On Fri, Mar 18, 2005 at 12:27:35PM -0500, Christopher Weimann wrote:
> So this [^\s]+ means match one or more of any char that
> isn't whitespace.
Could be just \S+
Greetings, J"o!
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On 03/17/2005-10:15AM, Mike Hall wrote:
>
> Very nice sir. I'm interested in what you're doing here with
> the caret metacharacter. For one thing, why enclose it and the
> whitespace flag within a character class?
A caret as the first charachter in a class is a negation.
So this [^\s]+ means
On Mar 18, 2005, at 1:02 PM, Christopher Weimann wrote:
On 03/18/2005-10:35AM, Mike Hall wrote:
A caret as the first charachter in a class is a negation.
So this [^\s]+ means match one or more of any char that
isn't whitespace.
Ok, so the context of metas change within a class. That makes sense,
bu
On 03/18/2005-10:35AM, Mike Hall wrote:
> >
> > A caret as the first charachter in a class is a negation.
> > So this [^\s]+ means match one or more of any char that
> > isn't whitespace.
> >
>
> Ok, so the context of metas change within a class. That makes sense,
> but I'm unclear on th
On Mar 18, 2005, at 9:27 AM, Christopher Weimann wrote:
On 03/17/2005-10:15AM, Mike Hall wrote:
Very nice sir. I'm interested in what you're doing here with
the caret metacharacter. For one thing, why enclose it and the
whitespace flag within a character class?
A caret as the first charachte
Mike Hall wrote:
On Mar 17, 2005, at 11:11 AM, Kent Johnson wrote:
The first one matches the space after 'in'. Without it the .+? will
match the single space, then \b matches the *start* of the next word.
I think I understand. Basically the first dot advances the pattern
forward in order to perf
On Mar 17, 2005, at 11:11 AM, Kent Johnson wrote:
The first one matches the space after 'in'. Without it the .+? will
match the single space, then \b matches the *start* of the next word.
I think I understand. Basically the first dot advances the pattern
forward in order to perform a non-greedy m
Mike Hall wrote:
On Mar 16, 2005, at 8:32 PM, Kent Johnson wrote:
"in (.*?)\b" will match against "in " because you use .* which will
match an empty string. Try "in (.+?)\b" (or "(?<=\bin)..+?\b" )to
require one character after the space.
Another working example, excellent. I'm not too clear on
I don't have that script on my system, but I may put pythoncard on here
and run it through that:
http://pythoncard.sourceforge.net/samples/redemo.html
Although regexPlor looks like it has the same functionality, so I may
just go with that. Thanks.
On Mar 17, 2005, at 1:31 AM, Michael Dunn wrote
On Mar 16, 2005, at 8:32 PM, Kent Johnson wrote:
"in (.*?)\b" will match against "in " because you use .* which will
match an empty string. Try "in (.+?)\b" (or "(?<=\bin)..+?\b" )to
require one character after the space.
Another working example, excellent. I'm not too clear on why the back
to
Very nice sir. I'm interested in what you're doing here with
the caret metacharacter. For one thing, why enclose it and the
whitespace flag within a character class? Does this not traditionally
mean you want to strip a metacharacter of it's special meaning?
On Mar 16, 2005, at 8:00 PM, Christo
As Kent said, redemo.py is a script that you run (e.g. from the
command line), rather than something to import into the python
interpretor. On my OSX machine it's located in the directory:
/Applications/MacPython-2.3/Extras/Tools/scripts
Cheers, Michael
___
Mike Hall wrote:
Liam, "re.compile("in (.*?)\b")" will not find any match in the example
string I provided. I have had little luck with these non-greedy matchers.
"in (.*?)\b" will match against "in " because you use .* which will match an empty string. Try "in
(.+?)\b" (or "(?<=\bin)..+?\b" )to
On 03/16/2005-12:12PM, Mike Hall wrote:
> I'm having trouble getting re to stop matching after it's consumed
> what I want it to. Using this string as an example, the goal is to
> match "CAPS":
>
> >>> s = "only the word in CAPS should be matched"
>
jet% python
Python 2.4 (#2, Jan 5 2005,
On Mar 16, 2005, at 5:32 PM, Sean Perry wrote:
I know this does not directly help, but I have never successfully used
\b in my regexs. I always end up writing something like foo\s+bar or
something more intense.
I've had luck with the boundary flag in relation to lookbehinds. For
example, if I wa
Liam, "re.compile("in (.*?)\b")" will not find any match in the example
string I provided. I have had little luck with these non-greedy
matchers.
I don't appear to have redemo.py on my system (on OSX), as an import
returns an error. I will look into finding this module, thanks for
pointing me
> >>> x=re.compile(r"(?<=\bin).+\b")
Try
>>> x = re.compile("in (.*?)\b")
.*? is a non-greedy matcher I believe.
Are you using python24/tools/scripts/redemo.py? Use that to test regexes.
Regards,
Liam Clarke
On Wed, 16 Mar 2005 12:12:32 -0800, Mike Hall
<[EMAIL PROTECTED]> wrote:
> I'm hav
I'm having trouble getting re to stop matching after it's consumed what I want it to. Using this string as an example, the goal is to match "CAPS":
>>> s = "only the word in CAPS should be matched"
So let's say I want to specify when to begin my pattern by using a lookbehind:
>>> x = re.compile
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