From: John Bond<li...@asd-group.com>
Subject: Re: Why this result with the re module
To: "Yingjie Lan"<lany...@yahoo.com>
Cc: python-list@python.org
Date: Tuesday, November 2, 2010, 8:09 PM
On 2/11/2010 12:19 PM, Yingjie Lan
wrote:
From: John Bond<li...@asd-group.com>
Subject: Re: Why this result with the re module
Firstly, thanks a lot for your patient explanation.
this time I have understood all your points
perfectly.
Secondly, I'd like to clarify some of my points,
which
did not get through because of my poor presentation.
I suggested findall return a tuple of
re.MatchObject(s),
with each MatchObject instance representing a match.
This is consistent with the re.match() function
anyway.
And it will eliminate the need of returning tuples,
and it is much more precise and information rich.
If that's not possible, and a tuple must be returned,
then the whole match (not just subgroups) should
always be included as the first element in the tuple,
as that's group(0) or '\0'. Less surprise would
arise.
Finally, it seems to me the algo for findall is
WRONG.
To re.findall('(.a.)*', 'Mary has a lamb'),
by reason of greediness of '*', and the requirement
of non-overlapping, it should go like this
(suppose an '' is at the beginning and at the end,
and between two consecutive characters there is
one and only one empty string ''. To show the
match of empty strings clearly,
I am concatenating each repeated match below):
Steps for re.findall('(.a.)*', 'Mary has a lamb'):
step 1: Match '' + 'Mar' + '' (gready!)
step 2: skip 'y'
step 3: Match ''
step 4: skip ' '
step 5: Match ''+'has'+' a '+'lam'+'' (greedy!)
step 6: skip 'b'
step 7: Match ''
So there should be exactly 4 matches in total:
'Mar', '', 'has a lam', ''
Also, the matches above shows
that if a repeated subgroup only captures
the last match, the subgroup (.a.)*
should always capture '' here (see steps
1, 3, 5, 7) above.
Yet the execution in Python results in 6 matches!
And, the capturing subgroup with repetition
sometimes got the wrong guy.
So I believe the algorithm for findall must be WRONG.
Regards,
Yingjie
At a guess, I'd say what is happening is something like
this:
Steps for re.findall('(.a.)*', 'Mary has a lamb'):
step 1: Match 'Mar' at string index 0
step 2: Match '' at string index 3 (before 'y')
step 3: skip 'y'
step 4: Match '' at string index 4 (before ' ')
step 5: skip ' '
step 6: Match 'has a lam' at string index 5
step 7: Match '' at string index 14 (before 'b')
step 8: skip 'b'
step 9: Match '' at string index 15 (before EOS)
<EOS reached>
matches: ('Mar', '', '', 'has a lam', '', '')
returns: ['Mar', '', '', 'lam', '', ''] (*)
(*) "has a " lost due to not being last repetition at that
match point
Which seems about right to me! Greediness has nothing to do
with it, except that it causes 'has a lam' to be matched in
one match, instead of as three separate matches (of 'has', '
a ' and 'lam').
