Bug in re.findall?

[Marcin Krol]

Hello everyone,

Is there a bug in re.findall in Python 2.4? See:

subnetlist="192 .168.100.0 , 192.168.101.0"
ipre=re.compile ("([0-9]{1,3}\.){3}[0-9]{1,3}")

>>ipre.findall( subnetlist)

['100.', '101.']


But:

a=ipre.finditer (subnetlist)

>>a.next().grou p()

'192.168.100.0'

>>a.next().grou p()

'192.168.101.0'

>>a.next().grou p()

Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration

Also:

>>ipre.search(s ubnetlist).grou p()

'192.168.100.0'

Is this a bug or am I doing smth wrong?

[dwahli@gmail.com]

Re: Bug in re.findall?

On Jul 4, 12:33 pm, Marcin Krol <mrk...@gmail.c omwrote:

Hello everyone,
>
Is there a bug in re.findall in Python 2.4? See:
>
subnetlist="192 .168.100.0 , 192.168.101.0"
ipre=re.compile ("([0-9]{1,3}\.){3}[0-9]{1,3}")
>
 >>ipre.findall( subnetlist)
>
['100.', '101.']
>
But:
>
a=ipre.finditer (subnetlist)
>
 >>a.next().grou p()
'192.168.100.0'
 >>a.next().grou p()
'192.168.101.0'
 >>a.next().grou p()
Traceback (most recent call last):
   File "<stdin>", line 1, in ?
StopIteration
>
Also:
>
 >>ipre.search(s ubnetlist).grou p()
'192.168.100.0'
>
Is this a bug or am I doing smth wrong?

Look strange but match the Python documentation for 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"

you must use this RE for your example to match what you expect:
ipre=re.compile ("(?:[0-9]{1,3}\.){3}[0-9]{1,3}")

but using re.finditer is IMHO better.

Cheer,
Dom

[Peter Otten]

Re: Bug in re.findall?

Marcin Krol wrote:

Hello everyone,
>
Is there a bug in re.findall in Python 2.4? See:
>
subnetlist="192 .168.100.0 , 192.168.101.0"
ipre=re.compile ("([0-9]{1,3}\.){3}[0-9]{1,3}")
>

>>ipre.findall( subnetlist)

>
['100.', '101.']
>
>
But:
>
a=ipre.finditer (subnetlist)
>

>>a.next().grou p()

'192.168.100.0'

>>a.next().grou p()

'192.168.101.0'

>>a.next().grou p()

Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration
>
Also:
>

>>ipre.search(s ubnetlist).grou p()

'192.168.100.0'
>
Is this a bug or am I doing smth wrong?

From the doc:

"""
findall( pattern, string[, flags])
Return a list of all non-overlapping matches of pattern in string. 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.
"""

So findall()'s behaviour changes depending on the number of explicit groups

None:

[m.group() for m in re.finditer(... )]

One:

[m.group(1) for m in re.finditer(... )]

More than one:

[m.groups() for m in re.finditer(... )]

all in accordance with the documentation.

Peter