Start a command line tool in Python

How can I create a command line tool in Python that works like ls of Linux or dir of Windows?

How would I start a script like the one that will later be compiled as an executable?

By comment what are you looking for are functions to simulate the behavior of ls or dir correct? so I recommend you look at the os.listdir command to list the files in a directory, the os.path to check for items from listdir oq is a file ( os.path.isfile ) or what is a directory ( os.path.isdir ) and os.stat to return the attributes of a file.

>>> import os
>>> files = os.listdir()
>>> files
['foo.txt']
>>> os.path.isfile(files[0])
True
>>> os.stat(files[0])
os.stat_result(st_mode=33188, st_ino=30049365, st_dev=16777218, st_nlink=1, st_uid=501, st_gid=20, st_size=4, st_atime=1470085341, st_mtime=1470085341, st_ctime=1470085341)

On the part of creating a command-line tool, you should search on argparse , one example would be:

import argparse
import os


parser = argparse.ArgumentParser(description='ls clone')
subparsers = parser.add_subparsers()

ls_parser = subparsers.add_parser(
    'ls', help='lista os arquivos e diretórios do diretório atual'
)
ls_parser.set_defaults(command='ls')


if __name__ == '__main__':
    args = parser.parse_args()
    if args.command == 'ls':
        print(os.listdir())

To use this script would be something like

$ python3 foo.py ls
['foo.txt']

To run an external command, use the subprocess module:

import subprocess

subprocess.call(["ls", "-l"])

Or using the os.system() method:

import os

os.system("ls -l")

Editing : Use sys.platform: to check the operating system and execute the commands according to the system:

import sys, subprocess

def main():
    if sys.platform == "linux":
        subprocess.call(["ls", "-l"])
    elif sys.platform == "darwin":
        subprocess.call("ls")
    elif sys.platform == "win32":
        subprocess.call("dir")

if __name__ == "__main__":
    main()