How to convert sys.argv to argparse in python

Question

I have

file1 = sys.argv[1]
file2 = sys.argv[2] 
file2 = sys.argv[3]

How can I put this in argparse?

Answer 1

Ready-opened file objects:

import argparse, sys

parser = argparse.ArgumentParser(description='Process src to dst')
parser.add_argument('src', type=argparse.FileType('r'),
                    default=sys.stdin)
parser.add_argument('dst', type=argparse.FileType('w'),
                    default=sys.stdout)
options = parser.parse_args()

Then use options.src and options.dst as already-open file objects.

Prints the following when you use the --help command-line switch:

usage: somescript.py [-h] src dst

Process src to dst

positional arguments:
  src
  dst

optional arguments:
  -h, --help  show this help message and exit

Answer 2

If the FileType in Martijn's answer confuses you, a more basic setup is

import argparse
p=argparse.ArgumentParser()
p.add_argument('file1')
p.add_argument('file2')
p.add_argument('file3')

test with:

import sys
sys.argv.extend(['file1','file2','file3')
p.parse_args()
# Namespace(file1='file1', file2='file2', file3='file3')

Here you open the files yourself, whereas with FileType , arpgparse does that for you, and takes care of error messages if it has problems.