How to load an image in python without using disk storage?

Question

I am writing a flask application that receives two image URLs as parameters. Traditionally, I would download this images on the server local disk and carry out my image processing operations. I am downloading them using following code.

urllib.request.urlretrieve(image1url, image_id + '.jpg')

After this I read the image using :

original_image = Image.open(image_id + '.jpg')

and carry out my Image Processing operations like crop and applying a few filters.

original_crop = original_image.crop((x, y, x + width / 3, y + height / 3))

Also, I use ImageFilter operations on this image. Now this code will be deployed on a server. If i continue this way I will keep downloading and saving images on the disk of the server. Of course, I understand that deleting the images after I am done with my Image Processing operations is one option. But if I get a few 100 calls per second, I might at some point of time use up a lot of space. The application is multi threaded using the call

app.run(threaded=true)

which works like a charm.

I want to know if there is a way to load an image without using disk storage of the server. Thus reducing the hard disk space requirements of my service.

are you opposed to just finding the image on the filesystem based on its file path and then deleting it when you're done with it? — JacobIRR
– JacobIRR, Commented May 26, 2017 at 4:26

Azat Ibrakov · Accepted Answer · 2017-05-26 06:54:45Z

2

if you don't want to store images in temporary files you can wrap URL content in stream and pass it to Image.open

import io
import urllib.request

from PIL import Image

# your avatar URL as example
url = ('https://www.gravatar.com/avatar/3341748e9b07c9854d50799e0e247fa3'
       '?s=328&d=identicon&response=PG&f=1')
content = urllib.request.urlopen(url).read()
original_image = Image.open(io.BytesIO(content))

edited May 26, 2017 at 6:54

answered May 26, 2017 at 6:06

Azat Ibrakov

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4 Comments

Harshil Over a year ago

This looks beautiful! A silly question but, can i then carry out operations on the 'original_image' as usual ?

Azat Ibrakov Over a year ago

ofc you can, since we have Image.Image object we can work with it as usual

Azat Ibrakov Over a year ago

@user2326079: but you definitely should try it yourself and let me know if something went wrong

Harry Over a year ago

Let's call it a wrap. Works like a dream. Thanks, Azat

Chris Johnson · Accepted Answer · 2017-05-26 04:28:08Z

0

You could move them to a known remote location and fetch them back as needed. Using Amazon S3 or a hosted FTP service like BrickFTP are both easy. S3 is especially cheap since you only pay for what you use -- no monthly fee. Brick is a great service if you want to make access to the images as simple as possible for other applications and users but there is a minimum monthly fee.

answered May 26, 2017 at 4:28

Chris Johnson

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2 Comments

Harry Over a year ago

Agreed. But I was wondering if there is a way in python itself to load an image at runtime without storing it at all. For example getting a byte array of the image per say ?

Chris Johnson Over a year ago

You can store any Python object on disk using the pickle module. That way you can store the result of your conversion and just reload that rather than the image.

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