awk or grep? - print keyword and following columns

Question

I did some benchmarks and the output is a single file containing multiple results with the following format

[blaaa]
1st run

T/V   N  NB  P  Q   Time
==========================
2     3   4  1  2   1.02
[blaaaa...]
2nd run
T/V   N  NB  P  Q   Time
==========================
4     42 4  1  2    1.22

I want only to obtain the runtimes of the different runs I made. So is it possible to grep "Time" and print the next but one column? Or even to print out only

1.02
1.22

for my example above? I would prefer printing out only the three result columns without the blaa part :D

Answer 1

Assuming Time is always in the sixth field:

$ awk '$6 == "Time" {t = NR} t && NR == t+2 {print $6}' file
1.02
1.22

Alternatively you could use getline twice - quick'n'dirty:

awk '$6 == "Time" {getline; getline; print $6}' file

("dirty" because it doesn't check the getline return value).

Answer 2

$ grep -o 'Time\|[0-9\.]\+$' file
Time
1.02
Time
1.22
$ grep -o '[0-9\.]\+$' file
1.02
1.22
$ grep -A2 ^T/V file
T/V   N  NB  P  Q   Time
==========================
2     3   4  1  2   1.02
--
T/V   N  NB  P  Q   Time
==========================
4     42 4  1  2    1.22

The results are separated by -- in the last grep which can be removed with:

$ grep -A2 ^T/V file | grep -v '^--'

And with the ==== lines removed:

$ grep -A2 ^T/V file | grep -v '^--\|=\+'

Answer 3

You got two answers using awk and grep, you can also try sed:

sed -E '/===*/{N;s/.*([0-9]+\.[0-9]*)/\1/g};t;d' file_name

Here /===*/, will match ===.., then it will go to next line and find floating point numbers having decimal in between them, and will skip and delete all other lines.

It will print output like:

1.02
1.22