lets say you have a model called Book and on it a field called 'cover_pic', in that case, you can do the following to compress the image:

from PIL import Image
b = Book.objects.get(title='Into the wild')
image = Image.open(b.cover_pic.path)
image.save(b.image.path,quality=20,optimize=True)

hope it helps to anyone stumbling upon it.

Unfortunately this is a property of JPEG compression and there is no strict inverse function (you can calculate it mathematically if you want large scale efficiency).

However, you can inefficiently brute force the calculation and get it close with the following. If you don't want the same size, just change the quality to a scaling factor and it should work similarly.

import os
from PIL import Image

def compress_under_size(size, file_path):
    '''file_path is a string to the file to be custom compressed
    and the size is the maximum size in bytes it can be which this 
    function searches until it achieves an approximate supremum'''

    quality = 90 #not the best value as this usually increases size

    current_size = os.stat(file_path).st_size

    while current_size > size or quality == 0:
        if quality == 0:
            os.remove(file_path)
            print("Error: File cannot be compressed below this size")
            break

        compress_pic(file_path, quality)
        current_size = os.stat(file_path).st_size
        quality -= 5


def compress_pic(file_path, qual):
    '''File path is a string to the file to be compressed and
    quality is the quality to be compressed down to'''
    picture = Image.open(file_path)
    dim = picture.size

    picture.save(file_path,"JPEG", optimize=True, quality=qual) 

    processed_size = os.stat(file_path).st_size

    return processed_size

for file in os.listdir("pics"):
    try:
        pic = f"./pics/{file}"
        compress_under_size(10000,pic)
    except Exception:
        pass