Remove all lines that don't end with specific string - regex

working on a large text file and I'd like to remove all lines that don't contain the text "event":"click"}]
I've tried to do some regex within Sublime 3 and can't get it to stick.

I have not used sublime but you could select all line not containing the text "event":"click"}] with the regex:
^(.(?!"event":"click"\}\]))*$
I think you could replace them by nothing(empty string) or backspace

Use this one to get result to stdout
sed -n '/"event":"click"\}\]$/p' your_large_file
Use this one to keep only lines that end with "event":"click"}], your_large_file.old backup will be generated
sed -i.old -n '/"event":"click"\}\]$/p' your_large_file

Related

How to use sed to search and replace a pattern who appears multiple times in the same line?

Because the question can be misleading, here is a little example. I have this kind of file:
some text
some text ##some-text-KEY-some-other-text##
text again ##some-text-KEY-some-other-text## ##some-text-KEY-some-other-text##
again ##some-text-KEY-some-other-text-KEY-text##
some text with KEY ##KEY-some-text##
blabla ##KEY##
In this example, I want to replace each occurrence of KEY- inside a pair of ## by VALUE-. I started with this sed command:
sed -i 's/\(##[^#]*\)KEY-\([^#]*##\)/\1VALUE-\2/g'
Here is how it works:
\(##[^#]*\): create a first group composed of two # and any characters except # ...
KEY-: ... until the last occurrence of KEY- on that line
\([^#]*##\): and create a second group with all the characters except # until the next pair of #.
The problem is my command can't handle correctly the following line because there are multiple KEY- inside my pair of ##:
again ##some-text-KEY-some-other-text-KEY-text##
Indeed, I get this result:
again ##some-text-KEY-some-other-text-VALUE-text##
If I want to replace all the occurrences of KEY- in that line, I have to run my command multiple times and I prefer to avoid that. I also tried with lazy operators but the problem is the same.
How can I create a regex and a sed command who can handle correctly all my file?
The problem is rather complex: you need to replace all occurrences of some multicharacter text inside blocks of text between identical multicharacter delimiters.
The easiest and safest way to solve the task is using Perl:
perl -i -pe 's/(##)(.*?)(##)/$end_delim=$3; "$1" . $2=~s|KEY-|VALUE-|gr . "$end_delim"/ge' file
See the online demo.
The (##)(.*?)(##) pattern will match strings between two adjacent ## substrings capturing the start delimiter into Group 1, end delimiter in Group 3, and all text in between into Group 2. Since the regex substitution re-sets all placeholders, the temporary variable is used to keep the value of the end delimiter ($end_delim=$3), then, "$1" . $2=~s|KEY-|VALUE-|gr . "$end_delim" replaces the match with the value in the Group 1 of the first match (the first ##), then the Group 2 value with all KEY- replaced with VALUE-, and then the end delimiter.
If there are no KEY-s in between matches on the same line you may use a branch with sed by enclosing your command with :A and tA:
sed -i ':A; s/\(##[^#]*\)KEY-\([^#]*##\)/\1VALUE-\2/g; tA' file
Note you missed the first placeholder in \VALUE-\2, it should be \1VALUE-\2.
See the online demo:
s="some KEY- text
some text ##some-text-KEY-some-other-text##
text again ##some-text-KEY-some-other-text## ##some-text-KEY-some-other-text##
again ##some-text-KEY-some-other-text-KEY-text##
some text with KEY ##KEY-some-text##
blabla ##KEY##"
sed ':A; s/\(##[^#]*\)KEY-\([^#]*##\)/\1VALUE-\2/g; tA' <<< "$s"
Output:
some KEY- text
some text ##some-text-VALUE-some-other-text##
text again ##some-text-VALUE-some-other-text## ##some-text-VALUE-some-other-text##
again ##some-text-VALUE-some-other-text-VALUE-text##
some text with KEY ##VALUE-some-text##
blabla ##KEY##
More details:
sed allows the usage of loops and branches. The :A in the code above is a label, a special location marker that can be "jumped at" using the appropriate operator. t is used to create a branch, this "command jumps to the label only if the previous substitute command was successful". So, once the pattern matched and the replacement occurred, sed goes back to where it was and re-tries a match. If it is not successful, sed goes on to search for the matches further in the string. So, tA means go back to the location marked with A if there was a successful search-and-replace operation.
This might work for you (GNU sed):
sed -E 's/##/\n/g;:a;s/^([^\n]*(\n[^\n]*\n[^\n]*)*\n[^\n]*)KEY-/\1VALUE-/;ta;s/\n/##/g' file
Convert ##'s to newlines. Using a loop, replace VAL- between matched newlines to VALUE-. When all done replace newlines by ##'s.

sed command to delete text until match is found for each line of a csv

I have a csv file and I am trying to delete all characters from the beginning of the line till it finds the first occurrence of "2015". I want to do this for each line in the csv file.
My csv file structure is as follows:
Field1 , Field2 , Field3 , Field4
sometext1 , 2015-07-15 , sometext2, sometext3
sometext1 , 2015-07-14 , sometext2, sometext3
sometext1 , 2015-07-13 , sometext2, sometext3
I cannot use the cut command or sed for the first occurrence of a comma because the text in the Field1 sometimes has commas in them too, which is making it complicated for parsing. I figured if I search for the first occurrence of the text 2015 for each line and replace all the preceding characters with nothing, then that should work.
FYI I only want to do this for the FIRST occurrence of 2015 only. There is another text field with 2015 in it within another column and I don't any text prior to that to be affected.
For example, if my original line is:
sometext1,#015,2015-07-10,sometext2,2015,sometext3
I want it to return:
2015-07-10,sometext2,2015,sometext3
Does anyone know the sed command to do this?
Any help will be appreciated!
Thanks
Here is a way to do it with sed assuming "#####" never occurs in a line:
sed -e 's/2015/#####&/'|sed -e 's/.*#####//'
For example:
> echo sometext1,#015,2015-07-10,sometext2,2015,sometext3\
|sed -e 's/2015/#####&/'|sed -e 's/.*#####//'
2015-07-10,sometext2,2015,sometext3
The first sed command prefixes "#####" to the first occurence of 2015 and the second sed command removes everything from the beginning to the end of the "#####" prefix.
The basic reason for using this two stage method is that sed's regular expression matcher has only greedy wildcards that always pick the longest match and does not support lazy matching which picks the shortest match.
If "#####" may occur in a line a more unlikely string could be substituted for it such as "7z#dNjm_wG8a3!esu#Rhv=".
To do this with sed without Perl-style non-greedy operators, you need to mark the first instance with something you know won't be in the line, as Tris describes. However, that solution requires knowledge of what won't be in the file. Fortunately, you can guarantee that a newline won't be in the line because that's what terminated the line. Thus you can do something like:
sed 's/2015/\n&/;s/.*\n//' input.txt > output.txt
NOTE: this won't modify the header row which you would have to treat specially.

sed deletes differing portions of text

I am trying to manipulate a vocabulary list which is in ZDT format, that is: Traditional Characters \t Simplified Characters \t Pinyin \t English \n. I want to get rid of the Traditional Characters at the beginning of the line, so I tried to delete them with sed 's/^[^\t]*\t//g' input.txt > output.txt yet this gets me nowhere near my desired result, as in some lines everything up to somewhere in the English section is deleted and in other lines nothing at all is deleted and I cannot make out a pattern.
I think that the RegEx is correct, as I’ve tested it here and Sublime Text 2 also works with it as expected. What is the problem here?
Edit:
Beginning of input.txt http://pastebin.com/fRemVPyT
Beginning of output.txt http://pastebin.com/EJkszFNF
Not all sed version likes \t. Try to use a literal tab character. You can create a bash variable containing a tab like this:
export TAB=$'\t'
Maybe like this:
sed "s/^[^$TAB]*$TAB//g" input.txt > output.txt

how to rejoin words that are split accross lines with a hyphen in a text file

OCR texts often have words that flow from one line to another with a hyphen at the end of the first line. (ie: the word has '-\n' inserted in it).
I would like rejoin all such split words in a text file (in a linux environment).
I believe this should be possible with sed or awk, but the syntax for these is dark magic to me! I knew a text editor in windows that did regex search/replace with newlines in the search expression, but am unaware of such in linux.
Make sure to back up ocr_file before running as this command will modify the contents of ocr_file:
perl -i~ -e 'BEGIN{$/=undef} ($f=<>) =~ s#-\s*\n\s*(\S+)#$1\n#mg; print $f' ocr_file
This answer is relevant, because I want the words joined together... not just a removal of the dash character.
cat file| perl -CS -pe's/-\n//'|fmt -w52
is the short answer, but uses fmt to reform paragraphs after the paragraphs were mangled by perl.
without fmt, you can do
#!/usr/bin/perl
use open qw(:std :utf8);
undef $/; $_=<>;
s/-\n(\w+\W+)\s*/$1\n/sg;
print;
also, if you're doing OCR, you can use this perl one-liner to convert unicode utf-8 dashes to ascii dash characters. note the -CS option to tell perl about utf-8.
# 0x2009 - 0x2015 em-dashes to ascii dash
perl -CS -pe 'tr/\x{2009}\x{2010}\x{2011}\x{2012\x{2013}\x{2014}\x{2015}/-/'
cat file | perl -p -e 's/-\n//'
If the file has windows line endings, you'll need to catch the cr-lf with something like:
cat file | perl -p -e 's/-\s\n//'
Hey this is my first answer post, here goes:
'-\n' I suspect are the line-feed characters. You can use sed to remove these. You could try the following as a test:
1) create a test file:
echo "hello this is a test -\n" > testfile
2) check the file has the expected contents:
cat testfile
3) test the sed command, this sends the edited text stream to standard out (ie your active console window) without overwriting anything:
sed 's/-\\n//g' testfile
(you should just see 'hello this is a test file' printed to the console without the '-\n')
If I build up the command:
a) First off you have the sed command itself:
sed
b) Secondly the expression and sed specific controls need to be in quotations:
sed 'sedcontrols+regex' (the text in quotations isn't what you'll actually enter, we'll fill this in as we go along)
c) Specify the file you are reading from:
sed 'sedcontrols+regex' testfile
d) To delete the string in question, sed needs to be told to substitute the unwanted characters with nothing (null,zero), so you use 's' to substitute, forward-slash, then the unwanted string (more on that in a sec), then forward-slash again, then nothing (what it's being substituted with), then forward-slash, and then the scale (as in do you want to apply the edit to a single line or more). In this case I will select 'g' which represents global, as in the whole text file. So now we have:
sed 's/regex//g' testfile
e) We need to add in the unwanted string but it gets confusing because if there is a slash in your string, it needs to be escaped out using a back-slash. So, the unwanted string
-\n ends up looking like -\\n
We can output the edited text stream to stdout as follows:
sed 's/-\\n//g' testfile
To save the results without overwriting anything (assuming testfile2 doesn't exist) we can redirect the output to a file:
sed 's/-\\n//g' testfile >testfile2
sed -z 's/-\n//' file_with_hyphens

How can I remove the text before and after a particular character?

I have been trying to remove the text before and after a particular character in each line of a text. It would be very hard to do manually since it contain 5000 lines and I need to remove text before that keyword in each line. Any software that could do it, would be great or any Perl scripts that could run on Windows. I run Perl scripts in ActivePerl, so scripts that could do this and run on ActivePerl would be helpful.
Thanks
I'd use this:
$text =~ s/ .*? (keyword) .* /$1/gx;
You don't need software, you can make this part of your existing script. Multiline regex replace along the lines of /a(b)c/ then you can backref b in the replacer with $1. Without knowing more about the text you're working with it's hard to guess what the actual pattern would be.
Presuming that you have the following:
text1 text2 keyword text3 text4 text5 keyword text6 text7
and what you want is
s/.*?keyword(.*?)keyword.*/keyword$1keyword/;
otherwise you can just replace the whole line with keyword
An example of the data may help us be clearer
I'd say, that if $text contains your whole text, you can do :
$text =~ s/^.*(keyword1|keyword2).*$/$1/m;
The m modifier makes ^ and $ see a beginning and an ending of line, and not the beginning and ending of the string.
Assuming you want to remove all text to the left of keyword1 and all text to the right of keyword2:
while (<>) {
s/.*(keyword1)/$1/;
s/(keyword2).*/$1/;
print;
}
Put this into a perl script and run it like this:
fix.pl original.txt > new.txt
Or if you just want to do this inplace, perhaps on several files at once:
perl -i.bak -pe 's/.*(keyword1)/$1/; s/(keyword2).*/$1/;' original.txt original2.txt
This will do inplace editing, renaming the original to have a .bak extension, use an implicit while-loop with print and execute the search and replace pattern before each print.
To be safe, verify it without the -i option first, or at the very least on only one file...