#!/usr/bin/env python """ Python implementation of ASCII85/ASCIIHex decoder (Adobe version). This code is in the public domain. """ import re import struct # ascii85decode(data) def ascii85decode(data): """ In ASCII85 encoding, every four bytes are encoded with five ASCII letters, using 85 different types of characters (as 256**4 < 85**5). When the length of the original bytes is not a multiple of 4, a special rule is used for round up. The Adobe's ASCII85 implementation is slightly different from its original in handling the last characters. The sample string is taken from: http://en.wikipedia.org/w/index.php?title=Ascii85 >>> ascii85decode('9jqo^BlbD-BleB1DJ+*+F(f,q') 'Man is distinguished' >>> ascii85decode('E,9)oF*2M7/c~>') 'pleasure.' """ n = b = 0 out = '' for c in data: if '!' <= c and c <= 'u': n += 1 b = b*85+(ord(c)-33) if n == 5: out += struct.pack('>L', b) n = b = 0 elif c == 'z': assert n == 0 out += '\0\0\0\0' elif c == '~': if n: for _ in range(5-n): b = b*85+84 out += struct.pack('>L', b)[:n-1] break return out # asciihexdecode(data) hex_re = re.compile(r'([a-f\d]{2})', re.IGNORECASE) trail_re = re.compile(r'^(?:[a-f\d]{2}|\s)*([a-f\d])[\s>]*$', re.IGNORECASE) def asciihexdecode(data): """ ASCIIHexDecode filter: PDFReference v1.4 section 3.3.1 For each pair of ASCII hexadecimal digits (0-9 and A-F or a-f), the ASCIIHexDecode filter produces one byte of binary data. All white-space characters are ignored. A right angle bracket character (>) indicates EOD. Any other characters will cause an error. If the filter encounters the EOD marker after reading an odd number of hexadecimal digits, it will behave as if a 0 followed the last digit. >>> asciihexdecode('61 62 2e6364 65') 'ab.cde' >>> asciihexdecode('61 62 2e6364 657>') 'ab.cdep' >>> asciihexdecode('7>') 'p' """ decode = (lambda hx: chr(int(hx, 16))) out = map(decode, hex_re.findall(data)) m = trail_re.search(data) if m: out.append(decode("%c0" % m.group(1))) return ''.join(out) if __name__ == '__main__': import doctest doctest.testmod()