Re: one more question on regex
2016-01-22 23:47 GMT+01:00 mg : > Il Fri, 22 Jan 2016 21:10:44 +0100, Vlastimil Brom ha scritto: > >> [...] > > You explanation of re.findall() results is correct. My point is that the > documentation states: > > re.findall(pattern, string, flags=0) > Return all non-overlapping matches of pattern in string, as a list of > strings > > and this is not what re.findall does. IMHO it should be more reasonable > to get back the whole matches, since this seems to me the most useful > information for the user. In any case I'll go with finditer, that returns > in match object all the infos that anyone can look for. > -- > https://mail.python.org/mailman/listinfo/python-list Hi, I don't know the reasoning for this special behaviour of findall, but it seems to be documented explicitly: https://docs.python.org/3/library/re.html#re.findall "... If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. finditer is clearly much more robust for general usage. I only use findall for quick one-line tests (and there one has to account for this specificities - either by using non capturing groups or enclosing the whole pattern in a "main" group and use the first items in the resulting tuples. vbr -- https://mail.python.org/mailman/listinfo/python-list
Re: one more question on regex
Il Fri, 22 Jan 2016 21:10:44 +0100, Vlastimil Brom ha scritto:
> 2016-01-22 16:50 GMT+01:00 mg :
>> Il Fri, 22 Jan 2016 15:32:57 +, mg ha scritto:
>>
>>> python 3.4.3
>>>
>>> import re re.search('(ab){2}','abzzabab')
>>> <_sre.SRE_Match object; span=(4, 8), match='abab'>
>>>
>> re.findall('(ab){2}','abzzabab')
>>> ['ab']
>>>
>>> Why for search() the match is 'abab' and for findall the match is
>>> 'ab'?
>>
>> finditer seems to be consistent with search:
>> regex = re.compile('(ab){2}')
>>
>> for match in regex.finditer('abzzababab'):
>> print ("%s: %s" % (match.start(), match.span() ))
>> ...
>> 4: (4, 8)
>>
>> -- https://mail.python.org/mailman/listinfo/python-list
>
> Hi,
> as was already pointed out, findall "collects" the content of the
> capturing groups (if present), rather than the whole matching text;
>
> for repeated captures the last content of them is taken discarding the
> previous ones; cf.:
>
re.findall('(?i)(a)x(b)+','axbB')
> [('a', 'B')]
> (for multiple capturing groups in the pattern, a tuple of captured parts
> are collected)
>
> or with your example with differenciated parts of the string using
> upper/lower case:
re.findall('(?i)(ab){2}','aBzzAbAB')
> ['AB']
> hth,
>vbr
You explanation of re.findall() results is correct. My point is that the
documentation states:
re.findall(pattern, string, flags=0)
Return all non-overlapping matches of pattern in string, as a list of
strings
and this is not what re.findall does. IMHO it should be more reasonable
to get back the whole matches, since this seems to me the most useful
information for the user. In any case I'll go with finditer, that returns
in match object all the infos that anyone can look for.
--
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Re: one more question on regex
2016-01-22 16:50 GMT+01:00 mg :
> Il Fri, 22 Jan 2016 15:32:57 +, mg ha scritto:
>
>> python 3.4.3
>>
>> import re re.search('(ab){2}','abzzabab')
>> <_sre.SRE_Match object; span=(4, 8), match='abab'>
>>
> re.findall('(ab){2}','abzzabab')
>> ['ab']
>>
>> Why for search() the match is 'abab' and for findall the match is 'ab'?
>
> finditer seems to be consistent with search:
> regex = re.compile('(ab){2}')
>
> for match in regex.finditer('abzzababab'):
> print ("%s: %s" % (match.start(), match.span() ))
> ...
> 4: (4, 8)
>
> --
> https://mail.python.org/mailman/listinfo/python-list
Hi,
as was already pointed out, findall "collects" the content of the
capturing groups (if present), rather than the whole matching text;
for repeated captures the last content of them is taken discarding the
previous ones; cf.:
>>> re.findall('(?i)(a)x(b)+','axbB')
[('a', 'B')]
>>>
(for multiple capturing groups in the pattern, a tuple of captured
parts are collected)
or with your example with differenciated parts of the string using
upper/lower case:
>>> re.findall('(?i)(ab){2}','aBzzAbAB')
['AB']
>>>
hth,
vbr
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Re: one more question on regex
Il Fri, 22 Jan 2016 15:32:57 +, mg ha scritto:
> python 3.4.3
>
> import re re.search('(ab){2}','abzzabab')
> <_sre.SRE_Match object; span=(4, 8), match='abab'>
>
re.findall('(ab){2}','abzzabab')
> ['ab']
>
> Why for search() the match is 'abab' and for findall the match is 'ab'?
finditer seems to be consistent with search:
regex = re.compile('(ab){2}')
for match in regex.finditer('abzzababab'):
print ("%s: %s" % (match.start(), match.span() ))
...
4: (4, 8)
--
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Re: one more question on regex
mg wrote:
> python 3.4.3
>
> import re
> re.search('(ab){2}','abzzabab')
> <_sre.SRE_Match object; span=(4, 8), match='abab'>
>
re.findall('(ab){2}','abzzabab')
> ['ab']
>
> Why for search() the match is 'abab' and for findall the match is 'ab'?
I suppose someone thought it was convenient for findall to return the
explicit groups if there are any. If you want the whole match aka group(0)
you can get that with
>>> re.findall('(?:ab){2}','abzzabab')
['abab']
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one more question on regex
python 3.4.3
import re
re.search('(ab){2}','abzzabab')
<_sre.SRE_Match object; span=(4, 8), match='abab'>
>>> re.findall('(ab){2}','abzzabab')
['ab']
Why for search() the match is 'abab' and for findall the match is 'ab'?
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Re: Question on regex
Prabhu Gurumurthy schrieb:
> to fix this problem, i used negative lookahead with ip pattern:
> so the ip pattern now changes to:
> \d{1,3}(\.\d{1,3}){3}(?!/\d+)
>
> now the problem is 10.150.100.0 works fine, 10.100.4.64 subnet gets
> matched with ip pattern with the following result:
>
> 10.100.4.6
>
> Is there a workaround for this or what should change in ip regex pattern.
>
I think what you want is that neither /d+ nor another digit nor a . follows:
\d{1,3}(\.\d{1,3}){3}(?!(/\d)|\d|\.)
This way 10.0.0.1234 won't be recognized as ip. Neither will 23.12.
which could be a problem if an ip is at the end of a sentence, so you
might want to omit that.
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Question on regex
Hello all -
I have a file which has IP address and subnet number and I use regex to extract
the IP separately from subnet.
pattern used for IP: \d{1,3}(\.\d{1,3}){3}
pattern used for subnet:((\d{1,3})|(\d{1,3}(\.\d{1,3}){1,3}))/(\d{1,2})
so I have list of ip/subnets strewn around like this
10.200.0.34
10.200.4.5
10.178.9.45
10.200/22
10.178/16
10.100.4.64/26,
10.150.100.0/28
10/8
with that above examples:
ip regex pattern works for all IP address
subnet regex pattern works for all subnets
problem now is ip pattern also matches the last 2 subnet numbers, because it
falls under ip regex.
to fix this problem, i used negative lookahead with ip pattern:
so the ip pattern now changes to:
\d{1,3}(\.\d{1,3}){3}(?!/\d+)
now the problem is 10.150.100.0 works fine, 10.100.4.64 subnet gets matched
with ip pattern with the following result:
10.100.4.6
Is there a workaround for this or what should change in ip regex pattern.
python script:
#!/usr/bin/env python
import re, sys
fh = 0
try:
fh = open(sys.argv[1], "r")
except IOError, message:
print "cannot open file: %s" %message
else:
for lines in fh.readlines():
lines = lines.strip()
pattIp = re.compile("(\d{1,3}(\.\d{1,3}){3})(?!/\d+)")
pattNet = re.compile("((\d{1,3})|(\d{1,3}(\.\d{1,3}){1,3}))/(\d{1,2})")
match = pattIp.search(lines)
if match is not None:
print "ipmatch: %s" %match.groups()[0]
match = pattNet.search(lines)
if match is not None:
print "subnet: %s" %match.groups()[0]
fh.close()
output with that above ip/subnet in a file
ipmatch: 10.200.0.34
ipmatch: 10.200.4.5
ipmatch: 10.178.9.45
subnet: 10.200
subnet: 10.178
ipmatch: 10.100.4.6
subnet: 10.100.4.64
subnet: 10.150.100.0
subnet: 10
TIA
Prabhu
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n:Gurumurthy;Prabhu
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title:Network Engineer
tel;work:(650) 357 8770 x134
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