I have a file with lines :
1
apple
2
orange
3
banana
4
mango
5
coconut
I have a list of patterns :
1
3
5
I want to delete the lines (and one additional line after) using the list of patterns, in order to get :
2
orange
4
mango
Since I have a large number of patterns that I could not type manually one by one in a single command, a list must be used.
sed + paste solution:
sed -E "/\<(`paste -s -d'|' patterns`)\>/{N;d;}" file
The output:
2
orange
4
mango
paste -s -d'|' patterns - merge lines of file patterns using | as a dilimiter to get the regex alternation group (1|3|5)
N - append the next line to the pattern space
d - delete lines of pattern space
10 lemon as well.
sed versions supporting \< \>.
Let's do some grep option juggling:
You could grep for all lines in your fruit file except for those in your list file with numbers to be excluded:
grep -v -x -f numbers fruits
(-f to give a pattern file, -x to prevent partly matches and -v to invert the match)
Unfortunally, the numbers and the fruits are on separate lines, so the fruits survive the grep. Let's get rid of them with an additional -e pattern:
grep -v -x -e "[a-z]*" -f numbers fruits
Now all lines consisting of letters only are removed, so no fruits anymore, not even the wanted ones. But we can recover them from the context:
grep -v -x -A1 -e "[a-z]*" -f numbers fruits
(-A1 gives an additional line after every match) finally gives what you wanted.
Try this .
The pattern_file has all your patterns and the rest content in file_content. Here I tried grep with option -A 1 which can change depending on number of lines after patter match.
#!/bin/bash
while read pattern
do
for line_num in `grep $pattern file_content -A 1 -n | awk -F":|-" '{print $1}' | paste -sd ","`
do
sed -i "${line_num}d" file_content
done
done < pattern_file
