Regular expression to match a #

[Tom Deco]

Hi,

I'm trying to use a regular expression to match a string containing a #
(basically i'm looking for #include ...)

I don't seem to manage to write a regular expression that matches this.

My (probably to naive) approach is: p = re.compile(r'\b#include\b)
I also tried p = re.compile(r'\b\#include\b) in a futile attempt to use
a backslash as escape character before the #
None of the above return a match for a string like "#include <stdio>".

I know a # is used for comments, hence my attempt to escape it...

Any suggestion on how to get a regular expression to find a #?

Thanks

[Dan]

> My (probably to naive) approach is: p = re.compile(r'\b#include\b)

I think your problem is the \b at the beginning. \b matches a word break
(defined as \w\W or \W\w). There would only be a word break before the #
if the preceding character were a \w (that is, [A-Za-z0-9_], and maybe
some other characters depending on your locale).

However, the \b after the "include" is exactly what you want.

--
I had picked out the theme of the baby's room and done other
things. I decided to let Jon have this.
- Jamie Cusack (of the Netherlands), whose husband Jon
finally talked her into letting him name their son Jon 2.0

[Tom Deco]

Thanks,
That did the trick...

[Duncan Booth]

Dan wrote:

My (probably to naive) approach is: p = re.compile(r'\b#include\b)

I think your problem is the \b at the beginning. \b matches a word break
(defined as \w\W or \W\w). There would only be a word break before the #
if the preceding character were a \w (that is, [A-Za-z0-9_], and maybe
some other characters depending on your locale).

However, the \b after the "include" is exactly what you want.

So the OP probably wanted '\B' the exact opposite of '\b' for the start of
the string, i.e. only match the # if it is NOT preceded by a wordbreak.

Alternatively for C style #includes search for r'^\s*#\s*include\b'.

[John Machin]

Tom Deco wrote:

Hi,

I'm trying to use a regular expression to match a string containing a #
(basically i'm looking for #include ...)

I don't seem to manage to write a regular expression that matches this.

My (probably to naive) approach is: p = re.compile(r'\b#include\b)
I also tried p = re.compile(r'\b\#include\b) in a futile attempt to use
a backslash as escape character before the #
None of the above return a match for a string like "#include <stdio>".

I know a # is used for comments, hence my attempt to escape it...

Any suggestion on how to get a regular expression to find a #?

Thanks

You definitely shouldn't have the first \b -- match() works only at the
beginning of the target string, so it is impossible for there to be a
word boundary just before the "#".

You probably shouldn't have the second \b.

You probably should read section A12 of K&R2.

You probably should be using a parser, but if you persist in using
regular expressions:

(a) read the manual.

(b) try something like this:

pat1 = re.compile(r'\s*#\s*include\s*<\s*([^>\s]+)\s*>\s*$')
pat1.match(" # include < fubar.h > ").group(1)

'fubar.h'

N.B. this is based the assumption that sane programmers don't have
whitespace embedded in the names of source files ;-)

HTH,
John

[John Machin]

Duncan Booth wrote:

Dan wrote:

My (probably to naive) approach is: p = re.compile(r'\b#include\b)

I think your problem is the \b at the beginning. \b matches a word break
(defined as \w\W or \W\w). There would only be a word break before the #
if the preceding character were a \w (that is, [A-Za-z0-9_], and maybe
some other characters depending on your locale).

However, the \b after the "include" is exactly what you want.

So the OP probably wanted '\B' the exact opposite of '\b' for the start of
the string, i.e. only match the # if it is NOT preceded by a wordbreak.

Alternatively for C style #includes search for r'^\s*#\s*include\b'.

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which Python's
re is NOT] it could be much worse. So please don't tell newbies to
search for r'^something'.

[Jeff Schwab]

John Machin wrote:

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which Python's
re is NOT] it could be much worse. So please don't tell newbies to
search for r'^something'.

How else would you match the beginning of a line in a multi-line string?

[Duncan Booth]

John Machin wrote:

Alternatively for C style #includes search for r'^\s*#\s*include\b'.

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which
Python's re is NOT] it could be much worse. So please don't tell
newbies to search for r'^something'.

Search for r'^something' is always better than searching for r'something'
when the spec requires the search to match only at the start of a line (on
the principle that code that works is better than code which doesn't).

It appears that this may be something the original poster wanted, so I
stand by my suggestion.

[Aahz]

In article <42********@news.eftel.com>,
John Machin <sj******@lexicon.net> wrote:

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which Python's
re is NOT] it could be much worse. So please don't tell newbies to
search for r'^something'.

You're somehow getting mixed up in thinking that "^" is some kind of
"not" operator -- it's the start of line anchor in this context.
--
Aahz (aa**@pythoncraft.com) <*> http://www.pythoncraft.com/

The way to build large Python applications is to componentize and
loosely-couple the hell out of everything.

[John Machin]

Jeff Schwab wrote:

John Machin wrote:

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which
Python's re is NOT] it could be much worse. So please don't tell
newbies to search for r'^something'.

How else would you match the beginning of a line in a multi-line string?

I beg your pardon -- I should have qualified that:

"""
So please don't tell newbies to search for r'^something' when match of
r'something' does the job.
"""

[John Machin]

Duncan Booth wrote:

John Machin wrote:

Alternatively for C style #includes search for r'^\s*#\s*include\b'.

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which
Python's re is NOT] it could be much worse. So please don't tell
newbies to search for r'^something'.

Search for r'^something' is always better than searching for r'something'
when the spec requires the search to match only at the start of a line (on
the principle that code that works is better than code which doesn't).

It appears that this may be something the original poster wanted, so I
stand by my suggestion.

We could well be lost in a semantic fog where at least one of us is
using "match" to mean "the match() method" and at least one of us is
using match to mean soemthing like "the outcome of using a search()
method [or a match() method]".

So I'll stand by my suggestion, too.

[John Machin]

Aahz wrote:

In article <42********@news.eftel.com>,
John Machin <sj******@lexicon.net> wrote:

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which Python's
re is NOT] it could be much worse. So please don't tell newbies to
search for r'^something'.

You're somehow getting mixed up in thinking that "^" is some kind of
"not" operator -- it's the start of line anchor in this context.

I can't imagine where you got that idea from.

If I change "[which Python's re is NOT]" to "[Python's re's search() is
not dopey]", does that help you?

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).

[Devan L]

John Machin wrote:

Aahz wrote:

In article <42********@news.eftel.com>,
John Machin <sj******@lexicon.net> wrote:

Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which Python's
re is NOT] it could be much worse. So please don't tell newbies to
search for r'^something'.

You're somehow getting mixed up in thinking that "^" is some kind of
"not" operator -- it's the start of line anchor in this context.

I can't imagine where you got that idea from.

If I change "[which Python's re is NOT]" to "[Python's re's search() is
not dopey]", does that help you?

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).

I don't see much difference.
Python 2.4.1 (#65, Mar 30 2005, 09:13:57) [MSC v.1310 32 bit (Intel)]
on win32
Type "copyright", "credits" or "license()" for more information.

************************************************** **************
Personal firewall software may warn about the connection IDLE
makes to its subprocess using this computer's internal loopback
interface. This connection is not visible on any external
interface and no data is sent to or received from the Internet.
************************************************** **************

IDLE 1.1.1

import timeit
t1 = timeit.Timer('re.search("^\w"," will not work")','import re')
t1.timeit() 34.938577109660628 t2 = timeit.Timer('re.match("\w"," will not work")','import re')
t2.timeit() 31.381461330979164 3.0/1000000 3.0000000000000001e-006 t1.timeit() 35.282282524734228 t2.timeit()

31.403153752781463

~4 second difference after a million times through seems to be trivial.
Then again, I haven't tested it for larger patterns and strings.

[John Machin]

Devan L wrote:

John Machin wrote:

Aahz wrote:

In article <42********@news.eftel.com>,
John Machin <sj******@lexicon.net> wrote:
Search for r'^something' can never be better/faster than match for
r'something', and with a dopey implementation of search [which Python's
re is NOT] it could be much worse. So please don't tell newbies to
search for r'^something'.
You're somehow getting mixed up in thinking that "^" is some kind of
"not" operator -- it's the start of line anchor in this context.

I can't imagine where you got that idea from.

If I change "[which Python's re is NOT]" to "[Python's re's search() is
not dopey]", does that help you?

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).

I don't see much difference.

and I didn't expect that you would -- like I wrote above: "Python's re's
search() is not dopey".

[Devan L]

John Machin wrote:

Devan L wrote:

John Machin wrote:

Aahz wrote:

In article <42********@news.eftel.com>,
John Machin <sj******@lexicon.net> wrote:
>Search for r'^something' can never be better/faster than match for
>r'something', and with a dopey implementation of search [which Python's
>re is NOT] it could be much worse. So please don't tell newbies to
>search for r'^something'.
You're somehow getting mixed up in thinking that "^" is some kind of
"not" operator -- it's the start of line anchor in this context.

I can't imagine where you got that idea from.

If I change "[which Python's re is NOT]" to "[Python's re's search() is
not dopey]", does that help you?

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).

I don't see much difference.

and I didn't expect that you would -- like I wrote above: "Python's re's
search() is not dopey".

Your wording makes it hard to distinguish what exactly is "dopey".

[John Machin]

John Machin wrote:

Devan L wrote:

John Machin wrote:

Aahz wrote:

In article <42********@news.eftel.com>,
John Machin <sj******@lexicon.net> wrote:
> Search for r'^something' can never be better/faster than match for
> r'something', and with a dopey implementation of search [which
> Python's
> re is NOT] it could be much worse. So please don't tell newbies to
> search for r'^something'.

You're somehow getting mixed up in thinking that "^" is some kind of
"not" operator -- it's the start of line anchor in this context.
I can't imagine where you got that idea from.

If I change "[which Python's re is NOT]" to "[Python's re's search() is
not dopey]", does that help you?

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).

I don't see much difference.

and I didn't expect that you would -- like I wrote above: "Python's re's
search() is not dopey".

*ahem*

C:\junk>python
Python 2.4.1 (#65, Mar 30 2005, 09:13:57) [MSC v.1310 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information.

import timeit
t1 = timeit.Timer('re.search("^\w"," will not work")','import re')
t2 = timeit.Timer('re.match("\w"," will not work")','import re')
t3 = timeit.Timer('obj(" will not work")','import re;obj=re.compile("^\w").s
earch') t4 = timeit.Timer('obj(" will not work")','import re;obj=re.compile("\w").ma
tch') t5 = timeit.Timer('obj(" will not work qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq")'
,'import re;obj=re.compile("^\w").search') t6 = timeit.Timer('obj(" will not work qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq")'
,'import re;obj=re.compile("\w").match') ["%.3f" % t.timeit() for t in t1, t2, t3, t4] ['5.510', '4.835', '1.588', '1.178'] ["%.3f" % t.timeit() for t in t1, t2, t3, t4] ['5.512', '4.808', '1.584', '1.170']

Observation: factoring out the compile step makes the difference much
more apparent.
["%.3f" % t.timeit() for t in t3, t4, t5, t6] ['1.578', '1.175', '2.283', '1.174'] ["%.3f" % t.timeit() for t in t3, t4, t5, t6] ['1.582', '1.179', '2.284', '1.172']

Conclusion: search time depends on length of searched string.

Meta-conclusion: Either I have to retract my
based-on-hope-rather-than-on-experimentation assertion, or redefine "not
dopey" to mean "surely nobody would search for ^x when match x would do,
so it would be dopey to optimise re for that" :-)

So, back to the original point:

If re.match("something") does the job you want, don't use
re.search("^something") instead.

[John Machin]

Devan L wrote:

John Machin wrote:

Devan L wrote:

John Machin wrote:
Aahz wrote:
>In article <42********@news.eftel.com>,
>John Machin <sj******@lexicon.net> wrote:
>
>
>
>>Search for r'^something' can never be better/faster than match for
>>r'something', and with a dopey implementation of search [which Python's
>>re is NOT] it could be much worse. So please don't tell newbies to
>>search for r'^something'.
>
>
>You're somehow getting mixed up in thinking that "^" is some kind of
>"not" operator -- it's the start of line anchor in this context.

I can't imagine where you got that idea from.

If I change "[which Python's re is NOT]" to "[Python's re's search() is
not dopey]", does that help you?

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).
I don't see much difference.

and I didn't expect that you would -- like I wrote above: "Python's re's
search() is not dopey".

Your wording makes it hard to distinguish what exactly is "dopey".

"""
dopey implementations of search() (which apply match() at offsets 0, 1,
2, .....).
"""

The "dopiness" is that the ^ operator means that the pattern cannot
possibly match starting at 1, 2, 3, etc but a non-optimised search will
not recognise that and will try all possibilities, so the failing case
takes time dependant on the length of the string.

[Bryan Olson]

John Machin wrote:
[...]

Observation: factoring out the compile step makes the difference much
more apparent.

>>> ["%.3f" % t.timeit() for t in t3, t4, t5, t6] ['1.578', '1.175', '2.283', '1.174'] >>> ["%.3f" % t.timeit() for t in t3, t4, t5, t6] ['1.582', '1.179', '2.284', '1.172'] >>>

To make it even more apparent, try:

import re
import profile

startsz = re.compile('^z')

for s in ('x' * 1000, 'x' * 100000, 'x'*10000000):
profile.run('startsz.search(s)')

Profile report is below.

Conclusion: search time depends on length of searched string.

Meta-conclusion: Either I have to retract my
based-on-hope-rather-than-on-experimentation assertion, or redefine "not
dopey" to mean "surely nobody would search for ^x when match x would do,
so it would be dopey to optimise re for that" :-)

No question, there's some dopiness to searching for the
beginning of the string at places other than beginning of the
string.

The tricky part would be optimizing '$'.
--
--Bryan
4 function calls in 0.003 CPU seconds

Ordered by: standard name

ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.000 0.000 :0(search)
1 0.003 0.003 0.003 0.003 :0(setprofile)
1 0.000 0.000 0.000 0.000 <string>:1(?)
0 0.000 0.000 profile:0(profiler)
1 0.000 0.000 0.003 0.003 profile:0(startsz.search(s))
4 function calls in 0.002 CPU seconds

Ordered by: standard name

ncalls tottime percall cumtime percall filename:lineno(function)
1 0.002 0.002 0.002 0.002 :0(search)
1 0.000 0.000 0.000 0.000 :0(setprofile)
1 0.000 0.000 0.002 0.002 <string>:1(?)
0 0.000 0.000 profile:0(profiler)
1 0.000 0.000 0.002 0.002 profile:0(startsz.search(s))
4 function calls in 0.228 CPU seconds

Ordered by: standard name

ncalls tottime percall cumtime percall filename:lineno(function)
1 0.228 0.228 0.228 0.228 :0(search)
1 0.000 0.000 0.000 0.000 :0(setprofile)
1 0.000 0.000 0.228 0.228 <string>:1(?)
0 0.000 0.000 profile:0(profiler)
1 0.000 0.000 0.228 0.228 profile:0(startsz.search(s))

[Duncan Booth]

John Machin wrote:

The point was made in a context where the OP appeared to be reading a
line at a time and parsing it, and re.compile(r'something').match()
would do the job; re.compile(r'^something').search() will do the job too
-- BECAUSE ^ means start of line anchor -- but somewhat redundantly, and
very inefficiently in the failing case with dopey implementations of
search() (which apply match() at offsets 0, 1, 2, .....).

Answering the question you think should have been asked rather than the
question which was actually asked is a great newsnet tradition, and often
more helpful to the poster than a straight answer would have been. However,
you do have to be careful to make it clear that is what you are doing.

The OP did not use the word 'line' once in his post. He simply said he was
searching a string. You didn't use the word 'line' either. If you are going
to read more into the question than was actually asked, please try to say
what question it is you are actually answering.

If he is using individual lines and re.match then the presence or absence
of a leading ^ makes virtually no difference. If he is looking for all
occurences in a multiline string then re.search with an anchored match is a
correct way to do it (splitting the string into lines and using re.match is
an alternative which may or may not be appropriate).

Either way, putting the focus on the ^ was inappropriate: the issue is
whether to use re.search or re.match. If you assume that the search fails
on an 80 character line, then I get timings of 6.48uS (re.search), 4.68uS
(re.match with ^), 4.66uS (re.match without ^). A failing search on a
10,000 character line shows how performance will degrade (225uS for search,
no change for match), but notice that searching 1 10,000 character string
is more than twice as fast as matching 125 80 character lines.

I don't understand what you think an implementation of search() can do in
this case apart from trying for a match at offsets 0, 1, 2, ...? It could
find a match at any starting offset within the string, so it must scan the
string in some form. A clever regex implementation will use Boyer-Moore
where it can to avoid checking every index in the string, but for the
pattern I suggested it would suprise me if any implementations actually
manage much of an optimisation.

[Duncan Booth]

John Machin wrote:

Your wording makes it hard to distinguish what exactly is "dopey".

"""
dopey implementations of search() (which apply match() at offsets 0, 1,
2, .....).
"""

The "dopiness" is that the ^ operator means that the pattern cannot
possibly match starting at 1, 2, 3, etc but a non-optimised search will
not recognise that and will try all possibilities, so the failing case
takes time dependant on the length of the string.

The ^ operator can match at any position in the string if the preceding
character was a newline. 'Dopey' would be failing to take this into
account.