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Home/ Questions/Q 7975447
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T08:38:25+00:00 2026-06-04T08:38:25+00:00

I have the following function which takes an image and then returns it in

  • 0

I have the following function which takes an image and then returns it in three sizes. Another function then uploads those images to Amazon S3. It seems to me like there is some redundancy in how the file is being saved –

def resize_image(image, size_as_tuple):
    """
    Example usage: resize_image(image, (100,200))
    """

    image_as_string=""
    for c in image.chunks(): 
        image_as_string += c

    imagefile = cStringIO.StringIO(image_as_string)
    image = Image.open(imagefile)

    if image.mode not in ("L", "RBG"):
        image = image.convert("RGB")

    filename = ''.join(random.choice(string.ascii_uppercase + string.digits) for x in range(14)) + ".jpg"
    height, width = size_as_tuple[0], size_as_tuple[1]
    image.thumbnail((height, width), Image.ANTIALIAS)

    imagefile = open(os.path.join('/tmp', filename), 'w')
    image.save(imagefile, 'JPEG')

    imagefile = open(os.path.join('/tmp', filename), 'r')
    content = File(imagefile)

    return (filename, content)

Is there a way to improve this?

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  1. Editorial Team
    Editorial Team
    2026-06-04T08:38:27+00:00Added an answer on June 4, 2026 at 8:38 am

    You could replace:

    height, width = size_as_tuple[0], size_as_tuple[1]
    image.thumbnail((height, width), Image.ANTIALIAS)
    

    with

    image.thumbnail(size_as_tuple, Image.ANTIALIAS)
    

    (especially since width and height are swapped; it should be width, height = size_as_tuple)

    And you don’t need the open(). image.save(os.path.join('/tmp', filename)) is enough.

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