Python – How to reduce the image file size using PIL

compressionimagepythonpython-imaging-library

I am using PIL to resize the images there by converting larger images to smaller ones. Are there any standard ways to reduce the file size of the image without losing the quality too much, lets say the original size of the image is 100KB, i want to get it down to like 5 or 10 KB especially for png and jpeg formats.

Best Answer

A built-in parameter for saving JPEGs and PNGs is optimize.

 >>> from PIL import Image
 # My image is a 200x374 jpeg that is 102kb large
 >>> foo = Image.open("path\\to\\image.jpg")
 >>> foo.size
  (200,374)
 # I downsize the image with an ANTIALIAS filter (gives the highest quality)
 >>> foo = foo.resize((160,300),Image.ANTIALIAS)
 >>> foo.save("path\\to\\save\\image_scaled.jpg",quality=95)
 # The saved downsized image size is 24.8kb
 >>> foo.save("path\\to\\save\\image_scaled_opt.jpg",optimize=True,quality=95)
 # The saved downsized image size is 22.9kb

The optimize flag will do an extra pass on the image to find a way to reduce its size as much as possible. 1.9kb might not seem like much, but over hundreds/thousands of pictures, it can add up.

Now to try and get it down to 5kb to 10 kb, you can change the quality value in the save options. Using a quality of 85 instead of 95 in this case would yield: Unoptimized: 15.1kb Optimized : 14.3kb Using a quality of 75 (default if argument is left out) would yield: Unoptimized: 11.8kb Optimized : 11.2kb

I prefer quality 85 with optimize because the quality isn't affected much, and the file size is much smaller.

Related Solutions

Python – How to check whether a file exists without exceptions

If the reason you're checking is so you can do something like if file_exists: open_it(), it's safer to use a try around the attempt to open it. Checking and then opening risks the file being deleted or moved or something between when you check and when you try to open it.

If you're not planning to open the file immediately, you can use os.path.isfile

Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.

import os.path
os.path.isfile(fname)

if you need to be sure it's a file.

Starting with Python 3.4, the pathlib module offers an object-oriented approach (backported to pathlib2 in Python 2.7):

from pathlib import Path

my_file = Path("/path/to/file")
if my_file.is_file():
    # file exists

To check a directory, do:

if my_file.is_dir():
    # directory exists

To check whether a Path object exists independently of whether is it a file or directory, use exists():

if my_file.exists():
    # path exists

You can also use resolve(strict=True) in a try block:

try:
    my_abs_path = my_file.resolve(strict=True)
except FileNotFoundError:
    # doesn't exist
else:
    # exists

Python – How to copy a file in Python

shutil has many methods you can use. One of which is:

from shutil import copyfile
copyfile(src, dst)

# 2nd option
copy(src, dst)  # dst can be a folder; use copy2() to preserve timestamp

Copy the contents of the file named src to a file named dst. Both src and dst need to be the entire filename of the files, including path.
The destination location must be writable; otherwise, an IOError exception will be raised.
If dst already exists, it will be replaced.
Special files such as character or block devices and pipes cannot be copied with this function.
With copy, src and dst are path names given as strs.

Another shutil method to look at is shutil.copy2(). It's similar but preserves more metadata (e.g. time stamps).

If you use os.path operations, use copy rather than copyfile. copyfile will only accept strings.

Best Answer

Related Solutions

Python – How to check whether a file exists without exceptions

Python – How to copy a file in Python

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