On Mon, Dec 19, 2011 at 4:15 PM, wrote:
> On Dec 16, 11:49 am, John Gordon wrote:
>> I'm working with IPv6 CIDR strings, and I want to replace the longest
>> match of "(:|$)+" with ":". But when I use re.sub() it replaces
>> the leftmost match, even if there is a longer match later in t
On Dec 16, 11:49 am, John Gordon wrote:
> I'm working with IPv6 CIDR strings, and I want to replace the longest
> match of "(:|$)+" with ":". But when I use re.sub() it replaces
> the leftmost match, even if there is a longer match later in the string.
Typically this means that your regu
MRAB wrote:
> On 16/12/2011 21:04, John Gordon wrote:
>> In Devin
>> Jeanpierre writes:
>>
>>> You could use re.finditer to find the longest match, and then
>>> replace it manually by hand (via string slicing).
>>
>>> (a match is the longest if (m.end() - m.start()) is the largest --
>>> s
On 12/16/2011 1:36 PM, Roy Smith wrote:
What you want is an IPv6 class which represents an address in some
canonical form. It would have constructors which accept any of the
RFC-2373 defined formats. It would also have string formatting methods
to convert the internal form into any of these fo
On 16/12/2011 21:04, John Gordon wrote:
In Devin
Jeanpierre writes:
You could use re.finditer to find the longest match, and then replace
it manually by hand (via string slicing).
(a match is the longest if (m.end() - m.start()) is the largest --
so, max(re.finditer(...), key=3Dlambda
In Roy Smith writes:
> Having done quite a bit of IPv6 work, my opinion here is that you're
> trying to do The Wrong Thing.
> What you want is an IPv6 class which represents an address in some
> canonical form. It would have constructors which accept any of the
> RFC-2373 defined formats.
In Ian Kelly
writes:
> >>> 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.
> pattern = r'\b0{1,3}([1-9a-f][0-9a-f]*|0)\b'
> re.sub(pattern, r'\1', string, flags=re.IGNORECASE)
Perfect
In Devin Jeanpierre
writes:
> You could use re.finditer to find the longest match, and then replace
> it manually by hand (via string slicing).
> (a match is the longest if (m.end() - m.start()) is the largest --
> so, max(re.finditer(...), key=3Dlambda m: (m.end() =3D m.start()))
I ended up
In article ,
John Gordon wrote:
> I'm working with IPv6 CIDR strings, and I want to replace the longest
> match of "(:|$)+" 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 wi
On 16/12/2011 17:57, Ian Kelly wrote:
On Fri, Dec 16, 2011 at 10:36 AM, MRAB 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 nece
On Fri, Dec 16, 2011 at 10:57 AM, Ian Kelly wrote:
> On Fri, Dec 16, 2011 at 10:36 AM, MRAB 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* matchin
On Fri, Dec 16, 2011 at 10:36 AM, MRAB 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.
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
You could use re.finditer to find the longest match, and then replace
it manually by hand (via string slicing).
(a match is the longest if (m.end() - m.start()) is the largest --
so, max(re.finditer(...), key=lambda m: (m.end() = m.start()))
-- Devin
P.S. does anyone else get bothered by how it'
On 17 Sep, 19:59, Peter Otten <__pete...@web.de> wrote:
> Jon Clements wrote:
> > (I reckon this is probably a question for MRAB and is not really
> > Python specific, but anyhow...)
>
> > Absolutely basic example: re.sub(r'(\d+)', r'\1', 'string1')
>
> > I've been searching around and I'm sure it'
Jon Clements wrote:
> (I reckon this is probably a question for MRAB and is not really
> Python specific, but anyhow...)
>
> Absolutely basic example: re.sub(r'(\d+)', r'\1', 'string1')
>
> I've been searching around and I'm sure it'll be obvious when it's
> pointed out, but how do I use the abo
On 17/09/2010 19:21, Jon Clements wrote:
Hi All,
(I reckon this is probably a question for MRAB and is not really
Python specific, but anyhow...)
Absolutely basic example: re.sub(r'(\d+)', r'\1', 'string1')
I've been searching around and I'm sure it'll be obvious when it's
pointed out, but how
Steven D'Aprano wrote:
On Thu, 12 Aug 2010 14:33:28 -0700, fuglyducky wrote:
if anyone happens to know about
passing a variable into a regex that would be great.
The same way you pass anything into any string.
Regexes are ordinary strings. If you want to construct a string from a
variable t
On Thu, 12 Aug 2010 14:33:28 -0700, fuglyducky wrote:
> if anyone happens to know about
> passing a variable into a regex that would be great.
The same way you pass anything into any string.
Regexes are ordinary strings. If you want to construct a string from a
variable t = "orl", you can do an
On Aug 13, 7:33 am, fuglyducky wrote:
> On Aug 12, 2:06 pm, fuglyducky wrote:
>
>
>
> > I have a function that I am attempting to call from another file. I am
> > attempting to replace a string using re.sub with another string. The
> > problem is that the second string is a variable. When I get t
On Aug 12, 2:06 pm, fuglyducky wrote:
> I have a function that I am attempting to call from another file. I am
> attempting to replace a string using re.sub with another string. The
> problem is that the second string is a variable. When I get the
> output, it shows the variable name rather than t
Thanks for your answers. They helped me to realize that I was
mistakenly using match.string (the whole string) when I should be
using math.group(0) (the whole match).
Best regards,
Javier
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http://mail.python.org/mailman/listinfo/python-list
On Tue, 06 Jul 2010 19:10:17 +0200, Javier Collado wrote:
> Hello,
>
> Let's imagine that we have a simple function that generates a
> replacement for a regular expression:
>
> def process(match):
> return match.string
>
> If we use that simple function with re.sub using a simple pattern an
On 07/06/2010 07:10 PM, Javier Collado wrote:
> Hello,
>
> Let's imagine that we have a simple function that generates a
> replacement for a regular expression:
>
> def process(match):
> return match.string
>
> If we use that simple function with re.sub using a simple pattern and
> a string
On Oct 16, 9:51 am, MRAB wrote:
> What do you mean "blow up"? It worked for me in Python v2.6.2.
My bad. False alarm. This was one of those cases where a bug in
another area appears like a bug in a different area.
Thank for the help.
cs
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http://mail.python.org/mailman/listinfo/python-list
Chris Seberino wrote:
What does this line do?...
input_ = re.sub("([a-zA-Z]+)", '"\\1"', input_)
Why don't you try it?
Does it remove parentheses from words?
e.g. (foo) -> foo ???
No, it puts quotes around them.
I'd like to replace [a-zA-Z] with \w but \w makes it blow up.
In other word
Chris Seberino wrote:
What does this line do?...
input_ = re.sub("([a-zA-Z]+)", '"\\1"', input_)
Does it remove parentheses from words?
e.g. (foo) -> foo ???
I'd like to replace [a-zA-Z] with \w but \w makes it blow up.
In other words, re.sub("(\w+)", '"\\1"', input_) blows up.
Why?
cs
J Wolfe wrote:
> Hi,
>
> Is there a way to flag re.sub not to replace a portion of the string?
>
> I have a very long string that I want to add two new line's to rather
> than one, but keep the value X:
>
> string = "testX.\n.today" <-- note X is a value
> string = re.sub("test...
Thanks Duncan,
I did look at that, but it was kinda greek to me. Thanks for pulling
out the part I was looking for that should do the trick.
Jonathan
> http://www.python.org/doc/current/library/re.html#re.sub
>
> > Backreferences, such as \6, are replaced with the substring matched by
> > group
On Wed, 11 Feb 2009 21:05:53 -, Paul McGuire
wrote:
On Feb 4, 10:51 am, "Emanuele D'Arrigo" wrote:
Hi everybody,
I'm having a ball with the power of regular expression
Don't forget the ball you can have with the power of ordinary Python
strings, string methods, and string interpolati
On Feb 4, 10:51 am, "Emanuele D'Arrigo" wrote:
> Hi everybody,
>
> I'm having a ball with the power of regular expression
Don't forget the ball you can have with the power of ordinary Python
strings, string methods, and string interpolation!
originalString = "spam:%(first)s ham:%(second)s"
print
>
> Book recommendation: _Mastering Regular Expressions_, Jeffrey Friedl
> --
> Aahz (a...@pythoncraft.com) <*> http://www.pythoncraft.com/
I wholeheartedly second this! The third edition is out now.
--
http://mail.python.org/mailman/listinfo/python-list
In article <4c7158d2-5663-46b9-b950-be81bd799...@z6g2000pre.googlegroups.com>,
Emanuele D'Arrigo wrote:
>
>I'm having a ball with the power of regular expression but I stumbled
>on something I don't quite understand:
Book recommendation: _Mastering Regular Expressions_, Jeffrey Friedl
--
Aahz (a
> Hi everybody,
>
> I'm having a ball with the power of regular expression but I stumbled
> on something I don't quite understand:
>
> theOriginalString = "spam:(?P.*) ham:(?P.*)"
> aReplacementPattern = "\(\?P.*\)"
> aReplacementString= "foo"
> re.sub(aReplacementPattern , aReplacementString, the
On Feb 4, 5:17 pm, MRAB wrote:
> You could use the lazy form "*?" which tries to match as little as
> possible, eg "\(\?P.*?\)" where the ".*?" matches:
> spam:(?P.*) ham:(?P.*)
> giving "spam:foo ham:(?P.*)".
A-ha! Of course! That makes perfect sense! Thank you! Problem solved!
Ciao!
Manu
--
Emanuele D'Arrigo wrote:
> Hi everybody,
>
> I'm having a ball with the power of regular expression but I stumbled
> on something I don't quite understand:
>
> theOriginalString = "spam:(?P.*) ham:(?P.*)"
> aReplacementPattern = "\(\?P.*\)"
> aReplacementString= "foo"
> re.sub(aReplacementPattern
On Dec 13, 9:00 pm, Davy <[EMAIL PROTECTED]> wrote:
>
> What's "\1" and the whole re.sub() mean?
>
Read about backreferences here:
http://www.regular-expressions.info/brackets.html
Also see the entry on parentheses here:
http://docs.python.org/lib/re-syntax.html
rick
--
http://mail.python.or
Thank you this answers my question. I wanted to make sure it was actually
designed this way.
Massimo
From: Tim Chase [EMAIL PROTECTED]
Sent: Tuesday, October 16, 2007 1:38 PM
To: DiPierro, Massimo
Cc: python-list@python.org; Berthiaume, Andre
Subject: Re
> Let me show you a very bad consequence of this...
>
> a=open('file1.txt','rb').read()
> b=re.sub('x',a,'x')
> open('file2.txt','wb').write(b)
>
> Now if file1.txt contains a \n or \" then file2.txt is not the
> same as file1.txt while it should be.
That's functioning as designed. If you want
> Even stranger
>
> >>> re.sub('a', '\\n','bab')
> 'b\nb'
> >>> print re.sub('a', '\\n','bab')
> b
> b
That's to be expected. When not using a print statement, the raw
evaluation prints the representation of the object. In this
case, the representation is 'b\nb'. When you use the print
st
d be.
Massimo
From: Tim Chase [EMAIL PROTECTED]
Sent: Tuesday, October 16, 2007 1:20 PM
To: DiPierro, Massimo
Cc: python-list@python.org; Berthiaume, Andre
Subject: Re: re.sub
> Even stranger
>
> >>> re.sub('a', '\\n',
It is the fisrt line that is wrong, the second follows from the first, I agree.
From: Tim Chase [EMAIL PROTECTED]
Sent: Tuesday, October 16, 2007 1:20 PM
To: DiPierro, Massimo
Cc: python-list@python.org; Berthiaume, Andre
Subject: Re: re.sub
> E
_
From: [EMAIL PROTECTED] [EMAIL PROTECTED] On Behalf Of Chris Mellon [EMAIL
PROTECTED]
Sent: Tuesday, October 16, 2007 1:12 PM
To: python-list@python.org
Subject: Re: re.sub
On 10/16/07, Massimo Di Pierro <[EMAIL PROTECTED]> wrote:
> Even stranger
>
> >>> re.sub(&#
On 10/16/07, Massimo Di Pierro <[EMAIL PROTECTED]> wrote:
> Even stranger
>
> >>> re.sub('a', '\\n','bab')
> 'b\nb'
> >>> print re.sub('a', '\\n','bab')
> b
> b
>
You called print, so instead of getting an escaped string literal, the
string is being printed to your terminal, which is printing th
Even stranger
>>> re.sub('a', '\\n','bab')
'b\nb'
>>> print re.sub('a', '\\n','bab')
b
b
Massimo
On Oct 16, 2007, at 1:54 AM, DiPierro, Massimo wrote:
> Shouldn't this
>
print re.sub('a','\\n','bab')
> b
> b
>
> output
>
> b\nb
>
> instead?
>
> Massimo
>
> On Oct 16, 2007, at 1:34 AM, G
Neil Cerutti schrieb:
> In other words, the fourth argument to sub is count, not a set of
> re flags.
I knew it had to be something very stupid.
Thanks a lot.
--
http://mail.python.org/mailman/listinfo/python-list
On 2007-08-07, Christoph Krammer <[EMAIL PROTECTED]> wrote:
> Hello everybody,
>
> I wanted to use re.sub to strip all HTML tags out of a given string. I
> learned that there are better ways to do this without the re module,
> but I would like to know why my code is not working. I use the
> followi
On Tue, 07 Aug 2007 10:28:24 -0700, Christoph Krammer wrote:
> Hello everybody,
>
> I wanted to use re.sub to strip all HTML tags out of a given string. I
> learned that there are better ways to do this without the re module,
> but I would like to know why my code is not working. I use the
> foll
Hugo Ferreira wrote:
> Hi!
>
> I'm trying to do a search-replace in places where some groups are
> optional... Here's an example:
>
> >> re.match(r"Image:([^\|]+)(?:\|(.*))?", "Image:ola").groups()
> ('ola', None)
>
> >> re.match(r"Image:([^\|]+)(?:\|(.*))?", "Image:ola|").groups()
> ('ola', '')
Hugo Ferreira wrote:
> Hi!
>
> I'm trying to do a search-replace in places where some groups are
> optional... Here's an example:
>
> >> re.match(r"Image:([^\|]+)(?:\|(.*))?", "Image:ola").groups()
> ('ola', None)
>
> >> re.match(r"Image:([^\|]+)(?:\|(.*))?", "Image:ola|").groups()
> ('ola', '')
Paddy wrote:
> Check the arguments to re.sub.
>
> >>> re.sub('(?m)^foo', 'bar', '\nfoo', count=0)
> '\nbar'
>
> - Paddy.
Duh! :) I appreciate it, thanks.
-- nyenyec
--
http://mail.python.org/mailman/listinfo/python-list
nyenyec wrote:
> I feel like a complete idiot but I can't figure out why re.sub won't
> match multiline strings:
>
> This works:
> >>> re.search("^foo", "\nfoo", re.MULTILINE)
> <_sre.SRE_Match object at 0x6c448>
>
> This doesn't. No replacement:
> >>> re.sub("^foo", "bar", "\nfoo", re.MULTILINE)
thanks - that's the trick.
On 8/17/06, Tim Chase <[EMAIL PROTECTED]> wrote:
> Looks like you need to be using "raw" strings for your
> replacements as well:
>
> s = re.sub(r'([A-Z]+)([A-Z][a-z])', r"\1_\2", s)
> s = re.sub(r'([a-z\d])([A-Z])', r"\1_\2", s)
>
> This should allow the backslashes to
> Tim's given you the solution to the problem: with the re module,
> *always* use raw strings in regexes and substitution strings.
"always" is so...um...carved in stone. One can forego using raw
strings if one prefers having one's strings looked like they were
trampled by a stampede of creatu
[EMAIL PROTECTED] wrote:
> using this code:
>
> import re
> s = 'HelloWorld19-FooBar'
> s = re.sub(r'([A-Z]+)([A-Z][a-z])', "\1_\2", s)
> s = re.sub(r'([a-z\d])([A-Z])', "\1_\2", s)
> s = re.sub('-', '_', s)
> s = s.lower()
> print "s: %s" % s
>
> i expect to get:
> hello_world19_foo_bar
>
> but i
> s = re.sub(r'([A-Z]+)([A-Z][a-z])', "\1_\2", s)
> s = re.sub(r'([a-z\d])([A-Z])', "\1_\2", s)
> i expect to get:
> hello_world19_foo_bar
>
> but instead i get:
> hell☺_☻orld19_fo☺_☻ar
Looks like you need to be using "raw" strings for your
replacements as well:
s = re.sub(r'([A-Z]+)([A-Z][a-z
Glad I could help.
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http://mail.python.org/mailman/listinfo/python-list
Thanks a lot! Compiling with re.DOTALL did fix my problem for the most
part; there still are a few problems with my code, but I think I can
fix those myself.
Again, thanks!
> Okay I just woke up and haven't had enough coffee so if I'm off here
> please forgive me. Are you saying that if there is
Okay I just woke up and haven't had enough coffee so if I'm off here
please forgive me. Are you saying that if there is an emptly line then
it borks? If so just use re.S ( re.DOTALL ) and that should take care
of it. It will treat the ( . ) special. Otherwise it ignores new
lines.
--
http://m
Actually, it happens in general when there is more than one linebreak
between the open and close statements; not only when there are empty
lines.
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