Porting to Python3: PyPDF2 mergePage() gives TypeError - python-3.4

Question

I'm using Python 3.4.2 and PyPDF2 1.24 (also using reportlab 3.1.44 in case that helps) on windows 7.
I recently upgraded from Python 2.7 to 3.4, and am in the process of porting my code. This code is used to create a blank pdf page with links embedded in it (using reportlab) and merge it (using PyPDF2) with an existing pdf page. I had an issue with reportlab in that saving the canvas used StringIO which needed to be changed to BytesIO, but after doing that I ran into this error:
Traceback (most recent call last):
File "C:\cms_software\pdf_replica\builder.py", line 401, in merge_pdf_files
input_page.mergePage(link_page)
File "C:\Python34\lib\site-packages\PyPDF2\pdf.py", line 2013, in mergePage
self.mergePage(page2)
File "C:\Python34\lib\site-packages\PyPDF2\pdf.py", line 2059, in mergePage
page2Content = PageObject._pushPopGS(page2Content, self.pdf)
File "C:\Python34\lib\site-packages\PyPDF2\pdf.py", line 1973, in _pushPopGS
stream = ContentStream(contents, pdf)
File "C:\Python34\lib\site-packages\PyPDF2\pdf.py", line 2446, in __init
stream = BytesIO(b_(stream.getData()))
File "C:\Python34\lib\site-packages\PyPDF2\generic.py", line 826, in getData
decoded._data = filters.decodeStreamData(self)
File "C:\Python34\lib\site-packages\PyPDF2\filters.py", line 326, in decodeStreamData
data = ASCII85Decode.decode(data)
File "C:\Python34\lib\site-packages\PyPDF2\filters.py", line 264, in decode
data = [y for y in data if not (y in ' \n\r\t')]
File "C:\Python34\lib\site-packages\PyPDF2\filters.py", line 264, in
data = [y for y in data if not (y in ' \n\r\t')]
TypeError: 'in <string>' requires string as left operand, not int
Here is the line and the line above where the traceback mentions:
link_page = self.make_pdf_link_page(pdf, size, margin, scale_factor, debug_article_links)
if link_page != None:
input_page.mergePage(link_page)
Here are the relevant parts of that make_pdf_link_page function:
packet = io.BytesIO()
can = canvas.Canvas(packet, pagesize=(size['width'], size['height']))
....# left out code here is just reportlab specifics for size and url stuff
can.linkURL(url, r1, thickness=1, color=colors.green)
can.rect(x1, y1, width, height, stroke=1, fill=0)
# create a new PDF with Reportlab that has the url link embedded
can.save()
packet.seek(0)
try:
new_pdf = PdfFileReader(packet)
except Exception as e:
logger.exception('e')
return None
return new_pdf.getPage(0)
I'm assuming it's a problem with using BytesIO, but I can't create the page with reportlab with StringIO. This is a critical feature that used to work perfectly with Python 2.7, so I'd appreciate any kind of feedback on this. Thanks!
UPDATE:
I've also tried changing from using BytesIO to just writing to a temp file, then merging. Unfortunately I got the same error.
Here is tempfile version:
import tempfile
temp_dir = tempfile.gettempdir()
temp_path = os.path.join(temp_dir, "tmp.pdf")
can = canvas.Canvas(temp_path, pagesize=(size['width'], size['height']))
....
can.showPage()
can.save()
try:
new_pdf = PdfFileReader(temp_path)
except Exception as e:
logger.exception('e')
return None
return new_pdf.getPage(0)
UPDATE:
I found an interesting bit of information on this. It seems if I comment out the can.rect and can.linkURL calls it will merge. So drawing anything on a page, then trying to merge it with my existing pdf is causing the error.

Answer 1

After digging in to PyPDF2 library code, I was able to find my own answer. For python 3 users, old libraries can be tricky. Even if they say they support python 3, they don't necessarily test everything. In this case, the problem was with the class ASCII85Decode in filters.py in PyPDF2. For python 3, this class needs to return bytes. I borrowed the code for this same type of function from pdfminer3k, which is a port for python 3 of pdfminer. If you exchange the ASCII85Decode() class for this code, it will work:
import struct
class ASCII85Decode(object):
def decode(data, decodeParms=None):
if isinstance(data, str):
data = data.encode('ascii')
n = b = 0
out = bytearray()
for c in data:
if ord('!') <= c and c <= ord('u'):
n += 1
b = b*85+(c-33)
if n == 5:
out += struct.pack(b'>L',b)
n = b = 0
elif c == ord('z'):
assert n == 0
out += b'\0\0\0\0'
elif c == ord('~'):
if n:
for _ in range(5-n):
b = b*85+84
out += struct.pack(b'>L',b)[:n-1]
break
return bytes(out)