I have a "srt" file(like standard movie-subtitle format) like shown in below link:http://pastebin.com/3k8a53SC
Excerpt:
1
00:00:53,000 --> 00:00:57,000
<any text that may span multiple lines>
2
00:01:28,000 --> 00:01:35,000
<any text that may span multiple lines>
But right now the subtitles timing is all wrong, as it lags behind by 9 seconds.
Is it possible to add 9 seconds(+9) to every time entry with regex ?
Even if the milliseconds is set to 000 then it's fine, but the addition of 9 seconds should adhere to "60 seconds = 1 minute & 60 minutes = 1 hour" rules.
Also the subtitle text after timing entry must not get altered by regex.
By the way the time format for each time string is "Hours:Minutes:Seconds.Milliseconds".
Quick answer is "no", that's not an application for regex. A regular expression lets you MATCH text, but not change it. Changing things is outside the scope of the regex itself, and falls to the language you're using -- perl, awk, bash, etc.
For the task of adjusting the time within an SRT file, you could do this easily enough in bash, using the date command to adjust times.
#!/usr/bin/env bash
offset="${1:-0}"
datematch="^(([0-9]{2}:){2}[0-9]{2}),[0-9]{3} --> (([0-9]{2}:){2}[0-9]{2}),[0-9]{3}"
os=$(uname -s)
while read line; do
if [[ "$line" =~ $datematch ]]; then
# Gather the start and end times from the regex
start=${BASH_REMATCH[1]}
end=${BASH_REMATCH[3]}
# Replace the time in this line with a printf pattern
linefmt="${line//[0-2][0-9]:[0-5][0-9]:[0-5][0-9]/%s}\n"
# Calculate new times
case "$os" in
Darwin|*BSD)
newstart=$(date -v${offset}S -j -f "%H:%M:%S" "$start" '+%H:%M:%S')
newend=$(date -v${offset}S -j -f "%H:%M:%S" "$end" '+%H:%M:%S')
;;
Linux)
newstart=$(date -d "$start today ${offset} seconds" '+%H:%M:%S')
newend=$(date -d "$end today ${offset} seconds" '+%H:%M:%S')
;;
esac
# And print the result
printf "$linefmt" "$newstart" "$newend"
else
# No adjustments required, print the line verbatim.
echo "$line"
fi
done
Note the case statement. This script should auto-adjust for Linux, OSX, FreeBSD, etc.
You'd use this script like this:
$ ./srtadj -9 < input.srt > output.srt
Assuming you named it that, of course. Or more likely, you'd adapt its logic for use in your own script.
No, sorry, you can’t. Regex are a context free language (see Chomsky e.g. https://en.wikipedia.org/wiki/Chomsky_hierarchy) and you cannot calculate.
But with a context sensitive language like perl it will work.
It could be a one liner like this ;-)))
perl -n -e 'if(/^(\d\d:\d\d:\d\d)([-,\d\s\>]*)(\d\d:\d\d:\d\d)(.*)/) {print plus9($1).$2.plus9($3).$4."\n";}else{print $_} sub plus9{ ($h,$m,$s)=split(/:/,shift); $t=(($h*60+$m)*60+$s+9); $h=int($t/3600);$r=$t-($h*3600);$m=int($r/60);$s=$r-($m*60);return sprintf "%02d:%02d:%02d", $h, $m, $s;}‘ movie.srt
with move.srt like
1
00:00:53,000 --> 00:00:57,000
hello
2
00:01:28,000 --> 00:01:35,000
I like perl
3
00:02:09,000 --> 00:02:14,000
and regex
you will get
1
00:01:02,000 --> 00:01:06,000
hello
2
00:01:37,000 --> 00:01:44,000
I like perl
3
00:02:18,000 --> 00:02:23,000
and regex
You can change the +9 in the "sub plus9{...}", if you want another delta.
How does it work?
We are looking for lines that matches
dd:dd:dd something dd:dd:dd something
and then we call a sub, which add 9 seconds to the matched group one ($1) and group three ($3). All other lines are printed unchanged.
added
If you want to put the perl oneliner in a file, say plus9.pl, you can add newlines ;-)
if(/^(\d\d:\d\d:\d\d)([-,\d\s\>]*)(\d\d:\d\d:\d\d)(.*)/) {
print plus9($1).$2.plus9($3).$4."\n";
} else {
print $_
}
sub plus9{
($h,$m,$s)=split(/:/,shift);
$t=(($h*60+$m)*60+$s+9);
$h=int($t/3600);
$r=$t-($h*3600);
$m=int($r/60);
$s=$r-($m*60);
return sprintf "%02d:%02d:%02d", $h, $m, $s;
}
Regular expressions strictly do matching and cannot add/substract. You can match each datetime string using python, for example, add 9 seconds to that, and then rewrite the string in the appropriate spot. The regular expression I would use to match it would be the following:
(?<hour>\d+):(?<minute>\d+):(?<second>\d+),(?<msecond>\d+)
It has labeled capture groups so it's really easy to get each section (you won't need msecond but it's there for visualization, I guess)
Regex101
Related
I have the following string:
load Add 20 percent
to accommodate
I want to get to:
load Add 20 percent to accommodate
With, e.g., regex in sublime, this is easily done by:
Regex:
([a-z])\n\s([a-z])
Replace:
$1 $2
However, in Perl, if I input this command, (adapted to test if I can match the pattern in any case):
perl -pi.orig -e 's/[a-z]\n.+to/TEST/g' file
It doesn't match anything.
Does anyone know why Perl would be different in this case, and what the correct formulation of the Perl command should be?
By default, Perl -p flag read input lines one by one. You can't thus expect your regex to match anything after \n.
Instead, you want to read the whole input at once. You can do this by using the flag -0777 (this is documented in perlrun):
perl -0777 -pi.orig -e 's/([a-z])\n\s(to)/$1 $2/' file
Just trying to help and reminding below your initial proposal for perl regex:
perl -pi.orig -e 's/[a-z]\n.+to/TEST/g' file
Note that in perl regex, [a-z] will match only one character, NOT including any whitespace. Then as a start please include a repetition specifier and include capability to also 'eat' whitespaces. Also to keep the recognized (but 'eaten') 'to' in the replacement, you must put it again in the replacement string, like finally in the below example perl program:
$str = "load Add 20 percent
to accommodate";
print "before:\n$str\n";
$str =~ s/([ a-z]+)\n\s*to/\1 to/;
print "after:\n$str\n";
This program produces the below input:
before:
load Add 20 percent
to accommodate
after:
load Add 20 percent to accommodate
Then it looks like that if I understood well what you want to do, your regexp should better look like:
s/([ a-z]+)\n\s*to/\1 to/ (please note the leading whitespace before 'a-z').
I have a very large file, containing the following blocks of lines throughout:
start :234
modify 123 directory1/directory2/file.txt
delete directory3/file2.txt
modify 899 directory4/file3.txt
Each block starts with the pattern "start : #" and ends with a blank line. Within the block, every line starts with "modify # " or "delete ".
I need to modify the path in each line, specifically appending a directory to the front. I would just use a general regex to cover the entire file for "modify #" or "delete ", but due to the enormous amount of other data in that file, there will likely be other matches to this somewhat vague pattern. So I need to use multi-line matching to find the entire block, and then perform edits within that block. This will likely result in >10,000 modifications in a single pass, so I'm also trying to keep the execution down to less than 30 minutes.
My current attempt is a sed one-liner:
sed '/^start :[0-9]\+$/ { :a /^[modify|delete] .*$/ { N; ba }; s/modify [0-9]\+ /&Appended_DIR\//g; s/delete /&Appended_DIR\//g }' file_to_edit
Which is intended to find the "start" line, loop while the lines either start with a "modify" or a "delete," and then apply the sed replacements.
However, when I execute this command, no changes are made, and the output is the same as the original file.
Is there an issue with the command I have formed? Would this be easier/more efficient to do in perl? Any help would be greatly appreciated, and I will clarify where I can.
I think you would be better off with perl
Specifically because you can work 'per record' by setting $/ - if you're records are delimited by blank lines, setting it to \n\n.
Something like this:
#!/usr/bin/env perl
use strict;
use warnings;
local $/ = "\n\n";
while (<>) {
#multi-lines of text one at a time here.
if (m/^start :\d+/) {
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
Each iteration of the loop will pick out a blank line delimited chunk, check if it starts with a pattern, and if it does, apply some transforms.
It'll take data from STDIN via a pipe, or myscript.pl somefile.
Output is to STDOUT and you can redirect that in the normal way.
Your limiting factor on processing files in this way are typically:
Data transfer from disk
pattern complexity
The more complex a pattern, and especially if it has variable matching going on, the more backtracking the regex engine has to do, which can get expensive. Your transforms are simple, so packaging them doesn't make very much difference, and your limiting factor will be likely disk IO.
(If you want to do an in place edit, you can with this approach)
If - as noted - you can't rely on a record separator, then what you can use instead is perls range operator (other answers already do this, I'm just expanding it out a bit:
#!/usr/bin/env perl
use strict;
use warnings;
while (<>) {
if ( /^start :/ .. /^$/)
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
We don't change $/ any more, and so it remains on it's default of 'each line'. What we add though is a range operator that tests "am I currently within these two regular expressions" that's toggled true when you hit a "start" and false when you hit a blank line (assuming that's where you would want to stop?).
It applies the pattern transformation if this condition is true, and it ... ignores and carries on printing if it is not.
sed's pattern ranges are your friend here:
sed -r '/^start :[0-9]+$/,/^$/ s/^(delete |modify [0-9]+ )/&prepended_dir\//' filename
The core of this trick is /^start :[0-9]+$/,/^$/, which is to be read as a condition under which the s command that follows it is executed. The condition is true if sed currently finds itself in a range of lines of which the first matches the opening pattern ^start:[0-9]+$ and the last matches the closing pattern ^$ (an empty line). -r is for extended regex syntax (-E for old BSD seds), which makes the regex more pleasant to write.
I would also suggest using perl. Although I would try to keep it in one-liner form:
perl -i -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit
Or you can use redirection of stdout:
perl -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit > new_file
with gnu sed (with BRE syntax):
sed '/^start :[0-9][0-9]*$/{:a;n;/./{s/^\(modify [0-9][0-9]* \|delete \)/\1NewDir\//;ba}}' file.txt
The approach here is not to store the whole block and to proceed to the replacements. Here, when the start of the block is found the next line is loaded in pattern space, if the line is not empty, replacements are performed and the next line is loaded, etc. until the end of the block.
Note: gnu sed has the alternation feature | available, it may not be the case for some other sed versions.
a way with awk:
awk '/^start :[0-9]+$/,/^$/{if ($1=="modify"){$3="newdirMod/"$3;} else if ($1=="delete"){$2="newdirDel/"$2};}{print}' file.txt
This is very simple in Perl, and probably much faster than the sed equivalent
This one-line program inserts Appended_DIR/ after any occurrence of modify 999 or delete at the start of a line. It uses the range operator to restrict those changes to blocks of text starting with start :999 and ending with a line containing no printable characters
perl -pe"s<^(?:modify\s+\d+|delete)\s+\K><Appended_DIR/> if /^start\s+:\d+$/ .. not /\S/" file_to_edit
Good grief. sed is for simple substitutions on individual lines, that is all. Once you start using constructs other than s, g, and p (with -n) you are using the wrong tool. Just use awk:
awk '
/^start :[0-9]+$/ { inBlock=1 }
inBlock { sub(/^(modify [0-9]+|delete) /,"&Appended_DIR/") }
/^$/ { inBlock=0 }
{ print }
' file
start :234
modify 123 Appended_DIR/directory1/directory2/file.txt
delete Appended_DIR/directory3/file2.txt
modify 899 Appended_DIR/directory4/file3.txt
There's various ways you can do the above in awk but I wrote it in the above style for clarity over brevity since I assume you aren't familiar with awk but should have no trouble following that since it reuses your own sed scripts regexps and replacement text.
I'm writing a bash shell script that I hope to ultimately use to automate the naming and 'attachment' of scanned documents to our db. The script OCR's a section of the first page of the pdf and outputs a text file containing three lines; a name, unique id, and a datetime string:
Smith, John
Case #: 234567 ( )
09/04/2013 11:34 AM
What I'd like to do is end up with two seperate strings as variables, "Smith, John" and "234567". I'm looking for help using regex with sed/awk/etc to extract this number. One issue is that the OCR will rarely output strings like:
"Case #2 234567 ( )"
or
"Ca$e # 2234567 ( 7"
So I'm thinking to take the only last 6-digits in the string, since only maybe 1 in 10,000+ of these ever get the last 6-digits read incorrectly. This unique ID is only 6 digits, and is always between 200000-999999. I'm learning regex, but it's slow going. Any help is greatly appreciated.
Edit:
For now I am using:
casename="$(cat test.txt | sed '1!d')"
casenum="$(cat test.txt | sed -n -r 's/.*([0-9]{6}).*/\1/p')"
echo ${casenum} ${casename}
234567 Smith, John
Any input for why this might not be a good way to do it, or what could be improved is (very) welcomed.
you might try this Regex (BRE):
[2-9][0-9]\{5\}\>
You could use the following regex for the second line:
^.*(\d{6})[^\d].*$
Here, the first named sub-group would denote the digits of interest.
For example, using Notepad++,
Orginal text:
Replace options:
Resulting text:
The regex should remain more or less the same across environments. You might need to simply change the way the named subexpression ($1 here) is referenced.
You could probably use something like this untested but somewhat-syntactically-valid snippet:
shopt -s extglob
declare -a cases
for casefile in casefiles/*
do
name=""
while read l
do
if [[ -z "$name" ]]
then
[[ "$l" == #(*, *) ]] && name=$l
elif [[ "$l" == +([0-9]) ]]
then
after=${l#*[2-9][0-9][0-9][0-9][0-9][0-9]}
l=${l%$after}
l=${l#${l%[2-9][0-9][0-9][0-9][0-9][0-9]}}
if [[ "$l" == #([2-9][0-9][0-9][0-9][0-9][0-9]) ]]
then
cases[$l]=$name
fi
name=""
fi
done < $casefile
done
The "hard part" prunes the first 6-digit number in your range and everything after it, then removes what's left (the stuff before the number) from the line. Then it removes the number from the beginning of the string, and removes what's left (the part after the number) from the end. If what's left is a 6-digit number in your range, it uses that as the index and the case name as the value in an array which you can later iterate over.
The rest should be pretty straightforward. :) If this doesn't quite work as expected, I blame the fact that I mostly use ksh, not bash. ;)
Custom batch rename files
Hello, Mac OS X takes screen shot's in a very long format of filename. I would like to rename any of them that sit at path /Users/me/desktop.
Here are some examples of the filenames:
Screen Shot 2012-08-02 at 1.15.29 AM.png
Screen Shot 2012-08-02 at 1.22.12 AM.png
Screen Shot 2012-08-02 at 1.22.14 PM.png
Screen Shot 2012-08-02 at 1.22.16 PM.png
I was once told, not to do a for loop against an ls so I am trying globbing this time around. So far, this is all I can come up with, but done know how to karen wrap the expression and then get that to a file rename in the format I desire:
for i in *; do
screen_name=$(echo $i | grep --only-matching --extended-regexp '(Screen\ Shot)\ [0-9]+-[0-9]+-[0-9]+\ at\ [0-9]+\.[0-9]+.[0-9]+.[AP]M\.png');
echo $screen_name;
done
I am not sure about the hour of the time, it may be safest to assume possible 2 digits on all chunks of the time, so 1.14.29 and 01.15.29
ss.08-02-12-01.15.29-AM.png
ss.08-02-12-01.22.12-AM.png
ss.08-02-12-01.22.14-PM.png
ss.08-02-12-01.22.16-PM.png
The end goal, is a bash script that when run will rename ALL files at the above mentioned path to the new format listed.
Thank you for any help.
for i in "Screen Shot"*.png; do
new=`echo $i |awk '
{
split($3,a,"-")
split($5,b,".")
printf("ss.%s-%s-%s-%02d.%02d.%02d-%s",a[2],a[3],a[1],b[1],b[2],b[3],$6)
}
'`
mv "$i" $new
done
Before:
Screen Shot 2012-08-02 at 1.22.16 PM.png
Screen Shot 2012-09-02 at 13.42.06 PM.png
After:
ss.08-02-2012-01.22.16-PM.png
ss.09-02-2012-13.42.06-PM.png
EDIT:
as suggested by steve
printf("ss.%s-%s-%s-%02d.%02d.%02d-%s",a[2],a[3],substr(a[1]3,2),b[1],b[2],b[3],$6)
which yields
ss.08-02-12-01.22.16-PM.png
ss.09-02-12-13.42.06-PM.png
You can use stream editor sed to match and substitute using regular expressions. You would do something like this
echo $i | sed "s/PATTERN/REPLACE/"
to genereate the filename out of $i. sed will read from stdin, search (s command) for pattern and replace it with REPLACE.
In your REGEXP pattern you can mark seperate groups by surrounding them with brackets (), in most situations you will have to escape them by () and access these parts in the replace pattern by using #, where # is the number of the subgroup starting from 1. Here's a simple example:
echo "ScreenShotXYZ.png" | sed "s/ScreenShot\(.*\)\.png/\1.png/"
Here, the XYZ is matched by the expression in brackets and can be accessed using \1 in the replacment string. The whole pattern in thus replaced by XYZ.png.
So use your regexp for matching, put brackets around the relevant blocks and do something like
ss.\1.\2.(and so on)
for your replacement pattern. There's still some way to optimize the process by first using sed to replace dashes by dots, then grouping the whole time block in just one pattern but for a start it's easier to code like that.
Consider the following:
var="text more text and yet more text"
echo $var | egrep "yet more (text)"
It should be possible to get the result of the regex as the string: text
However, I don't see any way to do this in bash with grep or its siblings at the moment.
In perl, php or similar regex engines:
$output = preg_match('/yet more (text)/', 'text more text yet more text');
$output[1] == "text";
Edit: To elaborate why I can't just multiple-regex, in the end I will have a regex with multiple of these (Pictured below) so I need to be able to get all of them. This also eliminates the option of using lookahead/lookbehind (As they are all variable length)
egrep -i "([0-9]+) +$USER +([0-9]+).+?(/tmp/Flash[0-9a-z]+) "
Example input as requested, straight from lsof (Replace $USER with "j" for this input data):
npviewer. 17875 j 11u REG 8,8 59737848 524264 /tmp/FlashXXu8pvMg (deleted)
npviewer. 17875 j 17u REG 8,8 16037387 524273 /tmp/FlashXXIBH29F (deleted)
The end goal is to cp /proc/$var1/fd/$var2 ~/$var3 for every line, which ends up "Downloading" flash files (Flash used to store in /tmp but they drm'd it up)
So far I've got:
#!/bin/bash
regex="([0-9]+) +j +([0-9]+).+?/tmp/(Flash[0-9a-zA-Z]+)"
echo "npviewer. 17875 j 11u REG 8,8 59737848 524264 /tmp/FlashXXYOvS8S (deleted)" |
sed -r -n -e " s%^.*?$regex.*?\$%\1 \2 \3%p " |
while read -a array
do
echo /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
It cuts off the first digits of the first value to return, and I'm not familiar enough with sed to see what's wrong.
End result for downloading flash 10.2+ videos (Including, perhaps, encrypted ones):
#!/bin/bash
lsof | grep "/tmp/Flash" | sed -r -n -e " s%^.+? ([0-9]+) +$USER +([0-9]+).+?/tmp/(Flash[0-9a-zA-Z]+).*?\$%\1 \2 \3%p " |
while read -a array
do
cp /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
Edit: look at my other answer for a simpler bash-only solution.
So, here the solution using sed to fetch the right groups and split them up. You later still have to use bash to read them. (And in this way it only works if the groups themselves do not contain any spaces - otherwise we had to use another divider character and patch read by setting $IFS to this value.)
#!/bin/bash
USER=j
regex=" ([0-9]+) +$USER +([0-9]+).+(/tmp/Flash[0-9a-zA-Z]+) "
sed -r -n -e " s%^.*$regex.*\$%\1 \2 \3%p " |
while read -a array
do
cp /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
Note that I had to adapt your last regex group to allow uppercase letters, and added a space at the beginning to be sure to capture the whole block of numbers. Alternatively here a \b (word limit) would have worked, too.
Ah, I forget mentioning that you should pipe the text to this script, like this:
./grep-result.sh < grep-result-test.txt
(provided your files are named like this). Instead you can add a < grep-result-test after the sed call (before the |), or prepend the line with cat grep-result-test.txt |.
How does it work?
sed -r -n calls sed in extended-regexp-mode, and without printing anything automatically.
-e " s%^.*$regex.*\$%\1 \2 \3%p " gives the sed program, which consists of a single s command.
I'm using % instead of the normal / as parameter separator, since / appears inside the regex and I don't want to escape it.
The regex to search is prefixed by ^.* and suffixed by .*$ to grab the whole line (and avoid printing parts of the rest of the line).
Note that this .* grabs greedy, so we have to insert a space into our regexp to avoid it grabbing the start of the first digit group too.
The replacement text contains of the three parenthesed groups, separated by spaces.
the p flag at the end of the command says to print out the pattern space after replacement. Since we grabbed the whole line, the pattern space consists of only the replacement text.
So, the output of sed for your example input is this:
5 11 /tmp/FlashXXu8pvMg
5 17 /tmp/FlashXXIBH29F
This is much more friendly for reuse, obviously.
Now we pipe this output as input to the while loop.
read -a array reads a line from standard input (which is the output from sed, due to our pipe), splits it into words (at spaces, tabs and newlines), and puts the words into an array variable.
We could also have written read var1 var2 var3 instead (preferably using better variable names), then the first two words would be put to $var1 and $var2, with $var3 getting the rest.
If read succeeded reading a line (i.e. not end-of-file), the body of the loop is executed:
${array[0]} is expanded to the first element of the array and similarly.
When the input ends, the loop ends, too.
This isn't possible using grep or another tool called from a shell prompt/script because a child process can't modify the environment of its parent process. If you're using bash 3.0 or better, then you can use in-process regular expressions. The syntax is perl-ish (=~) and the match groups are available via $BASH_REMATCH[x], where x is the match group.
After creating my sed-solution, I also wanted to try the pure-bash approach suggested by Mark. It works quite fine, for me.
#!/bin/bash
USER=j
regex=" ([0-9]+) +$USER +([0-9]+).+(/tmp/Flash[0-9a-zA-Z]+) "
while read
do
if [[ $REPLY =~ $regex ]]
then
echo cp /proc/${BASH_REMATCH[1]}/fd/${BASH_REMATCH[2]} ~/${BASH_REMATCH[3]}
fi
done
(If you upvote this, you should think about also upvoting Marks answer, since it is essentially his idea.)
The same as before: pipe the text to be filtered to this script.
How does it work?
As said by Mark, the [[ ... ]] special conditional construct supports the binary operator =~, which interprets his right operand (after parameter expansion) as a extended regular expression (just as we want), and matches the left operand against this. (We have again added a space at front to avoid matching only the last digit.)
When the regex matches, the [[ ... ]] returns 0 (= true), and also puts the parts matched by the individual groups (and the whole expression) into the array variable BASH_REMATCH.
Thus, when the regex matches, we enter the then block, and execute the commands there.
Here again ${BASH_REMATCH[1]} is an array-access to an element of the array, which corresponds to the first matched group. ([0] would be the whole string.)
Another note: Both my scripts accept multi-line input and work on every line which matches. Non-matching lines are simply ignored. If you are inputting only one line, you don't need the loop, a simple if read ; then ... or even read && [[ $REPLY =~ $regex ]] && ... would be enough.
echo "$var" | pcregrep -o "(?<=yet more )text"
Well, for your simple example, you can do this:
var="text more text and yet more text"
echo $var | grep -e "yet more text" | grep -o "text"