re.sub(): replace longest match instead of leftmost match?

[MRAB]

On 16/12/2011 17:57, Ian Kelly wrote:
> On Fri, Dec 16, 2011 at 10:36 AM, MRAB<python at mrabarnett.plus.com>  wrote:
>>  On 16/12/2011 16:49, John Gordon wrote:
>>>
>>>  According to the documentation on re.sub(), it replaces the leftmost
>>>  matching pattern.
>>>
>>>  However, I want to replace the *longest* matching pattern, which is
>>>  not necessarily the leftmost match.  Any suggestions?
>>>
>>>  I'm working with IPv6 CIDR strings, and I want to replace the longest
>>>  match of "(0000:|0000$)+" with ":".  But when I use re.sub() it replaces
>>>  the leftmost match, even if there is a longer match later in the string.
>>>
>>>  I'm also looking for a regexp that will remove leading zeroes in each
>>>  four-digit group, but will leave a single zero if the group was all
>>>  zeroes.
>>>
>>  How about this:
>>
>>  result = re.sub(r"\b0+(\d)\b", r"\1", string)
>
> Close.
>
> pattern = r'\b0+([1-9a-f]+|0)\b'
> re.sub(pattern, r'\1', string, flags=re.IGNORECASE)
>
Ah, OK.

The OP said "digit" instead of "hex digit". That's my excuse. :-)