To replace % marks but not \% in Perl Regex

Question

I am thinking how you can replace [^\]% marks but not \% marks in the sed -command of this answer. I think look-behind is not necessary.

My current Sed command but I think Perl is a must here

cat something | sed 's#%.*</#</#'                

which removes also everything after the % sign i.e. all comments in LaTeX but not percentage values.

My unsuccessful Perl attempt

cat something | perl 's#[^\]%.*</#</#'

where I do not know how I make Perl to take the standard output of cat.

Data

------------------------------
Protocol of pre-eclampsia
------------------------------
Monitoring in 90\% cases

Antihypertensives when % this is a comment, please, remove me!
$SBP/DBP > 160/110$; slowly.     
------------------------------

Desired output

------------------------------
Protocol of pre-eclampsia
------------------------------
Monitoring in 90\% cases

Antihypertensives when
$SBP/DBP > 160/110$; slowly.     
------------------------------

How can you replace % signs but not \%? If you can do this by Sed, please, comment.

Answer 1

Like many, if not most, text parsing tools, perl can take input from the command line, there's no need for cat. You just need -e which lets you pass a script as a command line parameter and -n which means "run the script on each line of input". ALternatively, you can use the -p switch which means "run the script on each line of input, then print that line". These two commands are equivalent (but the second is a classic useless use of cat, use the first) :

perl -pe 's/foo/bar/' file
cat file | perl -pe 's/foo/bar/'

Now, if I understand correctly, you want to delete all LaTeX comments (though that's not what your question states). If so, a lookbehind is the easiest way:

perl -pe 's/(?<!\\)%.*//' file 

Your regex should also work, you just need to keep the character you matched before the % and escape the backslash:

perl -pe 's/(^|[^\\]+)%.*/$1/' file

You can do the same thing with GNU sed:

sed -r 's/(^|[^\\])%.*/\1/' file

Answer 2

If you just want to replace what follows % but not what follows \%, in Perl, the easiest way is with a negative lookbehind: match %.* only if it isn't preceded by a backslash.

perl -pe 's/(?<!\\)%.*//'

However this won't match something like Hello world.\\%wibble. For that, you need to check that the % is preceded by an even number of backslashes. You can't do that with a lookbehind, because Perl's lookbehinds only support fixed-length patterns. Instead, match the backslashes in the regexp, and use a lookbehind to ensure that the regexp captures them all.

perl -pe 's/(?<!\\)((?:\\\\)*)%.*/$1/'

You can do that with tools that don't support lookbehind as well. In this case, you'll need to either use a tricky succession of replacement commands or match the backslashes and copy them to the replacement text.

sed -e 's/^\(\(\\\\\)*\)%.*/\1/' -e 's/\([^\\]\(\\\\\)*\)%.*/\1/'

Note that if you're processing a LaTeX document, there are other percent signs that might need to stay, for example in verbatim blocks. That can't be done with regexps alone.

Answer 3

A common idiom to replace unescaped characters in perl is with:

$ printf '%s\n' '% \% \\% \\\%' | perl -pe 's/(\\.)|%/$1||"<replacement>"/ge'
<replacement> \% \\<replacement> \\\%

So to remove everything starting from an unescaped %:

perl -pe 's/(\\.)|%.*/$1/g'

If your sed supports -E (FreeBSD/GNU):

sed -E 's/(\\.)|%.*/\1/g'

Or with GNU sed:

sed 's/\(\\.\)\|%.*/\1/g'

Where the alternation RE operator is not available (as in standard basic RE), you could often use \{0,1\} instead:

sed 's/\(\(\(\\.\)\{0,1\}[^\\%]*\)*\)\(%.*\)\{0,1\}/\1/'