How to use argparse arguments as function names

Question

I want to implement the example from the Argparse intro:

import argparse

parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('integers', metavar='N', type=int, nargs='+',
                   help='an integer for the accumulator')
parser.add_argument('--sum', dest='accumulate', action='store_const',
                   const=sum, default=max,
                   help='sum the integers (default: find the max)')

args = parser.parse_args()
print(args.accumulate(args.integers))

But in my case, I'd like to have a wider range of possible function names to choose from.

def a(): ...
def b(): ...
def c(): ...

parser.add_argument('-f', '--func', 
                     choices=[a, b, c], 
                     required=True,
                     help="""Choose one of the specified function to be run.""")
parser.func()

Using it like that doesn't work as intended, I get

$ python program.py -f=a
program.py: error: argument -f/--func: invalid choice: 'a' (choose from 
<function a at 0x7fa15f32f5f0>, 
<function b at 0x7fa15f32faa0>, 
<function c at 0x7ff1967099b0>)

I know I could solve it with basic string arguments and flow control, but it would be a lot less cluttered and easier to maintain if I could use parser arguments directly as function names.

Answer 1

You can use basic string arguments and minimize the clutter by using a dict to get the actual function from its name.

I'm on Python 2.6.6, so I can't use argparse, but this code should give you the general idea:

#!/usr/bin/env python

def a(): return 'function a'
def b(): return 'function b'
def c(): return 'function c'

func_list = [a, b, c]
func_names = [f.func_name for f in func_list]
funcs_dict = dict(zip(func_names, func_list))

f = funcs_dict['b']
print f()

output

function b

So you can pass func_names to argparse and compactly retrieve the desired function using funcs_dict.

---

In more recent Python versions, you should use f.__name__ instead of f.func_name. (That will also work in Python 2.6, but I was unfamiliar with the newer syntax when I wrote this answer).