CODEWITHSUNDEEP
python
wadda_wadda
wadda_wadda

Reputation: 1004

Get files from Directory Argument, Sorting by Size

I'm trying to write a program that takes a command line argument, scans through the directory tree provided by the argument and creating a list of every file in the directory, and then sorting by length of files.

I'm not much of a script-guy - but this is what I've got and it's not working:

import sys
import os
from os.path import getsize

file_list = []

#Get dirpath
dirpath = os.path.abspath(sys.argv[0])
if os.path.isdir(dirpath):
    #Get all entries in the directory
    for root, dirs, files in os.walk(dirpath):
        for name in files:
            file_list.append(name)
        file_list = sorted(file_list, key=getsize)
        for item in file_list:
            sys.stdout.write(str(file) + '\n')

else:
    print "not found"

Can anyone point me in the right direction?

Upvotes: 22

Views: 33902

Answers (6)

gndps
gndps

Reputation: 851

Plain old os.listdir but with extra params

import os
import operator

def oslistdir(directory='.', sort=True, size='smallest_first'):
    all_files = [file for file in os.listdir(directory) if os.path.isfile(os.path.join(directory, file))]
    files_and_sizes = ((os.path.join(directory, file), os.path.getsize(os.path.join(directory, file))) for file in all_files)
    if sort:
        sorted_files_with_size = sorted(files_and_sizes, key=operator.itemgetter(1), reverse=(size == 'largest_first'))
        sorted_files = [os.path.basename(file[0]) for file in sorted_files_with_size]
    else:
        sorted_files = [os.path.basename(file) for file in all_files]
    return sorted_files

Usage:

Smallest first:

[f for f in oslistdir(my_folder_path, sort=True, size='smallest_first')]

Largest first:

[f for f in oslistdir(my_folder_path, sort=True, size='largest_first')]

Upvotes: 0

Hoshang Zainadin
Hoshang Zainadin

Reputation: 31

I think this is what you are looking for:

import numpy as np
import os,glob

for file in sorted(glob.glob("*.file extension"),key=lambda file:os.stat(file).st_size,reverse=True):
    print(f'{file} is {np.around(os.stat(file).st_size/(1024),decimals=1)} KB')

Upvotes: 2

Raphvanns
Raphvanns

Reputation: 1894

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.

Upvotes: 2

koffein
koffein

Reputation: 1882

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) )

Upvotes: 23

Robert Jørgensgaard Engdahl
Robert Jørgensgaard Engdahl

Reputation: 3374

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))

Upvotes: 2

J. Owens
J. Owens

Reputation: 852

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

Upvotes: 17

Related Questions