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I am trying to convert the output of pyBarcode to a PIL Image file without first saving an image. First off, pyBarcode generates an image file like such: >>> import barcode >>> from barcode.writer import ImageWriter >>> ean = barcode.get_barcode('ean', '123456789102', writer=ImageWriter()) >>> filename = ean.save('ean13') >>> filename u'ean13.png' As you can see above, I don't want the image to be actually saved on my filesystem because I want the output to be processed into a PIL Image. So I did some modifications: i = StringIO() ean = barcode.get_barcode('ean', '123456789102', writer=ImageWriter()) ean.write(i) Now I have a StringIO file object and I want to PIL to "read" it and convert it to a PIL Image file. I wanted to use Image.new or Image.frombuffer but both these functions required me to enter a size...can't the size be determined from the barcode StringIO data? Image.open states this in their documentation: You can use either a string (representing the filename) or a file object. In the latter case, the file object must implement read, seek and tell methods, and be opened in binary mode Isn't a StringIO instance a file object as well? How do I open it as a binary file? Image.open(i, 'rb') >>> Image.open(i, 'rb') Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/home/mark/.virtualenvs/barcode/local/lib/python2.7/site-packages/PIL/Image.py", line 1947, in open

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