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regexp non-greedy matching bug?

P: n/a
I want to match one or two instances of a pattern in a string.

According to the docs for the 're' module
( http://python.org/doc/current/lib/re-syntax.html ) the '?' qualifier
is greedy by default, and adding a '?' after a qualifier makes it
non-greedy.
The "*", "+", and "?" qualifiers are all greedy...
Adding "?" after the qualifier makes it perform the match in
non-greedy or minimal fashion...


In the following example, though my re is intended to allow for 1 or 2
instinces of 'foo', there are 2 in the string I'm matching. So, I would
expect group(1) and group(3) to both be populated. (When I remove the
conditional match on the 2nd foo, the grouping is as I expect.)

$ python2.4
Python 2.4.1 (#2, Mar 31 2005, 00:05:10)
[GCC 3.3 20030304 (Apple Computer, Inc. build 1666)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
import re
foofoo = re.compile(r'^(foo)(.*?)(foo)?(.*?)$')
foofoo.match(s).group(0) 'foobarbazfoobar' foofoo.match(s).group(1) 'foo' foofoo.match(s).group(2) '' foofoo.match(s).group(3)
foofoo.match(s).group(4) 'barbazfoobar' foofoo = re.compile(r'^(foo)(.*?)(foo)(.*?)$')
foofoo.match(s).group(0) 'foobarbazfoobar' foofoo.match(s).group(1) 'foo' foofoo.match(s).group(2) 'barbaz' foofoo.match(s).group(3) 'foo' foofoo.match(s).group(4) 'bar'

So, is this a bug, or just a problem with my understanding? If it's my
brain that's broken, what's the proper way to do this with regexps?

And, if the above is expected behavior, should I submit a doc bug? It's
clear that the "?" qualifier (applied to the second foo group) is _not_
greedy in this situation.

-John
Dec 4 '05 #1
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8 Replies


P: n/a
My understanding of .*? and its ilk is that they will match as little
as is possible for the rest of the pattern to match, like .* will match
as much as possible. In the first instance, the first (.*?) did not
have to match anything, because all of the rest of the pattern can be
matched 0 or more times. I think that such a situation (non-greedy
operator followed by operators that match 0 or more times) will never
match. However, in the second instance, it has to match something later
on in the string, so it will capture something.

I believe that this is an operator precedence problem (greedy ? losing
to .*?), but is to be expected. So, if this is the case, by all means
it should be added in a note to the docs.

However, I am not a regular expression expert, so my analysis of the
situation may be well off the mark.

Dec 4 '05 #2

P: n/a
John Hazen <jo**@hazen.net> writes:
I want to match one or two instances of a pattern in a string.
Then you should be using the split() method of the match object on the
pattern in question.
According to the docs for the 're' module
( http://python.org/doc/current/lib/re-syntax.html ) the '?' qualifier
is greedy by default, and adding a '?' after a qualifier makes it
non-greedy.
The "*", "+", and "?" qualifiers are all greedy...
Adding "?" after the qualifier makes it perform the match in
non-greedy or minimal fashion...
The thing to understand is that regular expressions are *search*
functions, that return the first parsing that matches. They search a
space of possible matches to each term in the expression. If some term
fails to match, the preceeding term goes on to its next match, and you
try again. The "greedy" vs. "non-greedy" describes the order that the
term in question tries matches. If it's greedy, it will try the
longest possible match first. If it's non-greedy, it'll try the
shortest possible match first.
$ python2.4
Python 2.4.1 (#2, Mar 31 2005, 00:05:10)
[GCC 3.3 20030304 (Apple Computer, Inc. build 1666)] on darwin
Type "help", "copyright", "credits" or "license" for more information. import re
foofoo = re.compile(r'^(foo)(.*?)(foo)?(.*?)$')
foofoo.match(s).group(0) 'foobarbazfoobar' foofoo.match(s).group(1) 'foo' foofoo.match(s).group(2) '' foofoo.match(s).group(3)
foofoo.match(s).group(4) 'barbazfoobar'
First, this pattern doesn't look for one or two instances of "foo" in
a string. It looks for a string that starts with "foo" and maybe has a
second "foo" in it as well.

Here's what it does:

^ must match the beginning of the string (BTW, you can get the same
behavior by leaving off the ^ and using search instead of match).

(foo) must match the first foo.

(.*?) is non-greedy, so it tries the shortest thing that matches, the
empty string.

(foo)? matches what's next in the string by matching 0 occurrences of
(foo).

(.*?) tries the empty string as well.

$ fails to match the end of the string, so we backtrack, and the
second (.*?) tries again, adding a character. This keeps on until the
(.*?) matches the rest of the string and the $ succeeds.

Here, the only pattern that gets retried is the last one. It keeps
adding characters until it swallows the remainder of the string. This
is exactly what's expected. You can't change this behavior with
greedy/non-greedy, as that last term will always try searching the
entire string before it fails and the preceeding "(foo)?" gets to try
something else.
foofoo = re.compile(r'^(foo)(.*?)(foo)(.*?)$')
foofoo.match(s).group(0) 'foobarbazfoobar' foofoo.match(s).group(1) 'foo' foofoo.match(s).group(2) 'barbaz' foofoo.match(s).group(3) 'foo' foofoo.match(s).group(4) 'bar'


This time, the second (foo) keeps failing, forcing the first (.*?) to
add characters until the (foo) succeeds.
So, is this a bug, or just a problem with my understanding? If it's my
brain that's broken, what's the proper way to do this with regexps?


To do what you said you want to do, you want to use the split method:

foo = re.compile('foo')
if 2 <= len(foo.split(s)) <= 3:
print "We had one or two 'foo's"

As the founder of SPARE, I'd say the proper way to do this is with
string methods:

if 2 <= len(s.split('foo')) <= 3:
print "We had one or two 'foo's"

Where the two split's will produce exactly the same things.

You might be able to match "starts with 'foo' and has at most one
other 'foo'" with lookahead patternds, but I'm not going to go into
that. If you really need the string to start with foo, I'd just use
"s.startswith('foo')" to check it.

<mike
--
Mike Meyer <mw*@mired.org> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
Dec 4 '05 #3

P: n/a
[Mike Meyer]
The thing to understand is that regular expressions are *search*
functions, that return the first parsing that matches. They search a
space of possible matches to each term in the expression. If some term
fails to match, the preceeding term goes on to its next match, and you
try again. The "greedy" vs. "non-greedy" describes the order that the
term in question tries matches. If it's greedy, it will try the
longest possible match first. If it's non-greedy, it'll try the
shortest possible match first.
That's a good explanation. Thanks.

[John]
I want to match one or two instances of a pattern in a string.
> foofoo = re.compile(r'^(foo)(.*?)(foo)?(.*?)$')
[Mike] First, this pattern doesn't look for one or two instances of "foo" in
a string. It looks for a string that starts with "foo" and maybe has a
second "foo" in it as well.
Right. In simplifying the expression for public consumption, one of the
terms I dropped was r'^.*?(foo)...'.
To do what you said you want to do, you want to use the split method:

foo = re.compile('foo')
if 2 <= len(foo.split(s)) <= 3:
print "We had one or two 'foo's"
Well, this would solve my dumbed down example, but each foo in the
original expression was a stand-in for a more complex term. I was using
match groups to extract the parts of the match that I wanted. Here's an
example (using Tim's correction) that actually demonstrates what I'm
doing:
s = 'zzzfoo123barxxxfoo456baryyy'
s2 = 'zzzfoo123barxxxfooyyy'
foobar2 = re.compile(r'^.*?foo(\d+)bar(.*foo(\d+)bar)?.*$')
print foobar2.match(s).group(1) 123 print foobar2.match(s).group(3) 456 print foobar2.match(s2).group(1) 123 print foobar2.match(s2).group(3) None

Looking at re.split, it doesn't look like it returns the actual matching
text, so I don't think that fits my need.
As the founder of SPARE...


Hmm, not a very effective name. A google search didn't fing any obvious
hits (even after adding the "python" qualifier, and removing "spare time"
and "spare parts" hits). (I couldn't find it off your homepage,
either.)

Thanks for your help. If you have any suggestions about a non-re way to
do the above, I'd be interested.

-John
Dec 4 '05 #4

P: n/a
Mike Meyer wrote:
^ must match the beginning of the string (BTW, you can get the same
behavior by leaving off the ^ and using search instead of match).


that's backwards, isn't it? using ^ with match is usually pointless (since
match only looks at the first position anyway), and using ^ with search
is also usually pointless...

(the only time you really need ^ is in certain multiline searches; depending
on what flags you use, ^ can match not only at the beginning of a string,
but also after each newline character in the target string)

</F>

Dec 4 '05 #5

P: n/a
John Hazen <jo**@hazen.net> writes:
To do what you said you want to do, you want to use the split method:

foo = re.compile('foo')
if 2 <= len(foo.split(s)) <= 3:
print "We had one or two 'foo's"
Well, this would solve my dumbed down example, but each foo in the
original expression was a stand-in for a more complex term.


That actually doesn't matter. Just replace 'foo' with your more
complex term.
foo2 = re.compile(r'foo(\d+)bar') I was using
match groups to extract the parts of the match that I wanted. Here's an
example (using Tim's correction) that actually demonstrates what I'm
doing:
s = 'zzzfoo123barxxxfoo456baryyy'
s2 = 'zzzfoo123barxxxfooyyy'
foobar2 = re.compile(r'^.*?foo(\d+)bar(.*foo(\d+)bar)?.*$')
print foobar2.match(s).group(1) 123 print foobar2.match(s).group(3) 456 foo2.split('zzzfoo123barxxxfoo456baryyy') ['zzz', '123', 'xxx', '456', 'yyy'] print foobar2.match(s2).group(1) 123 print foobar2.match(s2).group(3) None foo2.split('zzzfoo123barxxxfooyyy')

['zzz', '123', 'xxxfooyyy']
Looking at re.split, it doesn't look like it returns the actual matching
text, so I don't think that fits my need.


split() returns the text matched by groups in the pattern used to do
the split, and is documented as doing so. The solution you gave
doesn't return "the actual matching text" for the instances, but just
the text in the groups inside that text - which is exactly what
split() returns.

While on that topic, I'll note that the solution Tim gave you doesn't
solve the problem as I originally understood it, either. You said you
wanted to match one or two instances, which I read as only one or two
instances, so that more than two instances would be treated as a
failure. On rereading it, I can see where I was wrong.
As the founder of SPARE...

Hmm, not a very effective name. A google search didn't fing any obvious
hits (even after adding the "python" qualifier, and removing "spare time"
and "spare parts" hits). (I couldn't find it off your homepage,
either.)


That's the Society for the Prevention of Abuse of Regular
Expressions. One of these days, my proof-reader will get back to me
and I'll add a link about it to my home page.

<mike
--
Mike Meyer <mw*@mired.org> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
Dec 4 '05 #6

P: n/a
"Fredrik Lundh" <fr*****@pythonware.com> writes:
Mike Meyer wrote:
^ must match the beginning of the string (BTW, you can get the same
behavior by leaving off the ^ and using search instead of match).

that's backwards, isn't it? using ^ with match is usually pointless (since
match only looks at the first position anyway), and using ^ with search
is also usually pointless...


Yup, you're right. The '^' was redundant in the OP's post.

<mike
--
Mike Meyer <mw*@mired.org> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
Dec 4 '05 #7

P: n/a
In article <ma***************************************@python. org>,
Fredrik Lundh <fr*****@pythonware.com> wrote:

that's backwards, isn't it? using ^ with match is usually pointless (since
match only looks at the first position anyway), and using ^ with search
is also usually pointless...


While you're technically correct, I've been bitten too many times by
forgetting whether to use match() or search(). I've fixed that problem
by choosing to always use search() and combine with ^ as appropriate.
--
Aahz (aa**@pythoncraft.com) <*> http://www.pythoncraft.com/

"Don't listen to schmucks on USENET when making legal decisions. Hire
yourself a competent schmuck." --USENET schmuck (aka Robert Kern)
Dec 4 '05 #8

P: n/a
Aahz wrote:
While you're technically correct, I've been bitten too many times by
forgetting whether to use match() or search(). I've fixed that problem
by choosing to always use search() and combine with ^ as appropriate.


that's a bit suboptimal, though, at least for cases where failed matches
are rather common:

C:\>timeit -s "import re; p = re.compile('b')" "p.match('a'*100)"
100000 loops, best of 3: 6.14 usec per loop

C:\>timeit -s "import re; p = re.compile('^b')" "p.match('a'*100)"
100000 loops, best of 3: 6.25 usec per loop

C:\>timeit -s "import re; p = re.compile('^b')" "p.search('a'*100)"
100000 loops, best of 3: 15.4 usec per loop

(afaik, search doesn't have any heuristics for figuring out if it can skip
the search, so it'll check ^ against all available positions)

on the other hand, benchmarking RE:s always results in confusing
results:

C:\>timeit -s "import re; p = re.compile('b')" "p.search('a'*100)"
100000 loops, best of 3: 4.32 usec per loop

(should this really be *faster* than match for this case ?)

</F>

Dec 5 '05 #9

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