Re: Regular expression question -- exclude substring
On Mon, 7 Nov 2005 16:38:11 -0800, James Stroud <[EMAIL PROTECTED]> wrote:
>On Monday 07 November 2005 16:18, [EMAIL PROTECTED] wrote:
>> Ya, for some reason your non-greedy "?" doesn't seem to be taking.
>> This works:
>>
>> re.sub('(.*)(00.*?01) target_mark', r'\2', your_string)
>
>The non-greedy is actually acting as expected. This is because non-greedy
>operators are "forward looking", not "backward looking". So the non-greedy
>finds the start of the first start-of-the-match it comes accross and then
>finds the first occurrence of '01' that makes the complete match, otherwise
>the greedy operator would match .* as much as it could, gobbling up all '01's
>before the last because these match '.*'. For example:
>
>py> rgx = re.compile(r"(00.*01) target_mark")
>py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
>['00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01']
>py> rgx = re.compile(r"(00.*?01) target_mark")
>py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
>['00 noise1 01 noise2 00 target 01', '00 dowhat 01']
>
>My understanding is that backward looking operators are very resource
>expensive to implement.
>
If the delimiting strings are fixed, we can use plain python string methods,
e.g.,
(not tested beyond what you see ;-)
>>> s = "00 noise1 01 noise2 00 target 01 target_mark"
>>> def findit(s, beg='00', end='01', tmk=' target_mark'):
... start = 0
... while True:
... t = s.find(tmk, start)
... if t<0: break
... start = s.rfind(beg, start, t)
... if start<0: break
... e = s.find(end, start, t)
... if e+len(end)==t: # _just_ after
... yield s[start:e+len(end)]
... start = t+len(tmk)
...
>>> list(findit(s))
['00 target 01']
>>> s2 = s + ' garbage noise3 00 almost 01 target_mark 00 success 01
>>> target_mark'
>>> list(findit(s2))
['00 target 01', '00 success 01']
(I didn't enforce exact adjacency the first time, obviously it would be more
efficient
to search for end+tmk instead of tmk and back to beg and forward to end ;-)
If there can be spurious target_marks, and tricky matching spans, additional
logic may be needed.
Too lazy to think about it ;-)
Regards,
Bengt Richter
--
http://mail.python.org/mailman/listinfo/python-list
Re: Regular expression question -- exclude substring
On Monday 07 November 2005 17:31, Kent Johnson wrote:
> James Stroud wrote:
> > On Monday 07 November 2005 16:18, [EMAIL PROTECTED] wrote:
> >>Ya, for some reason your non-greedy "?" doesn't seem to be taking.
> >>This works:
> >>
> >>re.sub('(.*)(00.*?01) target_mark', r'\2', your_string)
> >
> > The non-greedy is actually acting as expected. This is because non-greedy
> > operators are "forward looking", not "backward looking". So the
> > non-greedy finds the start of the first start-of-the-match it comes
> > accross and then finds the first occurrence of '01' that makes the
> > complete match, otherwise the greedy operator would match .* as much as
> > it could, gobbling up all '01's before the last because these match '.*'.
> > For example:
> >
> > py> rgx = re.compile(r"(00.*01) target_mark")
> > py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat
> > 01') ['00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01']
> > py> rgx = re.compile(r"(00.*?01) target_mark")
> > py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat
> > 01') ['00 noise1 01 noise2 00 target 01', '00 dowhat 01']
>
> ??? not in my Python:
> >>> rgx = re.compile(r"(00.*01) target_mark")
> >>> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat
> >>> 01')
>
> ['00 noise1 01 noise2 00 target 01']
>
> >>> rgx = re.compile(r"(00.*?01) target_mark")
> >>> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat
> >>> 01')
>
> ['00 noise1 01 noise2 00 target 01']
>
> Since target_mark only occurs once in the string the greedy and non-greedy
> match is the same in this case.
Somehow my cutting and pasting got messed up. It should be:
py> rgx = re.compile(r"(00.*?01) target_mark")
py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01
target_mark')
['00 noise1 01 noise2 00 target 01', '00 dowhat 01']
py> rgx = re.compile(r"(00.*01) target_mark")
py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01
target_mark')
['00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01']
Sorry about that.
James
--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095
http://www.jamesstroud.com/
--
http://mail.python.org/mailman/listinfo/python-list
Re: Regular expression question -- exclude substring
James Stroud wrote:
> On Monday 07 November 2005 16:18, [EMAIL PROTECTED] wrote:
>
>>Ya, for some reason your non-greedy "?" doesn't seem to be taking.
>>This works:
>>
>>re.sub('(.*)(00.*?01) target_mark', r'\2', your_string)
>
>
> The non-greedy is actually acting as expected. This is because non-greedy
> operators are "forward looking", not "backward looking". So the non-greedy
> finds the start of the first start-of-the-match it comes accross and then
> finds the first occurrence of '01' that makes the complete match, otherwise
> the greedy operator would match .* as much as it could, gobbling up all '01's
> before the last because these match '.*'. For example:
>
> py> rgx = re.compile(r"(00.*01) target_mark")
> py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
> ['00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01']
> py> rgx = re.compile(r"(00.*?01) target_mark")
> py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
> ['00 noise1 01 noise2 00 target 01', '00 dowhat 01']
??? not in my Python:
>>> rgx = re.compile(r"(00.*01) target_mark")
>>> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
['00 noise1 01 noise2 00 target 01']
>>> rgx = re.compile(r"(00.*?01) target_mark")
>>> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
['00 noise1 01 noise2 00 target 01']
Since target_mark only occurs once in the string the greedy and non-greedy
match is the same in this case.
Kent
--
http://mail.python.org/mailman/listinfo/python-list
Re: Regular expression question -- exclude substring
On Monday 07 November 2005 16:18, [EMAIL PROTECTED] wrote:
> Ya, for some reason your non-greedy "?" doesn't seem to be taking.
> This works:
>
> re.sub('(.*)(00.*?01) target_mark', r'\2', your_string)
The non-greedy is actually acting as expected. This is because non-greedy
operators are "forward looking", not "backward looking". So the non-greedy
finds the start of the first start-of-the-match it comes accross and then
finds the first occurrence of '01' that makes the complete match, otherwise
the greedy operator would match .* as much as it could, gobbling up all '01's
before the last because these match '.*'. For example:
py> rgx = re.compile(r"(00.*01) target_mark")
py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
['00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01']
py> rgx = re.compile(r"(00.*?01) target_mark")
py> rgx.findall('00 noise1 01 noise2 00 target 01 target_mark 00 dowhat 01')
['00 noise1 01 noise2 00 target 01', '00 dowhat 01']
My understanding is that backward looking operators are very resource
expensive to implement.
James
--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095
http://www.jamesstroud.com/
--
http://mail.python.org/mailman/listinfo/python-list
Re: Regular expression question -- exclude substring
Ya, for some reason your non-greedy "?" doesn't seem to be taking.
This works:
re.sub('(.*)(00.*?01) target_mark', r'\2', your_string)
--
http://mail.python.org/mailman/listinfo/python-list
Re: Regular expression question -- exclude substring
[EMAIL PROTECTED] wrote: > Hi, > > I'm having trouble extracting substrings using regular expression. Here > is my problem: > > Want to find the substring that is immediately before a given > substring. For example: from > "00 noise1 01 noise2 00 target 01 target_mark", > want to get > "00 target 01" > which is before > "target_mark". > My regular expression > "(00.*?01) target_mark" > will extract > "00 noise1 01 noise2 00 target 01". If there is a character that can't appear in the bit between the numbers then use everything-but-that instead of . - for example if spaces can only appear as you show them, use "(00 [^ ]* 01) target_mark" or "(00 \S* 01) target_mark" Kent -- http://mail.python.org/mailman/listinfo/python-list
