Regex Question

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

Hi all, i've been trying to construct a regex to help with cleaning up names that can appear in some pretty nasty formats, it's raw data from government agencies. Anyway, one thing i do is look for various 'care of' variants and standardize them to the % symbol. That is done through this regex:

    $name =~ s/(^|\s)
                    C
                    (\/|\\|%)
                    O
                    (\s|$)
                / % /xgi;
[download]

(yes i am looking for the case of 'C%O' which i also see sometimes)

And then i want to replace forward of backslashes with &, as long as they do not have digits on both sides (which would be a fraction, which does sometimes appear in names i process). That is done through this regex:

    $field =~ s/(?<!\d)     # not preceeded by digits
                (?:\\|\/)   # back or forward slash
                (?!\d)      # not succeeded by digits
            / & /xg;        # replace with '&' (globally)
[download]

The problem is that sometimes i want to perform only the second transformation without having done the first one, but i could not come up with any decent way to accomplish the second part with the exception of cases that are care-of's, as defined by the first part.

Any thoughts?

Update: I got this working effectively, but then realized that my spec gets even worse, because if i see something like "D/B/A", i want to change that to "DBA" (Doing Business As), so this adds a completely new twist onto when and when not to replace the slashes. So i added this regex after the care of regex:

    $field =~ s/(?:^|[^A-Z])
                    ([A-Z])\/
                    ([A-Z])\/
                    ([A-Z])
                (?:[^A-Z]|$)
                / $1$2$3 /xig;
[download]

Also, i think that using the separate regexes will work fine, i have worked around the problem of only wanting to perform the final stage without messing the earlier stages. Thanks for all the suggestions!

I use the most powerful debugger available: print!

Comment on Regex Question Select or Download Code

Replies are listed 'Best First'.
Re: Regex Question by Aristotle (Chancellor) on Nov 15, 2005 at 20:08 UTC
Roy Johnson’s technique, used only for the skipping part, produces the desired result: `$name =~ s{ (?<!C(?=.O)) # doesn't match C-something-O starting at previous ch +aracter (?<!\d) # not preceeded by digits (?:\\\|\/) # back or forward slash (?!\d) # not succeeded by digits } { & }xg; # replace with '&' (globally)` [download] Makeshifts last the longest.	[reply] [d/l]
Re: Regex Question by Aristotle (Chancellor) on Nov 15, 2005 at 19:39 UTC
This should work: `$field =~ s/ (?<!C) # not preceeded by "C" (?:\\\|\/) # back or forward slash (?!O) # not succeeded by "O" \| (?<!\d) # not preceeded by digits (?:\\\|\/) # back or forward slash (?!\d) # not succeeded by digits / & /xg; # replace with '&' (globally)` [download] Nope; wrong. I realised my blunder right after posting. Makeshifts last the longest.	[reply] [d/l]
Re^2: Regex Question by Roy Johnson (Monsignor) on Nov 15, 2005 at 19:47 UTC
You were bitten by negative-over-or-logic. The left side of the \| will match what the right side is intended to skip. It has to be something like `$field =~ s/ (?<!C(?=.O)) # Look back a character and make sure it's not C-some +thing-O from there (?<!\d(?=.\d)) # Make sure it's also not a digit-something-digit (?:\\\|\/) # The something is a slash or backslash / & /xgi;` [download] Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re^3: Regex Question by Aristotle (Chancellor) on Nov 15, 2005 at 19:58 UTC
~~That won’t match the exact same things. The OP’s regex will replace the slash in `"/2"` or in `"1/"`, yours won’t. Not sure if this is part of the spec, but it’s something to be aware of.~~ Meh. That’s it, I’m going back to bed. Makeshifts last the longest.	[reply] [d/l] [select]
Re^4: Regex Question by Roy Johnson (Monsignor) on Nov 15, 2005 at 20:23 UTC
Re^4: Regex Question by shemp (Deacon) on Nov 15, 2005 at 20:07 UTC
Re^2: Regex Question by ikegami (Patriarch) on Nov 15, 2005 at 19:43 UTC
Nope. It replaces the slash in "9\\9" even though it shouldn't. Read more... (1193 Bytes)	[reply] [d/l]
Re: Regex Question by ikegami (Patriarch) on Nov 15, 2005 at 19:52 UTC
This might be easiest: `$field =~ s/((.)(?:\\\|\/)(.))/ if ($2 eq 'C' && $3 eq 'O') { $1 } elsif (ord($2) >= ord('0') && ord($2) <= ord('9')) { $1 } elsif (ord($3) >= ord('0') && ord($3) <= ord('9')) { $1 } else { "$2 & $3" } /xeg;` [download] Update: I had a "==" where I needed a "eq". Also fixed the problem Roy found.	[reply] [d/l]
Re^2: Regex Question by Aristotle (Chancellor) on Nov 15, 2005 at 20:02 UTC
That will only match slashes surrounded by a character on both sides. `(?!\d)` is not the same as `\D`. Makeshifts last the longest.	[reply] [d/l] [select]
Re^3: Regex Question by ikegami (Patriarch) on Nov 15, 2005 at 20:11 UTC
And it won't work when there are multiple slashes in a row. Only some of them will get converted. Sorry, I meant to mention the caveats. I knew about them, but I thought they might be acceptable to the OP.	[reply]
Re^2: Regex Question by Roy Johnson (Monsignor) on Nov 15, 2005 at 20:16 UTC
Update: `if ($1 eq 'C' && $2 eq 'O')` should be `if ($2 eq 'C' && $3 eq 'O')`, because `$1` is the entire matched expression. Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]