在python中过滤非法xml unicode字符的快速方法？

该XML规范列出一串Unicode字符要么是非法或"望而却步".给定一个字符串,如何从中删除所有非法字符？

我提出了以下正则表达式,但它有点满口.

illegal_xml_re = re.compile(u'[\x00-\x08\x0b-\x1f\x7f-\x84\x86-\x9f\ud800-\udfff\ufdd0-\ufddf\ufffe-\uffff]')
clean = illegal_xml_re.sub('', dirty)

(Python 2.5不知道0xFFFF以上的Unicode字符,所以不需要过滤那些.)

最近我们(Trac XmlRpcPlugin维护者)已经被告知上面的正则表达式在Python窄版本上剥离了代理对(参见:评论:13:ticket:11050).另一种方法是使用以下正则表达式(参见:changeset:13729).

_illegal_unichrs = [(0x00, 0x08), (0x0B, 0x0C), (0x0E, 0x1F), 
                        (0x7F, 0x84), (0x86, 0x9F), 
                        (0xFDD0, 0xFDDF), (0xFFFE, 0xFFFF)] 
if sys.maxunicode >= 0x10000:  # not narrow build 
        _illegal_unichrs.extend([(0x1FFFE, 0x1FFFF), (0x2FFFE, 0x2FFFF), 
                                 (0x3FFFE, 0x3FFFF), (0x4FFFE, 0x4FFFF), 
                                 (0x5FFFE, 0x5FFFF), (0x6FFFE, 0x6FFFF), 
                                 (0x7FFFE, 0x7FFFF), (0x8FFFE, 0x8FFFF), 
                                 (0x9FFFE, 0x9FFFF), (0xAFFFE, 0xAFFFF), 
                                 (0xBFFFE, 0xBFFFF), (0xCFFFE, 0xCFFFF), 
                                 (0xDFFFE, 0xDFFFF), (0xEFFFE, 0xEFFFF), 
                                 (0xFFFFE, 0xFFFFF), (0x10FFFE, 0x10FFFF)]) 

_illegal_ranges = ["%s-%s" % (unichr(low), unichr(high)) 
                   for (low, high) in _illegal_unichrs] 
_illegal_xml_chars_RE = re.compile(u'[%s]' % u''.join(_illegal_ranges)) 

ps请参阅这篇关于代理人的帖子,解释他们的用途.

更新以便不匹配(替换)0x0D这是一个有效的XML字符.