Re: python regex character group matches

[Fredrik Lundh]

christopher taylor wrote:

my issue, is that the pattern i used was returning:
>
[ '\\uAD0X', '\\u1BF3', ... ]
>
when i expected:
>
[ '\\uAD0X\\u1BF3 ', ]
>
the code looks something like this:
>
pat = re.compile("(\\ \u[0-9A-F]{4})+", re.UNICODE|re.L OCALE)
#print pat.findall(txt _line)
results = pat.finditer(tx t_line)
>
i ran the pattern through a couple of my colleagues and they were all
in agreement that my pattern should have matched correctly.

First, [0-9A-F] cannot match an "X". Assuming that's a typo, your next
problem is a precedence issue: (X)+ means "one or more (X)", not "one or
more X inside parens". In other words, that pattern matches one or more
X's and captures the last one.

Assuming that you want to find runs of \uXXXX escapes, simply use
non-capturing parentheses:

pat = re.compile(u"(? :\\\u[0-9A-F]{4})")

and use group(0) instead of group(1) to get the match.

</F>

[Steven D'Aprano]

Re: python regex character group matches

On Wed, 17 Sep 2008 15:56:31 +0200, Fredrik Lundh wrote:

Assuming that you want to find runs of \uXXXX escapes, simply use
non-capturing parentheses:
>
pat = re.compile(u"(? :\\\u[0-9A-F]{4})")

Doesn't work for me:

>>pat = re.compile(u"(? :\\\u[0-9A-F]{4})")

UnicodeDecodeEr ror: 'unicodeescape' codec can't decode bytes in position
5-7: truncated \uXXXX escape


Assuming that the OP is searching byte strings, I came up with this:

>>pat = re.compile('(\\ \u[0-9A-F]{4})+')
>>pat.search('a bcd\\u1234\\uAA 99\\u0BC4efg'). group(0)

'\\u1234\\uAA99 \\u0BC4'



--
Steven

[Fredrik Lundh]

Re: python regex character group matches

Steven D'Aprano wrote:

>Assuming that you want to find runs of \uXXXX escapes, simply use
>non-capturing parentheses:
>>
> pat = re.compile(u"(? :\\\u[0-9A-F]{4})")

>
Doesn't work for me:
>

>>>pat = re.compile(u"(? :\\\u[0-9A-F]{4})")

it helps if you cut and paste the right line... here's a better version:

pat = re.compile(r"(? :\\u[0-9A-F]{4})+")

</F>