Regexp dashes at boundaries

cormanaz has asked for the wisdom of the Perl Monks concerning the following question:

Good day monks. I'm trying to regularize references to U.S. government entities in some text. Taking the U.S. Army as an example:

$text = "U.S. Army\nthe Army has been berry-berry good to me\nUS-Army\
+nUS Army\n";
$text =~ s/\bArmy\b|U\.S\. Army/US-Army/g;
[download]

This yields the following

US-Army
the US-Army has been berry-berry good to me
US-US-Army
US US-Army
[download]

The last two entries are obviously not what I want and they are happening because the dash counts as a boundary. This is even worse in the context of a loop like

while ($text =~ s/\bArmy\b|U\.S\. Army/US-Army/g) {
    print "$text\n\n";
}
[download]

which results in an infinite loop. So my question is: How can I get the dash to not count as a boundary character? (because of external constraints, it's not possible to use another joining character).

Many thanks in advance...

Steve

Comment on Regexp dashes at boundaries Select or Download Code

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Re: Regexp dashes at boundaries by brian_d_foy (Abbot) on Mar 17, 2005 at 22:42 UTC
You don't always need alternation, especially if most of the stuff on either side is the same. You can use the ? quantifier to denote some parts as optional. This example works for the input you showed, but other forms you might find might require it to change a bit. `#!/usr/bin/perl $text = <<"HERE"; U.S. Army the Army has been berry-berry good to me US-Army US Army HERE $text =~ s/(?:U.?S.?[\s-])*Army/US-Army/g; print $text;` [download] -- brian d foy <bdfoy@cpan.org>	[reply] [d/l]
Re^2: Regexp dashes at boundaries by cormanaz (Deacon) on Mar 21, 2005 at 22:29 UTC
Yes that does it. Now can I ask why you use the (?:...) extension? I tried it without and it works. I assume you have a good reason for putting it in there, but I looked up that regexp extension in the Camel book and the explanation is...um...sort of cryptic: "This is for clustering, not capturing; it groups subexpressions like ``()'', but doesn't make backreferences as ``()'' does." Thanks for the help... Steve	[reply]
Re^3: Regexp dashes at boundaries by brian_d_foy (Abbot) on Mar 24, 2005 at 18:43 UTC
I used the non-capturing parentheses to create a group so I could apply a quantifier to it. I want the "U.S. " to be optional, so I want to apply a * to that whole group. I should probably have used a ? (zero or one) though. -- brian d foy <bdfoy@cpan.org>	[reply]
Re: Regexp dashes at boundaries by Roy Johnson (Monsignor) on Mar 18, 2005 at 12:47 UTC
To answer your question: the boundary character is a zero-width assertion like `(?:(?=\w)(?<=\W)\|(?=\W)(?<=\w))` (next char is word and prev is non-word, or vice-versa). If you want a different definition of a boundary, you would need to roll your own boundary expression. Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re: Regexp dashes at boundaries by crashtest (Curate) on Mar 17, 2005 at 22:10 UTC
Your regular expression doesn't do what you think it does (I think!). The alternation meta-character `\|` works on the two entities directly next to it. You need to use parentheses with your alternation, maybe like this: `$text =~ s/(\bArmy\b)\|(U\.S\. Army)/US-Army/g;` [download] Update: Arghh! Completely disregard. Alternation extends further than I said unless you limit it with parentheses. I seem to be living in opposite land today. ... but it is cleaner to "factor out" the "Army" string and do something similar to: `$text =~ s/(U\.S\.)?\bArmy/US-Army/g;` [download] Note: completely untested! Hope this helps.	[reply] [d/l] [select]
Re^2: Regexp dashes at boundaries by tlm (Prior) on Mar 17, 2005 at 23:22 UTC
The alternation meta-character \| works on the two entities directly next to it. Not in Perl: `$t = 'abcd'; $t =~ s/bc\|xy/pq/; print "$t\n"; ==> apqd` [download] If what you said were true, the match above would have failed, and the contents of `$t` would have remained unchanged. the lowliest monk	[reply] [d/l] [select]
Re^3: Regexp dashes at boundaries by ww (Archbishop) on Mar 18, 2005 at 15:25 UTC
and for possible further illumination: $t = 'abcd'; $t1 = 'ababxycdcd'; $t2 = 'ababxycdcd'; $t3 = 'ababxycdcd'; # same as $t2, which is identical to $t1 $t4 = 'ababxycdcd'; # still the same... $t =~ s/bc\|xy/pq/; print "\$t is: $t\n"; $t1 =~ s/(ab\|xy)/pq/; print "\$t1 is: $t1\n"; $t2 =~ s/(ab\|xy)/pq/g; print "\$t2 is: $t2\n"; $t3 =~ s/(ab)\|(xy)/pq/g; print "\$t3 is: $t3\n"; $t4 =~ s/((ab)\|(xy))/pq\1/g; #capture regex -outer (); grouping () in +side print "\$t4 is: $t4\n"; =head1 OUT Output of C:\_perl\pl_test>perl 440581.pl $t is: apqd matched the 'b' $t1 is: pqabxycdcd parenthesdized the alt; now matches the FIRST +'ab' $t2 is: pqpqpqcdcd added /g (match globally) replaces both 'ab's +and the xy $t3 is: pqpqpqcdcd Shifting parens ==> no change, here $t4 is: pqabpqabpqxycdcd Capturing regex -- 3 'pq' pairs, each pair f +ol by 'ab' or 'xy' =cut [download] additional example added	[reply] [d/l]
Re^2: Regexp dashes at boundaries by cormanaz (Deacon) on Mar 17, 2005 at 22:33 UTC
I suppose you are right about the alternation error; however the suggest regexp yields: `U.S. US-Army the US-Army has been berry-berry good to me US-US-Army US US-Army` [download] Which now does the first case wrong as well.	[reply] [d/l]
Re: Regexp dashes at boundaries by sh1tn (Priest) on Mar 18, 2005 at 18:35 UTC
`while(my $line=<DATA>){ $_ .= $line } s/(u)\W?(s)\W?\s\W?\s(army)/uc($1.$2).'-'.ucfirst$3/gieo; print # STDOUT: # # US-Army # the US-Army has been # US-Army # USA Army . US-Army __DATA__ U.S. Army the US army has been US - Army USA Army . U-S- Army` [download]	[reply] [d/l]