Print Nth Line – How to Print Nth Line Before Each Matching Pattern

awksearchtext processing

I'm trying to print only the <N>th line before a search pattern. grep -B<N> prints all the <N> lines before the search pattern. I saw the awk code here that can print only the <N>th line after the search pattern.

awk 'c&&!--c;/pattern/{c=N}' file

How to modify this to print only the <N>th line before each line that matches pattern ? For example, here is my input file

...
...
   0.50007496  0.42473932  0.01527831
   0.99997456  0.97033575  0.44364198
Direct configuration=     1
   0.16929051  0.16544726  0.16608723
   0.16984300  0.16855274  0.50171112
...
...
   0.50089841  0.42608090  0.01499159
   0.99982054  0.97154975  0.44403547
Direct configuration=     2
   0.16931296  0.16553376  0.16600890
   0.16999941  0.16847055  0.50170694  
...

I need a command that can give me back the 2nd line before the search string Direct configuration.
I'm trying to run this in SUSE-Linux

Best Answer

A buffer of lines needs to be used.

Give a try to this:

awk -v N=4 -v pattern="example.*pattern" '{i=(1+(i%N));if (buffer[i]&& $0 ~ pattern) print buffer[i]; buffer[i]=$0;}' file

Set N value to the Nth line before the pattern to print.

Set patternvalue to the regex to search.

buffer is an array of N elements. It is used to store the lines. Each time the pattern is found, the Nth line before the pattern is printed.

Related Solutions

Text Processing – How to Print File Content Only if the First Line Matches a Pattern

You could do that with ed:

ed -s infile <<\IN 2>/dev/null
1s/PATTERN/&/
,p
q
IN

the trick here is to try to replace PATTERN on 1st line with itself. ed will error out if it can't find the specified pattern so ,p (print whole file) will only be executed if 1s/PATTERN/&/ is successful.

Or with sed:

sed -n '1{
/PATTERN/!q
}
p' infile

this quits if the first line does not (!) match PATTERN, otherwise it prints all lines.
Or, as pointed out by Toby Speight, with GNU sed:

sed '1{/PATTERN/!Q}' infile

Q is the same as q but it does not print the pattern space.

Find each line matching a pattern but print only the line above it

If the pattern cannot occur on consecutive lines you can simply run

sed '$!N;/.*\n.*PATTERN.*/P;D' infile

I've explained here how the N;P;D cycle works. The difference is that here the first line in the pattern space is printed only if the second one matches, otherwise it's deleted.

If the pattern can occur on consecutive lines the above solution will print a line that matches if it's followed by another line that matches.
To ignore consecutive matches add a second condition to print the first line in the pattern space only if it doesn't match:

sed '$!N;/.*\n.*PATTERN.*/{/.*PATTERN.*\n.*/!P;};D' infile

Another way, using the hold buffer.
If you want to ignore consecutive matches:

sed '/PATTERN/!{              # if line doesn't match PATTERN
h                             # copy pattern space content over the hold buffer
d                             # delete pattern space
}
//{                           # if line matches PATTERN
x                             # exchange pattern space with hold space
//d                           # if line matches PATTERN delete it
}' infile

or, in one line

sed '/PATTERN/!{h;d;};//{x;//d;}' infile

If you don't want to ignore consecutive matches:

sed '/PATTERN/!{              # if line doesn't match PATTERN
h                             # copy pattern space content over the hold buffer
d                             # delete pattern space
}
//x                           # if line matches PATTERN exchange buffers
' infile

or, in one line

sed '/PATTERN/!{h;d;};//x' infile

Though keep in mind the solutions that use the hold buffer will print a leading empty line if the first line in your file matches.

Best Answer

Related Solutions

Text Processing – How to Print File Content Only if the First Line Matches a Pattern

Find each line matching a pattern but print only the line above it

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