Issue 1883 - Work around another PIL bug by using RGBA insted LA

imageConversion.py

 # coding: utf-8
 
 # This Source Code Form is subject to the terms of the Mozilla Public
 # License, v. 2.0. If a copy of the MPL was not distributed with this
 # file, You can obtain one at http://mozilla.org/MPL/2.0/.
 
 import os
 import re
 from StringIO import StringIO
 
 try:
   from PIL import Image
   from PIL import ImageOps
 except ImportError:
   import Image
   import ImageOps
 
 def get_alpha(image):
   if image.mode in ('RGBA', 'LA'):
     return image.split()[image.getbands().index('A')]
 
   # In order to generate an alpha channel for images using a palette, we
-  # convert the image to grayscale+alpha. Initially, we created an alpha
-  # channel by replacing opaque pixels with a high mark and transparent
+  # convert the image to RGBA. It's important to use RGBA, not LA (grayscale+alpha),
+  # since PIL can't reliably convert P to LA. Also initially, we created an
+  # alpha channel by replacing opaque pixels with a high mark and transparent
   # pixels with a low mark. However, it turned out that you can't rely on the
   # value of Image.info['transparency'] since in some cases it might be an
   # unparsed string instead an int indicating the value of transparent pixels.
   if image.mode == 'P' and 'transparency' in image.info:
-    return image.convert('LA').split()[1]
+    return image.convert('RGBA').split()[4]
 
 def load_image(path):
   image = Image.open(path)
   # Make sure the image is loaded, some versions of PIL load images lazily.
   image.load()
   return image
 
 def ensure_same_mode(im1, im2):
   # if both images already have the same mode (and palette, in
   # case of mode P), don't convert anything. Images with mode P,
@@ skipping 75 matching lines @@
 
     for step in steps:
       filter, args = re.match(r'([^(]+)(?:\((.*)\))?', step).groups()
       args = re.split(r'\s*,\s*', args) if args else ()
       image = globals()['filter_' + filter](image, baseDir, *args)
 
     f = StringIO()
     f.name = filename
     image.save(f)
     files[filename] = f.getvalue()