Re: [Tutor] stopping greedy matches

[Mike Hall]

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 towards it :)

On Mar 16, 2005, at 2:36 PM, Liam Clarke wrote:
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 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(r"(?<=\bin)") #this will simply match the spot in
front of "in"
So that's straight forward, but let's say I don't want to use a
lookahead to specify the end of my pattern, I simply want it to stop
after it has combed over the word following "in". I would expect this
to work, but it doesn't:
x=re.compile(r"(?<=\bin).+\b") #this will consume everything past
"in" all the way to the end of the string
In the above example I would think that the word boundary flag "\b"
would indicate a stopping point. Is ".+\b" not saying, "keep matching
characters until a word boundary has been reached"?
Even stranger are the results I get from:
x=re.compile(r"(?<=\bin).+\s") #keep matching characters until a
whitespace has been reached(?)
r = x.sub("[EMAIL PROTECTED]", s)
print r
only the word [EMAIL PROTECTED]
For some reason there it's decided to consume three words instead of
one.
My question is simply this:  after specifying a start point,  how do I
make a match stop after it has found one word, and one word only? As
always, all help is appreciated.
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