How to capture the first occurrence of a pattern using grep after 'n' numbers of lines in a large size file ?
For instance,
I have 1000 lines of code in which 'wire' occur before and after 451st line.
I want to grab the first occurrence of wire after 451st line .
You can use sed's range expressions to perform this task easily. For example:
sed -n '452,$ { /wire/ {p;q} }' /tmp/foo
This will skip the first 451 lines, then scan each line until EOF for "wire." When found, it will print the pattern space and then quit.
Related
Suppose I have a file which contains:
something
line=1
file=2
other
lines
ignore
something
line=2
file=3
other
lines
ignore
Eventually, I want a unique list of the line and file combinations in each section. In the first stage I am trying to get sed to output just those lines combined into one line, like
line=1file=2
line=2file=3
Then I can use sort and uniq.
So I am trying
sed -n -r 's/(line=)(.*?)(\r)(file=)(.*?)(\r)/\1\2\4\5/p' sample.txt
(It isn't necessarily just a number after each)
But it won't match across the lines. I have tried \n and \r\n but it doesn't seem to be the style of new line, since:
sed -n -r 's/(line=)(.*?)(\r)/\1\2/p' sample.txt
Will output the "line=" lines, but I just can't get it to span the new line, and collect the second line as well.
By default, sed will operate only on chunks separated by \n character, so you can never match across multiple lines. Some sed implementations support -z option which will make it to operate on chunks separated by ASCII NUL character instead of newline character (this could work for small files, assuming NUL character won't affect the pattern you want to match)
There are also some sed commands that can be used for multiline processing
sed -n '/line=/{N;s/\n//p}'
N command will add the next line to current chunk being processed (which has to match line= in this case)
s/\n//p then delete the newline character, so that you get the output as single line
If your input has dos style line ending, first convert it to unix style (see Why does my tool output overwrite itself and how do I fix it?) or take care of \r as well
sed -n '/line=/{N;s/\r\n//p}'
Note that these commands were tested on GNU sed, syntax may vary for other implementations
I have a file with a lot of text, but I want to print only words that contain "#" at the beginning. Ex:
My name is #Laura and I live in #London. Name=#Laura. City=#London
How can I print all words that start with #?.I did this the following and it worked, but I want to do it using sed. I tried several patters, but I cannot make it print anything.
grep -o -E "#\w+" file.txt
Thanks
Use this sed command:
sed 's/[^#]*\(#[^ .]*\)/\1\n/g' file.txt
Explanation: we invoke the substitution command of sed. This has following structure: sed 's/regex/replace/options'. We will search for a regex and replace it using the g option. g makes sure the match is made multiple times per line.
We look for a series of non at chars followed by an # and a number of non-spaces #[^ ]*. We put this last part in a group \(\) and sub it during the replacement \1.
Note that we add a newline at the end of each match, you can also get the output on a single line by omitting the \n.
Both grep and sed handle input line-by-line and, as far as I know, getting either of them to handle multiple lines isn't very straightforward. What I'm looking for is an alternative or alternatives to these two programs that treat newlines as just another character. Is there any tool that fits such a criteria
The tool you want is awk. It is record-oriented, not line-oriented, and you can specify your record-separator by setting the builtin variable RS. In particular, GNU awk lets you set RS to any regular expression, not just a single character.
Here is an example where awk uses one blank line to separate every record. If you show us what data you have, we can help you with it.
cat file
first line
second line
third line
fourth line
fifth line
sixth line
seventh line
eight line
more data
Running awk on this and reconstruct data using blank line as new record.
awk -v RS= '{$1=$1}1' file
first line second line third line
fourth line fifth line sixth line
seventh line eight line
more data
PS RS is not equal to file, is set to RS= blank, equal to RS=""
1) Sed can handle a block lines together, not always line by line.
In sed, normally I use :loop; $!{N; b loop}; to get all the lines available in pattern space delimited by newline.
Sample:
Productivity
Google Search\
Tips
"Web Based Time Tracking,
Web Based Todo list and
Reduce Key Stores etc"
result (remove the content between ")
sed -e ':loop; $!{N; b loop}; s/\"[^\"]*\"//g' thegeekstuff.txt
Productivity
Google Search\
Tips
You should read this URL (Unix Sed Tutorial: 6 Examples for Sed Branching Operation), it will give you detail how it works.
http://www.thegeekstuff.com/2009/12/unix-sed-tutorial-6-examples-for-sed-branching-operation/
2) For grep, check if your grep support -z option, which needn't handle input line by line.
-z, --null-data
Treat the input as a set of lines, each terminated by a zero
byte (the ASCII NUL character) instead of a newline. Like the
-Z or --null option, this option can be used with commands like
sort -z to process arbitrary file names.
I want to get a list of lines in a batch file which are greater than 120 characters length. For this I thought of using sed. I tried but I was not successful. How can i achieve this ?
Is there any other way to get a list other than using sed ??
Thanks..
Another way to do this using awk:
cat file | awk 'length($0) > 120'
You can use grep and its repetition quantifier:
grep '.\{120\}' script.sh
Using sed, you have some alternatives:
sed -e '/.\{120\}/!d'
sed -e '/^.\{,119\}$/d'
sed -ne '/.\{120\}/p'
The first option matches lines that don't have (at least) 120 characters (the ! after the expression is to execute the command on lines that don't match the pattern before it), and deletes them (ie. doesn't print them).
The second option matches lines that from start (^) to end ($) have a total of characters from zero to 119. These lines are also deleted.
The third option is to use the -n flag, which tells sed to not print lines by default, and only print something if we tell it to. In this case, we match lines that have (at least) 120 characters, and use p to print them.
I have a file that, occasionally, has split lines. The split is signaled by the fact that the line starts with '+' (possibly preceeded by spaces).
line 1
line 2
+ continue 2
line 3
...
I'd like join the split line back:
line 1
line 2 continue 2
line 3
...
using sed. I'm not clear how to join a line with the preceeding one.
Any suggestion?
This might work for you:
sed 'N;s/\n\s*+//;P;D' file
These are actually four commands:
N
Append line from the input file to the pattern space
s/\n\s*+//
Remove newline, following whitespace and the plus
P
print line from the pattern space until the first newline
D
delete line from the pattern space until the first newline, e.g. the part which was just printed
The relevant manual page parts are
Selecting lines by numbers
Addresses overview
Multiline techniques - using D,G,H,N,P to process multiple lines
Doing this in sed is certainly a good exercise, but it's pretty trivial in perl:
perl -0777 -pe 's/\n\s*\+//g' input
I'm not partial to sed so this was a nice challenge for me.
sed -n '1{h;n};/^ *+ */{s// /;H;n};{x;s/\n//g;p};${x;p}'
In awk this is approximately:
awk '
NR == 1 {hold = $0; next}
/^ *\+/ {$1 = ""; hold=hold $0; next}
{print hold; hold = $0}
END {if (hold) print hold}
'
If the last line is a "+" line, the sed version will print a trailing blank line. Couldn't figure out how to suppress it.
You can use Vim in Ex mode:
ex -sc g/+/-j -cx file
g global search
- select previous line
j join with next line
x save and close
Different use of hold space with POSIX sed... to load the entire file into the hold space before merging lines.
sed -n '1x;1!H;${g;s/\n\s*+//g;p}'
1x on the first line, swap the line into the empty hold space
1!H on non-first lines, append to the hold space
$ on the last line:
g get the hold space (the entire file)
s/\n\s*+//g replace newlines preceeding +
p print everything
Input:
line 1
line 2
+ continue 2
+ continue 2 even more
line 3
+ continued
becomes
line 1
line 2 continue 2 continue 2 even more
line 3 continued
This (or potong's answer) might be more interesting than a sed -z implementation if other commands were desired for other manipulations of the data you can simply stick them in before 1!H, while sed -z is immediately loading the entire file into the pattern space. That means you aren't manipulating single lines at any point. Same for perl -0777.
In other words, if you want to also eliminate comment lines starting with *, add in /^\s*\*/d to delete the line
sed -n '1x;/^\s*\*/d;1!H;${g;s/\n\s*+//g;p}'
versus:
sed -z 's/\n\s*+//g;s/\n\s*\*[^\n]*\n/\n/g'
The former's accumulation in the hold space line by line keeps you in classic sed line processing land, while the latter's sed -z dumps you into what could be some painful substring regexes.
But that's sort of an edge case, and you could always just pipe sed -z back into sed. So +1 for that.
Footnote for internet searches: This is SPICE netlist syntax.
A solution for versions of sed that can read NUL separated data, like here GNU Sed's -z:
sed -z 's/\n\s*+//g'
Compared to potong's solution this has the advantage of being able to join multiple lines that start with +. For example:
line 1
line 2
+ continue 2
+ continue 2 even more
line 3
becomes
line 1
line 2 continue 2 continue 2 even more
line 3