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

[Ian Kelly]

On Fri, Dec 16, 2011 at 10:57 AM, Ian Kelly <ian.g.kelly at gmail.com> 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)

Doh, that's still not quite right.

pattern = r'\b0{1,3}([1-9a-f][0-9a-f]*|0)\b'
re.sub(pattern, r'\1', string, flags=re.IGNORECASE)