2010/9/25 AON LAZIO :
> Hi,
> Say I have
> p = re.compile('a|b')
> text = 'a'
> d = p.findall(text)
> #d -> ['a']
> What is the way to find out which pattern p match (the former or latter)?
> I mean without knowing the outcome of p.findall
> Thanks
>
> --
> Aonlazio
> '
Hi,
Say I have
p = re.compile('a|b')
text = 'a'
d = p.findall(text)
#d -> ['a']
What is the way to find out which pattern p match (the former or latter)?
I mean without knowing the outcome of p.findall
Thanks
--
Aonlazio
'Peace is always the way.' NW
--
http://mail.p
"Johnny Lee" <[EMAIL PROTECTED]> writes:
> Fredrik Lundh wrote:
[...]
> To the HTMLParser, there is another problem (take my code for example):
>
> import urllib
> import formatter
> parser = htmllib.HTMLParser(formatter.NullFormatter())
> parser.feed(urllib.urlopen(baseUrl).read())
> parser.clos
"Fredrik Lundh" <[EMAIL PROTECTED]> writes:
[...]
> or, if you're going to parse HTML pages from many different sources, a
> real parser:
>
> from HTMLParser import HTMLParser
>
> class MyHTMLParser(HTMLParser):
>
> def handle_starttag(self, tag, attrs):
> if tag == "
Fredrik Lundh wrote:
> ".*" gives the longest possible match (you can think of it as searching back-
> wards from the right end). if you want to search for "everything until a
> given
> character", searching for "[^x]*x" is often a better choice than ".*x".
>
> in this case, I suggest using some
Johnny Lee wrote:
> I've met a problem in match a regular expression in python. Hope
> any of you could help me. Here are the details:
>
> I have many tags like this:
> xxxhttp://xxx.xxx.xxx"; xxx>xxx
> xx
> xxxhttp://xxx.xxx.xxx"; xxx>xxx
> .
> And I want to find
Hi,
I've met a problem in match a regular expression in python. Hope
any of you could help me. Here are the details:
I have many tags like this:
xxxhttp://xxx.xxx.xxx"; xxx>xxx
xx
xxxhttp://xxx.xxx.xxx"; xxx>xxx
.
And I want to find all the "http://xxx.xxx.
Thanks you...
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[EMAIL PROTECTED] wrote:
Hello,
For example I have a string : "Halo by by by"
Then I want to take and know the possition of every "by"
how can I do it in python?
[ match.start() for match in p.finditer("Helo by by by") ]
see:
http://mail.python.org/pipermail/python-list/2004-December/255013.html
--
[EMAIL PROTECTED] wrote:
Hello,
For example I have a string : "Halo by by by"
Then I want to take and know the possition of every "by"
how can I do it in python?
I tried to use:
p = re.compile(r"by")
m = p.search("Helo by by by")
print m.group() # result "by"
print m.span() # result (5,7)
<[EMAIL PROTECTED]> wrote:
> For example I have a string : "Halo by by by"
> Then I want to take and know the possition of every "by"
> how can I do it in python?
>
> I tried to use:
>
> p = re.compile(r"by")
> m = p.search("Helo by by by")
> print m.group() # result "by"
> print m.span() # r
Hello,
For example I have a string : "Halo by by by"
Then I want to take and know the possition of every "by"
how can I do it in python?
I tried to use:
p = re.compile(r"by")
m = p.search("Helo by by by")
print m.group() # result "by"
print m.span() # result (5,7)
How can I get the inf
Johann C. Rocholl wrote:
>
> How do you people handle this?
Usually we don't bothe too much. But it has been suggested to do something
like this:
class Matcher:
def __init__(self, rex):
self.rex = rex
def match(self, s):
self.m = self.rex.match(s)
return not self.m
Hello python-list,
I have a question about the match objects that are returned from the
match() method of compiled regular expression objects from the 're'
module. To parse Postscript T1 fonts that were disassembled into
plaintext, I came up with the following code:
import re
rmoveto = re.compile
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