On 16/12/2011 17:57, Ian Kelly wrote:
On Fri, Dec 16, 2011 at 10:36 AM, MRAB<pyt...@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. :-)
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