Sort files by size in a directory in Python

In this post, I will be explaining how to sort files by size in a directory in Python.

Basic Process/Algorithm for the Task: Sort files by size

• I’ll be using Python’s inbuilt ‘os’ module for the task

  import os

• you’ll need to enter the path to the folder where you want to perform the task.

  def path_allocation():
      dir_path = input('Enter the Path to a directory:\n')
      if os.path.isdir(dir_path):
          return dir_path
      else:
          dir_path = path_allocation()
          return dir_path

  And call this function from the main method to work with the directory path you have entered.

• In case you enter a wrong path, you’ll be asked to re-enter a correct path until you interrupt the execution or enter the right path.
• For better performance, I’ll first create a dictionary of {filenames: sizes}, then sort the dictionary by the value.

  # creating a dictionary with file sizes
      size_dic = {}
      for files in os.listdir(directory_location):
          if os.path.isfile(files): # restricting for the files only
              size_dic[files] = os.stat(files).st_size # this will give size in bytes

  sorting the diction by values, using key=lambda


  # sorting dictionary by size, --> value of the dictionary
      print(f'\tFile Name{6*"  "}\t\tFile Size\n{24*"--"}')
      for file,size in sorted(size_dic.items(), key = lambda kv:(kv[1], kv[0])):
          print(f'{file}\t\t----> {size/1000:.03f} Kb') # printing file size and names sorted by their size

   

Now If I sum up the whole process, the code will look like as below:

import os
def path_allocation():
    dir_path = input('Enter the Path to a directory:\n')
    if os.path.isdir(dir_path):
        return dir_path
    else:
        dir_path = path_allocation()
        return dir_path

if __name__ == '__main__':

    directory_location = path_allocation() # for getting the directory path from the user.
    os.chdir(directory_location) # changing the current directory to the given path

    # creating dictionary with file sizes
    size_dic = {}
    for files in os.listdir(directory_location):
        if os.path.isfile(files): # restricting for the files only
            size_dic[files] = os.stat(files).st_size # this will give size in bytes

    # sorting dictionary by size, --> value of the dictionary
    print(f'\tFile Name{6*"  "}\t\tFile Size\n{24*"--"}')
    for file,size in sorted(size_dic.items(), key = lambda kv:(kv[1], kv[0])):
        print(f'{file}\t\t----> {size/1000:.03f} Kb') # printing file size and names sorted by their size

**Note: os.path(‘file-path’).st_size return file size in bytes, thus to convert them into kilobytes, I have divided by 1000. And used :.03f to precise the file size up to 3 decimal places only.

The main advantage of this code is that this is a platform-independent code and is really fast to sort a large number of file by their sizes.

And the Output will look like this:

Also read:

• Python File Handling