How do I specify a range of unicode characters

Question

How do I specify a range of unicode characters from ' ' (space) to \u00D7FF?

I have a regular expression like r'[\u0020-\u00D7FF]' and it won't compile saying that it's a bad range. I am new to Unicode regular expressions so I haven't had this problem before.

Is there a way to make this compile or a regular expression that I'm forgetting or haven't learned yet?

Answer 1

If you're using Python 2.x, you should make sure you're specifying a unicode string (with u'', or the "unicode" built-in):

>>> r = re.compile(u'[\u0020-\uD7FF]')
>>> r.search(u'foo \uD7F0 bar')
<_sre.SRE_Match object at 0xb7084950>
r.search(u' ')
<_sre.SRE_Match object at 0xb7084b48>

Using raw strings (as you are, with r'') gives you the (ascii) string composed by "backstroke" + the letter "u" plus the number 0 plus...

Answer 2

The syntax of your unicode range will not do what you expect.

1. The raw r'' string prevents \u escapes from being parsed, and the regex engine will not do this. The only range in this set is [0-\]:

   >>> re.compile(r'[\u0020-\u00d7ff]', re.DEBUG)
   in
     literal 117
     literal 48
     literal 48
     literal 50
     range (48, 117)
     literal 48
     literal 48
     literal 100
     literal 55
     literal 102
     literal 102
   

2. Making it a Unicode literal causes \u parsing while leaving other backslashes alone (although that’s not a concern here), but the leading zeroes are messing it up. The syntax is \uxxxx or \Uxxxxxxxx, so it’s parsed as "\u00d7, f, f".

   >>> re.compile(ur'[\u0020-\u00d7ff]', re.DEBUG)
   in
     range (32, 215)
     literal 102
     literal 102
   

3. Removing the leading zeroes or switching to \U0000d7ff will fix it:

   >>> re.compile(ur'[\u0020-\ud7ff]', re.DEBUG)
   in
     range (32, 55295)