Python PIL, Image. Error after image.load()

Question

I'm trying to load my .jpg file and it raises error, but if I try it again, it's ok! Why??

My code and error:

>>> import Image
>>> im1 = Image.open('/tmp/test.jpg')
>>> im1.load()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib64/python2.7/site-packages/PIL/ImageFile.py", line 201, in load
raise IOError("image file is truncated (%d bytes not processed)" % len(b))
IOError: image file is truncated (0 bytes not processed)
>>> im1.load()
<PixelAccess object at 0x7feffc2a1170>
>>>

Thank you!

The message suggests the image you're tying to load is (probably) missing some trailing bytes. Why does it work intermitently, I don't know, as you don't say if the image is static, etc. — Ricardo Cárdenes
– Ricardo Cárdenes, Commented Feb 9, 2012 at 13:41

Community · Accepted Answer · 2017-05-23 11:33:14Z

21

I had this same issue and came up with a solution which I discuss here: https://stackoverflow.com/a/23575424/3622198.

Somewhere before your code block, simply add the following:

from PIL import ImageFile
ImageFile.LOAD_TRUNCATED_IMAGES = True

... and you should be good to go!

EDIT: It looks like this helps for the version of PIL bundled with Pillow ("pip install pillow"), but may not work for default installations of PIL

edited May 23, 2017 at 11:33

CommunityBot

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answered May 9, 2014 at 23:33

abhillman

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Comments

Mark Ransom · Accepted Answer · 2012-02-09 16:29:21Z

8

PIL uses lazy loading, which means the image isn't actually read from the file until you try to perform an action on it. The first call to load is that first action, so that's when the problem with the file format is detected. The second call to load doesn't read the file again, it just returns information that was cached when the file was loaded.

answered Feb 9, 2012 at 16:29

Mark Ransom

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3 Comments

Ricardo Cárdenes Over a year ago

Heh... Only now I noticed that the second call to load mentioned by OP was in the example...

Hare Over a year ago

Hm, sorry, I haven't told that the similar error happens when the first call is image.show. A second call image.show is ок.

Mark Ransom Over a year ago

@Hare, the same reasoning applies. The open request doesn't read the entire file so it doesn't get the error. The first operation whether it's load or show causes the entire file to be read, and the file defect is detected.

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