Ttykuu

Consider the following text file

one 1
two 2
three 3
four 4
five 5
six 6
seven 7
eight 8

I would like to access to the second line after the one which matched four. This would be the line

six 6

The resulting line (so the one above) would then be piped for further processing (say, a | cut -d' ' -f2).

Is there a way to do this in bash and other typical utilities? (otherwise I will script it in Python)

There is nothing wrong with the previous two answers, but I thought I would make you aware that finding the third line after a pattern can be done in a single sed call:

sed -n "/four/ n; n; p " SourceData.txt

Because a single program does the work, this is more efficient than running multiple filters. The above command outputs the third line after every instance of "four", except where this occurs again in one of the two lines following a match (the other solutions don't handle this case in the expected manner either); also, no output is generated if the pattern is in the last or second-last line of the file, which may or may not be what you want.

To match the first instance only:

sed -n "/four/ n; n; p; q " SourceData.txt

I add this solution because it is worth getting to know sed and, despite its rather off-putting syntax (regular expressions are bad enough!), it can often be extremely useful. This tutorial is a good introduction.

You can use this expression (input.txt):

grep "four" -A 2 input.txt | tail -n 1

Output is:

six 6

The grep option "-A 2" states that two lines after the matched line are outputted.

And the tail option "-n 1" states that only the last 1 lines of this result are returned.

For multiple occurences, and assumming that no lines start with --:

( grep -A 2 pattern data.txt; echo '--' ) | grep -E -B1 '^--' | grep -Ev '^--'

In slo-mo:

• 
  ( grep -A 2 pattern data.txt; echo '--' ) prints the pattern and the next two lines, and inserts a -- line between the groups. echo '--' makes sure that the last group is also followed by --.


• 
  grep -E -B1 '^--' print the separators and the lines just before (whoch are the ones we are looking for)


• 
  grep -Ev '^--' drops the separators, leaving only the lines we are looking for.

This solution prints the current line iff there was a match two lines ago. It is slightly different from other answers (so far) because it won't miss another match even if it occurs soon after the previous match.

awk -v delay=2 'for (i=delay; i>=0; i--) t[i]=t[i-1]; /four/ t[0]="m"; if (t[delay]) print'

Whenever there's a match, the information is stored in t[0]. With each line the t array is shifted (including shifting t[-1] to t[0] to reset the value of t[0]). The line is printed iff the array indicates there was a match two lines ago.

You can easily set a different delay (e.g. delay=7) or use another pattern (e.g. /sda[[:digit:]]/)