Delete n1 previous lines and n2 lines following with respect to a line containing a pattern - bash

Question

sed -e '/XXXX/,+4d' fv.out
I have to find a particular pattern in a file and delete 5 lines above and 4 lines below it simultaneously. I found out that the line above removes the line containing the pattern and four lines below it.
sed -e '/XXXX/,~5d' fv.out
In sed manual it was given that ~ represents the lines which is followed by the pattern. But when i tried it, it was the lines following the pattern that was deleted.
So, how do I delete 5 lines above and 4 lines below a line containing the pattern simultaneously?

Answer 1

One way using sed, assuming that the patterns are not close enough each other:
Content of script.sed:
## If line doesn't match the pattern...
/pattern/ ! {
## Append line to 'hold space'.
H
## Copy content of 'hold space' to 'pattern space' to work with it.
g
## If there are more than 5 lines saved, print and remove the first
## one. It's like a FIFO.
/\(\n[^\n]*\)\{6\}/ {
## Delete the first '\n' automatically added by previous 'H' command.
s/^\n//
## Print until first '\n'.
P
## Delete data printed just before.
s/[^\n]*//
## Save updated content to 'hold space'.
h
}
### Added to fix an error pointed out by potong in comments.
### =======================================================
## If last line, print lines left in 'hold space'.
$ {
x
s/^\n//
p
}
### =======================================================
## Read next line.
b
}
## If line matches the pattern...
/pattern/ {
## Remove all content of 'hold space'. It has the five previous
## lines, which won't be printed.
x
s/^.*$//
x
## Read next four lines and append them to 'pattern space'.
N ; N ; N ; N
## Delete all.
s/^.*$//
}
Run like:
sed -nf script.sed infile

Answer 2

A solution using awk:
awk '$0 ~ "XXXX" { lines2del = 5; nlines = 0; }
nlines == 5 { print lines[NR%5]; nlines-- }
lines2del == 0 { lines[NR%5] = $0; nlines++ }
lines2del > 0 { lines2del-- }
END { while (nlines-- > 0) { print lines[(NR - nlines) % 5] } }' fv.out
Update:
This is the script explained:
I remember the last 5 lines in the array lines using rotatory indexes (NR%5; NR is the record number; in this case lines).
If I find the pattern in the current line ($0 ~ "XXXX; $0 being the current record: in this case a line; and ~ being the Extended Regular Expression match operator), I reset the number of lines read and note that I have 5 lines to delete (including the current line).
If I already read 5 lines, I print the current line.
If I do not have lines to delete (which is also true if I had read 5 lines, I put the current line in the buffer and increment the number of lines. Note how the number of lines is decremented and then incremented if a line is printed.
If lines need to be deleted, I do not print anything and decrement the number of lines to delete.
At the end of the script, I print all the lines that are in the array.
My original version of the script was the following, but I ended up optimizing it to the above version:
awk '$0 ~ "XXXX" { lines2del = 5; nlines = 0; }
lines2del == 0 && nlines == 5 { print lines[NR%5]; lines[NR%5] }
lines2del == 0 && nlines < 5 { lines[NR%5] = $0; nlines++ }
lines2del > 0 { lines2del-- }
END { while (nlines-- > 0) { print lines[(NR - nlines) % 5] } }' fv.out
awk is a great tool ! I strongly recommend that you find a tutorial on the net and read it. One important thing: awk works with Extended Regular Expressions (ERE). Their syntax is a little different from Standard Regular Expression (RE) used in sed, but all that can be done with RE can be done with ERE.

Answer 3

The idea is to read 5 lines without printing them. If you find the pattern, delete the unprinted lines and the 4 lines bellow. If you do not find the pattern, remember the current line and print the 1st unprinted line. At the end, print what is unprinted.
sed -n -e '/XXXX/,+4{x;s/.*//;x;d}' -e '1,5H' -e '6,${H;g;s/\n//;P;s/[^\n]*//;h}' -e '${g;s/\n//;p;d}' fv.out
Of course, this only works if you have one occurrence of your pattern in the file. If you have many, you need to read 5 new lines after finding your pattern, and it gets complicated if you again have your pattern in those lines. In this case, I think sed is not the right tool.

Answer 4

This might work for you:
sed 'H;$!d;g;s/\([^\n]*\n\)\{5\}[^\n]*PATTERN\([^\n]*\n\)\{5\}//g;s/.//' file
or this:
awk --posix -vORS='' -vRS='([^\n]*\n){5}[^\n]*PATTERN([^\n]*\n){5}' 1 file
a more efficient sed solution:
sed ':a;/PATTERN/,+4d;/\([^\n]*\n\)\{5\}/{P;D};$q;N;ba' file

Answer 5

If you are happy to output the result to a file instead of stdout, vim can do it quite efficiently:
vim -c 'g/pattern/-5,+4d' -c 'w! outfile|q!' infile
or
vim -c 'g/pattern/-5,+4d' -c 'x' infile
to edit the file in-place.