Hi,
On 2014-08-16 09:01:57 +, Peter Otten said:
Philipp Kraus wrote:
The code works till last week correctly, I don't change the pattern.
Websites' contents and structure change sometimes.
My question is, can it be a problem with string encoding?
Your regex is all-ascii. So an
Philipp Kraus wrote:
I have create a short script:
-
#!/usr/bin/env python
import re, urllib2
def URLReader(url) :
f = urllib2.urlopen(url)
data = f.read()
f.close()
return data
print re.match( \small\ \.*\\/small\,
Philipp Kraus wrote:
The code works till last week correctly, I don't change the pattern.
Websites' contents and structure change sometimes.
My question is, can it be a problem with string encoding?
Your regex is all-ascii. So an encoding problem is very unlikely.
found = re.search( a
Hello,
I have defined a function with:
def URLReader(url) :
try :
f = urllib2.urlopen(url)
data = f.read()
f.close()
except Exception, e :
raise MyError.StopError(e)
return data
which get the HTML source code from an URL. I use this to get a part of
a HTML
In article lsm8ic$j90$1...@online.de,
Philipp Kraus philipp.kr...@flashpixx.de wrote:
found = re.search( a
href=\/projects/boost/files/latest/download\?source=files\
title=\/boost/(.*),
Utilities.URLReader(http://sourceforge.net/projects/boost/files/boost/;)
)
if found == None :
On 2014-08-16 00:48:46 +, Roy Smith said:
In article lsm8ic$j90$1...@online.de,
Philipp Kraus philipp.kr...@flashpixx.de wrote:
found = re.search( a
href=\/projects/boost/files/latest/download\?source=files\
title=\/boost/(.*),
In article lsmeej$49n$1...@online.de,
Philipp Kraus philipp.kr...@flashpixx.de wrote:
The code works till last week correctly, I don't change the pattern.
OK, so what did you change? Can you go back to last week's code and
compare it to what you have now to see what changed?
My question
Philipp Kraus wrote:
The code works till last week correctly, I don't change the pattern. My
question is, can it be
a problem with string encoding? Did I mask the question mark and quotes
correctly?
If you didn't change the code, how could the *exact same code* not mask the
question mark
Stefan Behnel wrote:
Wolfgang Rohdewald wrote:
I want to match a string only if a word (C1 in this example) appears
at most once in it.
def match(s):
if s.count(C1) 1:
return None
return s
If this doesn't fit your requirements, you may want to provide some
On Oct 4, 9:34 pm, Wolfgang Rohdewald wolfg...@rohdewald.de wrote:
Hi,
I want to match a string only if a word (C1 in this example) appears
at most once in it. This is what I tried:
re.match(r'(.*?C1)((?!.*C1))','C1b1b1b1 b3b3b3b3 C1C2C3').groups()
('C1b1b1b1 b3b3b3b3 C1', '')
On Monday 05 October 2009, Carl Banks wrote:
What you're not realizing is that if a regexp search comes to a
dead end, it won't simply return no match. Instead it'll throw
away part of the match, and backtrack to a previously-matched
variable-length subexpression, such as .*?, and try
Wolfgang Rohdewald wrote:
I want to match a string only if a word (C1 in this example) appears
at most once in it.
def match(s):
if s.count(C1) 1:
return None
return s
If this doesn't fit your requirements, you may want to provide some more
details.
Stefan
--
On Oct 4, 11:17 pm, Wolfgang Rohdewald wolfg...@rohdewald.de wrote:
On Monday 05 October 2009, Carl Banks wrote:
What you're not realizing is that if a regexp search comes to a
dead end, it won't simply return no match. Instead it'll throw
away part of the match, and backtrack to a
On Monday 05 October 2009, Stefan Behnel wrote:
Wolfgang Rohdewald wrote:
I want to match a string only if a word (C1 in this example)
appears at most once in it.
def match(s):
if s.count(C1) 1:
return None
return s
If this doesn't fit your
On Monday 05 October 2009, MRAB wrote:
(?!.*?(C1).*?\1) will succeed only if .*?(C1).*?\1 has failed,
in which case the group (group 1) will be undefined (no capture).
I see.
I should have moved the (C1) out of this expression anyway:
re.match(r'L(?Ptile..)(?!.*?(?P=tile).*?(?P=tile))(.*?
Hi,
I want to match a string only if a word (C1 in this example) appears
at most once in it. This is what I tried:
re.match(r'(.*?C1)((?!.*C1))','C1b1b1b1 b3b3b3b3 C1C2C3').groups()
('C1b1b1b1 b3b3b3b3 C1', '')
re.match(r'(.*?C1)','C1b1b1b1 b3b3b3b3 C1C2C3').groups()
('C1',)
but this should
Why not check it simply by count()?
s = '1234C156789'
s.count('C1')
1
--
http://mail.python.org/mailman/listinfo/python-list
Analog Kid wrote:
Hi guys:
Thanks for your responses. Points taken. Basically, I am looking for a
combination of the following ...
[^\w] and %(?!20) ... How do I do this in a single RE?
Thanks for all you help.
Regards,
AK
On Mon, Dec 15, 2008 at 10:54 PM, Steve Holden
Hi guys:
Thanks for your responses. Points taken. Basically, I am looking for a
combination of the following ...
[^\w] and %(?!20) ... How do I do this in a single RE?
Thanks for all you help.
Regards,
AK
On Mon, Dec 15, 2008 at 10:54 PM, Steve Holden st...@holdenweb.com wrote:
Analog Kid
Analog Kid wrote:
Hi All:
I am new to regular expressions in general, and not just re in python.
So, apologies if you find my question stupid :) I need some help with
forming a regex. Here is my scenario ...
I have strings coming in from a list, each of which I want to check
against a regular
Hi All:
I am new to regular expressions in general, and not just re in python. So,
apologies if you find my question stupid :) I need some help with forming a
regex. Here is my scenario ...
I have strings coming in from a list, each of which I want to check against
a regular expression and see
Analog Kid wrote:
Hi All:
I am new to regular expressions in general, and not just re in python.
So, apologies if you find my question stupid :) I need some help with
forming a regex. Here is my scenario ...
I have strings coming in from a list, each of which I want to check
against a
Hi John,
John Machin schrieb am 11/20/2007 09:40 PM:
On Nov 21, 8:05 am, Fabian Braennstroem [EMAIL PROTECTED] wrote:
Hi,
I would like to use re to search for lines in a files with
the word README_x.org, where x is any number.
E.g. the structure would look like this:
*):
print File Found
print str(files)
break
As soon as it finds the file, it should stop the searching
process; but there is the same matching problem like above.
Does anyone have any suggestions about the regex problem?
Greetings
On Nov 21, 8:05 am, Fabian Braennstroem [EMAIL PROTECTED] wrote:
Hi,
I would like to use re to search for lines in a files with
the word README_x.org, where x is any number.
E.g. the structure would look like this:
[[file:~/pfm_v99/README_1.org]]
I tried to use these kind of matchings:
#
import re
s1 ='nbsp;25000nbsp;'
s2 = 'nbsp;5.5910nbsp;'
mypat = re.compile('[0-9]*(\.[0-9]*|$)')
rate= mypat.search(s1)
print rate.group()
rate=mypat.search(s2)
print rate.group()
rate = mypat.search(s1)
price = float(rate.group())
print price
I get an error when it hits the whole number, that
On Sep 21, 5:04 pm, crybaby [EMAIL PROTECTED] wrote:
import re
s1 ='nbsp;25000nbsp;'
s2 = 'nbsp;5.5910nbsp;'
mypat = re.compile('[0-9]*(\.[0-9]*|$)')
rate= mypat.search(s1)
print rate.group()
rate=mypat.search(s2)
print rate.group()
rate = mypat.search(s1)
price = float(rate.group())
crybaby wrote:
import re
s1 ='nbsp;25000nbsp;'
s2 = 'nbsp;5.5910nbsp;'
mypat = re.compile('[0-9]*(\.[0-9]*|$)')
rate= mypat.search(s1)
print rate.group()
rate=mypat.search(s2)
print rate.group()
rate = mypat.search(s1)
price = float(rate.group())
print price
I get an error when
On Sep 21, 2007, at 4:04 PM, crybaby wrote:
import re
s1 ='nbsp;25000nbsp;'
s2 = 'nbsp;5.5910nbsp;'
mypat = re.compile('[0-9]*(\.[0-9]*|$)')
rate= mypat.search(s1)
print rate.group()
rate=mypat.search(s2)
print rate.group()
rate = mypat.search(s1)
price = float(rate.group())
print
Hi all,
line is am trying to match is
1959400|Q2BYK3|Q2BYK3_9GAMM Hypothetical outer membra29.90.00011 1
regex i have written is
re.compile
(r'(\d+?)\|((P|O|Q)\w{5})\|\w{3,6}\_\w{3,5}\s+?.{25}\s{3}(\d+?\.\d)\s+?(\d\.\d+?)')
I am trying to extract 0.0011 value from the above line.
why
line is am trying to match is
1959400|Q2BYK3|Q2BYK3_9GAMM Hypothetical outer membra29.90.00011 1
regex i have written is
re.compile
(r'(\d+?)\|((P|O|Q)\w{5})\|\w{3,6}\_\w{3,5}\s+?.{25}\s{3}(\d+?\.\d)\s+?(\d\.\d+?)')
I am trying to extract 0.0011 value from the above line.
why
HI Tim,
oof!
thats true!
thanks a lot.
Is there any tool to simplify building the regex ?
regards,
KM
On 11/23/06, Tim Chase [EMAIL PROTECTED] wrote:
line is am trying to match is
1959400|Q2BYK3|Q2BYK3_9GAMM Hypothetical outer membra29.90.00011
1
regex i have written is
line is am trying to match is
1959400|Q2BYK3|Q2BYK3_9GAMM Hypothetical outer membra29.90.00011 1
regex i have written is
re.compile
(r'(\d+?)\|((P|O|Q)\w{5})\|\w{3,6}\_\w{3,5}\s+?.{25}\s{3}(\d+?\.\d)\s+?(\d\.\d+?)')
I am trying to extract 0.0011 value from the above line.
On 2005-07-26, Duncan Booth [EMAIL PROTECTED] wrote:
rx1=re.compile(r\b\d{4}(?:-\d{4})?,)
rx1.findall(1234,-,4567,)
['1234,', '-,', '4567,']
Thanks all for good advice. However this last expression
also matches the first four digits when the input is more
than four digits. To
Input is a string of four digit sequences, possibly
separated by a -, for instance like this
1234,-,4567,
My regular expression is like this:
rx1=re.compile(r\A(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)*\Z)
When running rx1.findall(1234,-,4567,)
I only get the last match as the
Am Tue, 26 Jul 2005 09:57:23 + schrieb Odd-R.:
Input is a string of four digit sequences, possibly
separated by a -, for instance like this
1234,-,4567,
My regular expression is like this:
rx1=re.compile(r\A(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)*\Z)
Hi,
try it without \A and
Odd-R. wrote:
Input is a string of four digit sequences, possibly
separated by a -, for instance like this
1234,-,4567,
My regular expression is like this:
rx1=re.compile(r\A(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)*\Z)
When running rx1.findall(1234,-,4567,)
I only get
John Machin wrote:
So here's the mean lean no-flab version -- you don't even need the
parentheses (sorry, Thomas).
rx1=re.compile(r\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)
rx1.findall(1234,-,4567,)
['1234,', '-,', '4567,']
No flab? What about all that repetition of \d? A less
Duncan Booth wrote:
John Machin wrote:
So here's the mean lean no-flab version -- you don't even need the
parentheses (sorry, Thomas).
rx1=re.compile(r\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)
rx1.findall(1234,-,4567,)
['1234,', '-,', '4567,']
No flab? What about all that
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