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Home/ Questions/Q 511759
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T07:15:00+00:00 2026-05-13T07:15:00+00:00

I’m building a online font previewer, with following architecture. I wrapped preview creation function

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I’m building a online font previewer, with following architecture. I wrapped preview creation function in a standalone .py file and making system calls to it in a Django view in order to run them in parallel and maximum performance on multi-core CPU system.

preview.py

....
def make_preview(text, fontfile, imagefile, fontsize=30):
    try:
        font = ImageFont.truetype(fontfile, fontsize)
        text_width, text_height = font.getsize(text)
        img = Image.new('RGBA', (text_width, text_height))
        draw = ImageDraw.Draw(img)
        draw.text((0, 0), text, font=font, fill=(0, 0, 0))
        return True
    except:
        return False
if __name__=='__main__':
    make_preview(sys.argv[1], sys.argv[2], sys.argv[3], sys.argv[4])

views.py

...
def ajax_preview(resquest):
    ttf_file_list = [construct a list of .ttf files]
    image_list = []
    for ttf_file in ttf_file_list:
        image_file = ttf_file.replace('.ttf', '.png')
        image_list.append(image_file)
        subprocess.Popen(['python', 'preview.py', text, ttf_file, image_file, 30)
    return HttpResponse(image_list)

Is it possible to make above code even faster? Maybe there is a better architecture to do this job or any optimization I can do for the code?

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  1. Editorial Team
    Editorial Team
    2026-05-13T07:15:00+00:00Added an answer on May 13, 2026 at 7:15 am

    You should look into the multiprocessing module. You could create a pool of workers equal to your number of CPU cores and then send jobs to your make_preview function.

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