Re: re.findall() hangs in python

[Peter Otten]

[EMAIL PROTECTED] wrote:

 I have the following regular expression.
 It works when 'data' contains the pattern and I see 'match2' get print
 out.
 But when 'data' does not contain pattern, it just hangs at
 're.findall'
 
 pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?),
 re.S)
 
 print before find all
 
 match = re.findall(pattern, data)
 
 if (match):
print match2
 
 
 
 Can you please tell me why it that?

Could it be that it is just slow? If not, post a small example of data that
provokes findall() to hang.

Peter
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Re: re.findall() hangs in python

[7stud]

On Mar 31, 9:12 pm, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
 Hi,

 I have the following regular expression.
 It works when 'data' contains the pattern and I see 'match2' get print
 out.
 But when 'data' does not contain pattern, it just hangs at
 're.findall'

 pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?),
 re.S)

 print before find all

 match = re.findall(pattern, data)

 if (match):
print match2

 Can you please tell me why it that?

It doesn't hang when I try it.  Why don't you post a complete example
that hangs.

Also, you might consider using exterior single quotes around your
string so that you don't have to escape double quotes inside the
string.

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Re: re.findall() hangs in python

[Gabriel Genellina]

En Sun, 01 Apr 2007 03:58:51 -0300, Peter Otten [EMAIL PROTECTED]  
escribió:

 [EMAIL PROTECTED] wrote:

 I have the following regular expression.
 It works when 'data' contains the pattern and I see 'match2' get print
 out.
 But when 'data' does not contain pattern, it just hangs at
 're.findall'

 pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?),
 re.S)

 Could it be that it is just slow? If not, post a small example of data  
 that
 provokes findall() to hang.

I bet it is very slooow!
To the OP: do you actually need all those groups? Specially the first and  
last (.*), they match all the surrounding text.

-- 
Gabriel Genellina

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Re: re.findall() hangs in python

[irstas]

On Apr 1, 6:12 am, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
 But when 'data' does not contain pattern, it just hangs at
 're.findall'

 pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?),
 re.S)

That pattern is just really slow to evaluate. What you want is
probably something more like this:

re.compile(r'img [^]*src\s*=\s*([^]*img[^]*)')

dot is usually not so great. Prefer NOT end-character, like [^]
or [^].

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Re: re.findall() hangs in python

[[EMAIL PROTECTED]]

On Apr 1, 5:23 am, [EMAIL PROTECTED] wrote:
 On Apr 1, 6:12 am, [EMAIL PROTECTED]

 [EMAIL PROTECTED] wrote:
  But when 'data' does not contain pattern, it just hangs at
  're.findall'

  pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?),
  re.S)

 That pattern is just really slow to evaluate. What you want is
 probably something more like this:

 re.compile(r'img [^]*src\s*=\s*([^]*img[^]*)')

 dot is usually not so great. Prefer NOT end-character, like [^]
 or [^].

Thank you. Your suggestion solves my problem!

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re.findall() hangs in python

[[EMAIL PROTECTED]]

Hi,

I have the following regular expression.
It works when 'data' contains the pattern and I see 'match2' get print
out.
But when 'data' does not contain pattern, it just hangs at
're.findall'

pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?),
re.S)

print before find all

match = re.findall(pattern, data)

if (match):
   print match2



Can you please tell me why it that?

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