noted you may use the argparse module

parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('-a', type=int)
parser.add_argument('-b', type=int)

args = parser.parse_args()
fun(**dict(args._get_kwargs()))

but then you have to adjust the way you input your arguments

Answer 2

Use * to unpack a list into arguments:

fun(*sys.argv[1:])

Note that this will result in string arguments; sys.argv command line entries are always strings.

You may want to look into the argparse module to handle command line arguments for you, including automatic conversion:

import argparse

parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('-a', type=int, default=7)
parser.add_argument('-b', type=int, default=1)
args = parser.parse_args()

func(args.a, args.b)

Then use:

$ python my_program.py -a=5 -b=6

The added advantage here is that you get proper help feedback, and you follow standard command-line conventions:

$ python my_program.py --help
usage: [-h] [-a A] [-b B]

Process some integers.

optional arguments:
  -h, --help  show this help message and exit
  -a A
  -b B