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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.

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  • 1
    Please include an example of your input and your desired output. What should be done with \%%%%%%%%? How does < come into it? Commented May 30, 2015 at 10:13
  • I added an example input and output. \%%%%%... is a comment so everything after second % should be removed. Commented May 30, 2015 at 10:25
  • @terdon It should be replaced by \%, that's what TeX does. However, things like \\% and \\\% are more problematic. Commented May 30, 2015 at 10:25
  • @lcd047 I was asking because the OP's code would replace a %.*< with #< and I wasn't sure if he wanted comments removed or true replacement of all matching characters. I still don't know why there's s < in the code. Commented May 30, 2015 at 10:27
  • 2
    If this is about TeX comments pleas note that the behaviour of whitespace added when replacing % with a newline could pose a problem Commented May 30, 2015 at 15:44

3 Answers 3

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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
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  • This replaces all % signs with #, while it should just delete comment signs (%) and leave \% signs alone. Commented May 30, 2015 at 10:29
  • 1
    @Masi OK, see update. But please edit your question and clarify what you need. Commented May 30, 2015 at 10:37
  • perl -pe 's/([^\\]*)%.*/$1/' test.tex gives ... Monitoring in 90\ Antihypertensives when ... so no percent sign. My Perl version is 5.18.2. Commented May 30, 2015 at 10:44
  • @Masi well, yes, that's what you asked for. That's what you show in your desired output, so that's what I gave you. Commented May 30, 2015 at 10:48
  • Desired output is updated in the body, sorry for confusion. It should be Monitoring in 90\% cases by just removing the delete me comment. Commented May 30, 2015 at 10:50
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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.

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    Is there a wobble in your wibble in the sed version? I don't get the expected results. If I uderstand it correctly, it is intended to delete zero to all prefixed '\\' pairs + % + the rest of the line such as in the lines – %wibble x0, \\%wibble x2, \\\\%wibble x4, x0 \\%wbble, x2 \\%wbble, x4 \\\\%wobble. If that is so, then I believe this works: sed -e 's/^\(\\\\\)*%.*//' -e 's/\([^\\]\)\(\\\\\)*%.*/\1/' Commented May 31, 2015 at 5:25
  • @Peter.O Oops, yes, both the perl and the sed versions were missing some bits. Your sed version removes backslashes. Commented May 31, 2015 at 12:19
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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/'
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