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need some regular expression help

P: n/a
I need a pattern that matches a string that has the same number of '('
as ')':
findall( compile('...'), '42^((2x+2)sin(x)) + (log(2)/log(5))' ) = [
'((2x+2)sin(x))', '(log(2)/log(5))' ]
Can anybody help me out?

Thanks for any help!

Oct 7 '06 #1
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14 Replies


P: n/a

Chris wrote:
I need a pattern that matches a string that has the same number of '('
as ')':
findall( compile('...'), '42^((2x+2)sin(x)) + (log(2)/log(5))' ) = [
'((2x+2)sin(x))', '(log(2)/log(5))' ]
Can anybody help me out?
This is not possible with regular expressions - they can't "remember"
how many parens they already encountered.

You will need a real parser for this - pyparsing seems to be the most
popular choice today, I personally like spark. I'm sure you find an
example-grammar that will parse simple arithmetical expressions like
the one above.

Diez

Oct 7 '06 #2

P: n/a
Chris wrote:
I need a pattern that matches a string that has the same number of '('
as ')':
findall( compile('...'), '42^((2x+2)sin(x)) + (log(2)/log(5))' ) = [
'((2x+2)sin(x))', '(log(2)/log(5))' ]
Can anybody help me out?
No, there is so such pattern. You will have to code up a function.

Consider what your spec really is: '42^((2x+2)sin(x)) +
(log(2)/log(5))' has the same number of left and right parentheses; so
does the zero-length string; so does ') + (' -- perhaps you need to add
'and starts with a "("'

Consider what you are going to do with input like this:

print '(' + some_text + ')'

Maybe you need to do some lexical analysis and work at the level of
tokens rather than individual characters.

Which then raises the usual question: you have a perception that
regular expressions are the solution -- to what problem??

HTH,
John

Oct 7 '06 #3

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On 7 Oct 2006 15:00:29 -0700, Diez B. Roggisch <de***@web.dewrote:
>
Chris wrote:
I need a pattern that matches a string that has the same number of '('
as ')':
findall( compile('...'), '42^((2x+2)sin(x)) + (log(2)/log(5))' ) = [
'((2x+2)sin(x))', '(log(2)/log(5))' ]
Can anybody help me out?

This is not possible with regular expressions - they can't "remember"
how many parens they already encountered.
Remember that regular expressions are used to represent regular
grammars. Most regex engines actually aren't regular in that they
support fancy things like look-behind/ahead and capture groups...IIRC,
these cannot be part of a true regular expression library.

With that said, the quote-unquote regexes in Lua have a special
feature that supports balanced expressions. I believe Python has a
PCRE lib somewhere; you may be able to use the experimental ??{ }
construct in that case.

-- Theerasak
Oct 8 '06 #4

P: n/a
In article <11*********************@e3g2000cwe.googlegroups.c om>,
"Chris" <ch*********@gmail.comwrote:
I need a pattern that matches a string that has the same number of '('
as ')':
findall( compile('...'), '42^((2x+2)sin(x)) + (log(2)/log(5))' ) = [
'((2x+2)sin(x))', '(log(2)/log(5))' ]
Can anybody help me out?

Thanks for any help!
Why does it need to be a regex? There is a very simple and well-known
algorithm which does what you want.

Start with i=0. Walk the string one character at a time, incrementing i
each time you see a '(', and decrementing it each time you see a ')'. At
the end of the string, the count should be back to 0. If at any time
during the process, the count goes negative, you've got mis-matched
parentheses.

The algorithm runs in O(n), same as a regex.

Regex is a wonderful tool, but it's not the answer to all problems.
Oct 8 '06 #5

P: n/a
Why does it need to be a regex? There is a very simple and well-known
algorithm which does what you want.

Start with i=0. Walk the string one character at a time, incrementing i
each time you see a '(', and decrementing it each time you see a ')'. At
the end of the string, the count should be back to 0. If at any time
during the process, the count goes negative, you've got mis-matched
parentheses.

The algorithm runs in O(n), same as a regex.

Regex is a wonderful tool, but it's not the answer to all problems.
Following Roy's suggestion, one could use something like:
>>s = '42^((2x+2)sin(x)) + (log(2)/log(5))'
d = {'(':1, ')':-1}
sum(d.get(c, 0) for c in s)
0
If you get a sum() 0, then you have too many "(", and if you
have sum() < 0, you have too many ")" characters. A sum() of 0
means there's the same number of parens. It still doesn't solve
the aforementioned problem of things like ')))(((' which is
balanced, but psychotic. :)

-tkc


Oct 8 '06 #6

P: n/a

hanumizzle wrote:
On 7 Oct 2006 15:00:29 -0700, Diez B. Roggisch <de***@web.dewrote:

Chris wrote:
I need a pattern that matches a string that has the same number of '('
as ')':
findall( compile('...'), '42^((2x+2)sin(x)) + (log(2)/log(5))' ) = [
'((2x+2)sin(x))', '(log(2)/log(5))' ]
Can anybody help me out?
This is not possible with regular expressions - they can't "remember"
how many parens they already encountered.

Remember that regular expressions are used to represent regular
grammars. Most regex engines actually aren't regular in that they
support fancy things like look-behind/ahead and capture groups...IIRC,
these cannot be part of a true regular expression library.
Certainly true, and it always gives me a hard time because I don't know
to which extend a regular expression nowadays might do the job because
of these extensions. It was so much easier back in the old times....
With that said, the quote-unquote regexes in Lua have a special
feature that supports balanced expressions. I believe Python has a
PCRE lib somewhere; you may be able to use the experimental ??{ }
construct in that case.
Even if it has - I'm not sure if it really does you good, for several
reasons:

- regexes - even enhanced ones - don't build trees. But that is what
you ultimately want
from an expression like sin(log(x))

- even if they are more powerful these days, the theory of context
free grammars still applies.
so if what you need isn't LL(k) but LR(k), how do you specify that
to the regex engine?

- the regexes are useful because of their compact notations, parsers
allow for better structured outcome
Diez

Oct 8 '06 #7

P: n/a
On 8 Oct 2006 01:49:50 -0700, Diez B. Roggisch <de***@web.dewrote:
Even if it has - I'm not sure if it really does you good, for several
reasons:

- regexes - even enhanced ones - don't build trees. But that is what
you ultimately want
from an expression like sin(log(x))

- even if they are more powerful these days, the theory of context
free grammars still applies.
so if what you need isn't LL(k) but LR(k), how do you specify that
to the regex engine?

- the regexes are useful because of their compact notations, parsers
allow for better structured outcome
Just wait for Perl 6 :D

-- Theerasak
Oct 8 '06 #8

P: n/a
Tim Chase:
It still doesn't solve the aforementioned problem
of things like ')))(((' which is balanced, but psychotic. :)
This may solve the problem:

def balanced(txt):
d = {'(':1, ')':-1}
tot = 0
for c in txt:
tot += d.get(c, 0)
if tot < 0:
return False
return tot == 0

print balanced("42^((2x+2)sin(x)) + (log(2)/log(5))") # True
print balanced("42^((2x+2)sin(x) + (log(2)/log(5))") # False
print balanced("42^((2x+2)sin(x))) + (log(2)/log(5))") # False
print balanced(")))(((") # False

A possibile alternative for Py 2.5. The dict solution looks better, but
this may be faster:

def balanced2(txt):
tot = 0
for c in txt:
tot += 1 if c=="(" else (-1 if c==")" else 0)
if tot < 0:
return False
return tot == 0

Bye,
bearophile

Oct 8 '06 #9

P: n/a
be************@lycos.com wrote:
The dict solution looks better, but this may be faster:
it's slightly faster, but both your alternatives are about 10x slower
than a straightforward:

def balanced(txt):
return txt.count("(") == txt.count(")")

</F>

Oct 8 '06 #10

P: n/a
Thus spoke Diez B. Roggisch (on 2006-10-08 10:49):
Certainly true, and it always gives me a hard time because I don't know
to which extend a regular expression nowadays might do the job because
of these extensions. It was so much easier back in the old times....
Right, in perl, this would be a no-brainer,
its documented all over the place, like:

my $re;

$re = qr{
(?:
(?[^\\()]+ | \\. )
|
\( (??{ $re }) \)
)*
}xs;

where you have a 'delayed execution'
of the

(??{ $re })

which in the end makes the whole a thing
recursive one, it gets expanded and
executed if the match finds its way
to it.

Above regex will match balanced parens,
as in:

my $good = 'a + (b / (c - 2)) * (d ^ (e+f)) ';
my $bad1 = 'a + (b / (c - 2) * (d ^ (e+f)) ';
my $bad2 = 'a + (b / (c - 2)) * (d) ^ (e+f) )';

if you do:

print "ok \n" if $good =~ /^$re$/;
print "ok \n" if $bad1 =~ /^$re$/;
print "ok \n" if $bad2 =~ /^$re$/;
This in some depth documented e.g. in
http://japhy.perlmonk.org/articles/tpj/2004-summer.html
(topic: Recursive Regexes)

Regards

M.
Oct 8 '06 #11

P: n/a
Mirco Wahab schrieb:
Thus spoke Diez B. Roggisch (on 2006-10-08 10:49):
>Certainly true, and it always gives me a hard time because I don't know
to which extend a regular expression nowadays might do the job because
of these extensions. It was so much easier back in the old times....

Right, in perl, this would be a no-brainer,
its documented all over the place, like:

my $re;

$re = qr{
(?:
(?[^\\()]+ | \\. )
|
\( (??{ $re }) \)
)*
}xs;

where you have a 'delayed execution'
of the

(??{ $re })

which in the end makes the whole a thing
recursive one, it gets expanded and
executed if the match finds its way
to it.

Above regex will match balanced parens,
as in:

my $good = 'a + (b / (c - 2)) * (d ^ (e+f)) ';
my $bad1 = 'a + (b / (c - 2) * (d ^ (e+f)) ';
my $bad2 = 'a + (b / (c - 2)) * (d) ^ (e+f) )';

if you do:

print "ok \n" if $good =~ /^$re$/;
print "ok \n" if $bad1 =~ /^$re$/;
print "ok \n" if $bad2 =~ /^$re$/;
This in some depth documented e.g. in
http://japhy.perlmonk.org/articles/tpj/2004-summer.html
(topic: Recursive Regexes)
That clearly is a recursive grammar rule, and thus it can't be regular
anymore :) But first of all, I find it ugly - the clean separation of
lexical and syntactical analysis is better here, IMHO - and secondly,
what are the properties of that parsing? Is it LL(k), LR(k), backtracking?

Diez
Oct 8 '06 #12

P: n/a
Fredrik Lundh wrote:
it's slightly faster, but both your alternatives are about 10x slower
than a straightforward:
def balanced(txt):
return txt.count("(") == txt.count(")")
I know, but if you read my post again you see that I have shown those
solutions to mark ")))(((" as bad expressions. Just counting the parens
isn't enough.

Bye,
bearophile

Oct 8 '06 #13

P: n/a
"Diez B. Roggisch" <de***@web.dewrote:
Certainly true, and it always gives me a hard time because I don't know
to which extend a regular expression nowadays might do the job because
of these extensions. It was so much easier back in the old times....
What old times? I've been working with regex for mumble years and there's
always been the problem that every implementation supports a slightly
different syntax. Even back in the "good old days", grep, awk, sed, and ed
all had slightly different flavors.
Oct 8 '06 #14

P: n/a
On 10/8/06, Roy Smith <ro*@panix.comwrote:
"Diez B. Roggisch" <de***@web.dewrote:
Certainly true, and it always gives me a hard time because I don't know
to which extend a regular expression nowadays might do the job because
of these extensions. It was so much easier back in the old times....

What old times? I've been working with regex for mumble years and there's
always been the problem that every implementation supports a slightly
different syntax. Even back in the "good old days", grep, awk, sed, and ed
all had slightly different flavors.
Which grep? Which awk? :)

-- Theerasak
Oct 8 '06 #15

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