Get files from Directory Argument, Sorting by Size

You are extracting the command and not the first argument with argv[0]; use argv[1] for that:

dirpath = sys.argv[1]  # argv[0] contains the command itself.

For performance reasons I suggest you prefetch the file sizes instead of asking the OS about the size of the same file multiple times during the sorting (as suggested by Koffein, os.walk is the way to go):

files_list = []
for path, dirs, files in os.walk(dirpath)):
    files_list.extend([(os.path.join(path, file), getsize(os.path.join(path, file))) for file in files])

Assuming you don't need the unsorted list, we will use the in-place sort() method:

files_list.sort(key=operator.itemgetter(1))

How about using pandas?

import pandas as pd
import os

file_paths = [os.path.join(files_dir, file_name) for file_name in os.listdir(files_dir)]
file_sizes = [os.path.getsize(file_path) for file_path in file_paths]

df = pd.DataFrame({'file_path': file_paths, 'file_size': file_sizes}).sort_values('file_size', ascending = False)

You can then easily recuperate the list of values from the df.

Hopefully this function will help you out (I'm using Python 2.7):

import os    

def get_files_by_file_size(dirname, reverse=False):
    """ Return list of file paths in directory sorted by file size """

    # Get list of files
    filepaths = []
    for basename in os.listdir(dirname):
        filename = os.path.join(dirname, basename)
        if os.path.isfile(filename):
            filepaths.append(filename)

    # Re-populate list with filename, size tuples
    for i in xrange(len(filepaths)):
        filepaths[i] = (filepaths[i], os.path.getsize(filepaths[i]))

    # Sort list by file size
    # If reverse=True sort from largest to smallest
    # If reverse=False sort from smallest to largest
    filepaths.sort(key=lambda filename: filename[1], reverse=reverse)

    # Re-populate list with just filenames
    for i in xrange(len(filepaths)):
        filepaths[i] = filepaths[i][0]

    return filepaths

This is a approach using generators. Should be faster for large number of files…

This is the beginning of both examples:

import os, operator, sys
dirpath = os.path.abspath(sys.argv[0])
# make a generator for all file paths within dirpath
all_files = ( os.path.join(basedir, filename) for basedir, dirs, files in os.walk(dirpath) for filename in files   )

If you just want a list of the files without the size, you can use this:

sorted_files = sorted(all_files, key = os.path.getsize)

But if you want files and paths in a list, you can use this:

# make a generator for tuples of file path and size: ('/Path/to/the.file', 1024)
files_and_sizes = ( (path, os.path.getsize(path)) for path in all_files )
sorted_files_with_size = sorted( files_and_sizes, key = operator.itemgetter(1) )