$data=~s/((.*)\/.*\/)\2/$1/g;
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• perldoc perlre
• Mastering regular expressions
• Japhy's (upcoming) book

Hey, that's pretty good, arhuman. I will use that as a starting point...

To be more precise:

• There is a list of cities each separated by a slash
• There might be some duplicate cities anywhere in the list (ie 0, 1 or many duplicates)
• I want to form a list in any order without any duplicates, each separated by a slash (only one)
Examples:
• New York/Chicago/New York
• New York/Chicago/New York/Chicago/Buffalo
• Buffalo/New York/Chicago/New York/Chicago

I read Mastering Regular Expressions a while ago, and I'll read Japhy's book when it comes out, but regular expressions sometimes mystify me. Is it easier for people with the name Jeffrey for some reason?

-- iakobski

I would strongly discourage regexes in this case!! It's a lot trickier than the split/hash solution (and a lot trickier than you think :). Take as an example the solution $data=~s!((.*)/.*/)\2!$1!g; from arhuman which doesn't work correctly even for your first example (NY/Ch/NY), as it leaves a '/' at the end of the string. And much more important it breaks in nearly all other cases. The problems include

• greedyness of .*
• correct removal of '/'
• consider sth like 'abc/def/abcd' (not a real life example but a similar thing might come up) and 'abc' will be deleted
• speed: the regex solution makes a comparison for pairs, the first word with all others, then the second word with all others. This results in an O(n^2) algorithm (i.e. double the number of elements in the string, quadruple the execution time) whereas the split/hash solution is O(n)

-- Hofmator

I think that with those new elements Hofmator is right,
regexes aren't probably the good solution.

Especially with other simple answers available like :


print join'/',grep{if(!$u{$_}){$u{$_}=1}}split'/',$data;
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Note to hofmator: don't hit me too hard...
My (poor) regex was not meant to be a solution but rather to show a possible way to study
(wasn't approach bold enough ? ;-)
Anyway I++ your post beccause your rightly outline some of the problems that may arise with the regex approach...
and in particular in my poor regex approach

"Only Bad Coders Code Badly In Perl" (OBC2BIP)

/Jeff(?:rey)?/

Well, I have two comments: a regex is a cool approach to take, and a regex is the wrong approach to take. You said yourself you could split and use a hash. That's what I'd do. I'd use a regex if I were trying to impress someone (which I'm not). But if you are, then here's my offering:

$cities = "Here/There/Everywhere/There/Again/Here/Too";

$cities =~ s{
  ([^/]+)      # non-slashes (the city): \1
  /            # a /
  (?=          # look ahead for:...
    (?:        # group (not capture)
      [^/]* /  # city and /
    )*?        # 0 or more times, non-greedily
    \1         # is the city here again?
    (?: /|$ )  # a / or end-of-string
  )
}
{}gx;
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_____________________________________________________
Jeff japhy Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

OK, this is probably not the most beautiful and compact regex, but it works:

my $string = "New York/Chicago/New York/Chicago/Boston";

1 while $string =~ s!(^|/)([^/]+)(.*)/\2(/|$)!$1$2$3$4!g;

print $string;
#outputs "New York/Chicago/Boston"
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Update Well of course this is an abuse of regex. I would not use it myself nor recommend it, but it's a nice exercise.

Here's a hybrid solution. I'm inclined to agree that join(grep(split())) is going to be more robust, but this might be more economical:

$_ = '/New York/Chicago/New York/Boston/Houston/Chicago/Seattle/';
my %tally;
s/\/([^\/]+)/($tally{$1}++) ? '' : "\/$1"/ge;
print $_;

__END__
prints /New York/Chicago/Boston/Houston/Washington/
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Note that this requires a leading / if it's to remove duplicates of the first item in the list. Split handles that sort of thing better...

I couldn't help but thinking 'sort | uniq', so I did a search on uniq and came up with == uniq, which looks promising. I've read the comments warning against the regex solutions and have to agree, so you may wish to view MeowChow's array code for the same.