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py3: make tcr input/output work

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Fix tcr compression by unifying bytearrays on python2/python3 and acting
appropriately.
This commit is contained in:
Eli Schwartz 2019-09-12 19:51:20 -04:00
parent 8311b26205
commit 2876724325
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GPG Key ID: CEB167EFB5722BD6
1 changed files with 10 additions and 7 deletions
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17
src/calibre/ebooks/compression/tcr.py
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@ -6,7 +6,7 @@ __copyright__ = '2009, John Schember <john@nachtimwald.com>'
__docformat__ = 'restructuredtext en' __docformat__ = 'restructuredtext en'
import re import re
from polyglot.builtins import range, int_to_byte from polyglot.builtins import int_to_byte, is_py3, range
class TCRCompressor(object): class TCRCompressor(object):
@ -42,8 +42,10 @@ class TCRCompressor(object):
possible_codes.append(single_code.pop()) possible_codes.append(single_code.pop())
for code in possible_codes: for code in possible_codes:
if not is_py3:
code = bytearray(code)
self.coded_txt = self.coded_txt.replace(code, code[0:1]) self.coded_txt = self.coded_txt.replace(code, code[0:1])
self.codes[ord(code[0:1])] = b'%s%s' % (self.codes[ord(code[0:1])], self.codes[ord(code[1:2])]) self.codes[code[0]] = b'%s%s' % (self.codes[code[0]], self.codes[code[1]])
def _free_unused_codes(self): def _free_unused_codes(self):
''' '''
@ -79,8 +81,9 @@ class TCRCompressor(object):
self.codes = list(set(re.findall(b'(?ms).', txt))) self.codes = list(set(re.findall(b'(?ms).', txt)))
# Replace the text with their corresponding code # Replace the text with their corresponding code
for c in txt: # FIXME: python3 is native bytearray, but all we want are bytes
self.coded_txt += int_to_byte(self.codes.index(c)) for c in bytearray(txt):
self.coded_txt += int_to_byte(self.codes.index(int_to_byte(c)))
# Zero the unused codes and record which are unused. # Zero the unused codes and record which are unused.
for i in range(len(self.codes), 256): for i in range(len(self.codes), 256):
@ -95,9 +98,9 @@ class TCRCompressor(object):
unused_code = self.unused_codes.pop() unused_code = self.unused_codes.pop()
# Take the last possible codes and split it into individual # Take the last possible codes and split it into individual
# codes. The last possible code is the most often occurring. # codes. The last possible code is the most often occurring.
code1, code2 = possible_codes.pop() code = possible_codes.pop()
self.codes[unused_code] = b'%s%s' % (self.codes[ord(code1)], self.codes[ord(code2)]) self.codes[unused_code] = b'%s%s' % (self.codes[ord(code[0:1])], self.codes[ord(code[1:2])])
self.coded_txt = self.coded_txt.replace(b'%s%s' % (code1, code2), int_to_byte(unused_code)) self.coded_txt = self.coded_txt.replace(code, int_to_byte(unused_code))
self._combine_codes() self._combine_codes()
self._free_unused_codes() self._free_unused_codes()
possible_codes = self._new_codes() possible_codes = self._new_codes()