re.findall() hangs in python

[silverburgh.meryl@gmail.com]

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(patt ern, data)

if (match):
print "match2"



Can you please tell me why it that?

[Peter Otten]

Re: re.findall() hangs in python

silverburgh.mer yl@gmail.com 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(patt ern, 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

[7stud]

Re: re.findall() hangs in python

On Mar 31, 9:12 pm, "silverburgh.me ...@gmail.com"
<silverburgh.me ...@gmail.comwr ote:

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(patt ern, 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.

[Gabriel Genellina]

Re: re.findall() hangs in python

En Sun, 01 Apr 2007 03:58:51 -0300, Peter Otten <__peter__@web. de>
escribió:

silverburgh.mer yl@gmail.com 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 slooooooow!
To the OP: do you actually need all those groups? Specially the first and
last (.*), they match all the surrounding text.

--
Gabriel Genellina

[irstas@gmail.com]

Re: re.findall() hangs in python

On Apr 1, 6:12 am, "silverburgh.me ...@gmail.com"
<silverburgh.me ...@gmail.comwr ote:

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'<i mg [^>]*src\s*=\s*"([^"]*img[^"]*)"')

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

[silverburgh.meryl@gmail.com]

Re: re.findall() hangs in python

On Apr 1, 5:23 am, irs...@gmail.co m wrote:

On Apr 1, 6:12 am, "silverburgh.me ...@gmail.com"
>
<silverburgh.me ...@gmail.comwr ote:

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'<i mg [^>]*src\s*=\s*"([^"]*img[^"]*)"')
>
"dot" is usually not so great. Prefer "NOT end-character", like [^>]
or [^"].

Thank you. Your suggestion solves my problem!