如何仅匹配模式的前N个实例，然后在每个模式后面打印行直到空白行？

Question

I have a log file summarising calculation results that I need to prepare for analysis. Each result is given a heading, of the form:

 Excited State   1:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000"

Followed by an unknown number of data lines of the form:

"76 -> 81  0.36917" 

(an integer, an arrow, another integer, then a float). Each result is separated from the next result by a blank line. I want to be able to take the first two sets (including the data lines) of results where the heading contains the pattern "Triplet". Later, I need to be able to do the same for the "Singlet" pattern, so I can't just delete those.

Unfortunately, it is important for later analysis that the data lines be kept separated in some way, as I will need to order the data lines in decreasing order of magnitude (by the float column).

I have been able to use sed to return all instances of the Triplet headings and following data lines (until the blank line), as follows:

sed -n '/Triplet/,/^ *$/p' test.txt

But I don't know how to get only the first two instances.

Ideally, if the input file looks like the following:

 Excited State   1:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...

Excited State   2:      Singlet-A      3.3656 eV  379.43 nm  f=0.0029
76 -> 81         0.38068
76 ->101         0.10777
...

Excited State   3:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...
...

I'd like to be able to get:

Excited State   1:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...

Excited State   3:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...

And while, in this case, I could just remove the second data set, that won't generalise.

Answer 1

$ awk '/Triplet/ { n += 1 } n <= 2 && /Triplet/,/^ *$/' input.txt
 Excited State   1:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...

Excited State   3:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...
...

Answer 2

A gnu awk version (gnu due to RS with multiple characters)

awk -v RS='Excited State' '/Triplet/ {if (n++<2) printf "%s",RS$0}' file
Excited State   1:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...

Excited State   3:      Triplet-A      3.1118 eV  398.43 nm  f=0.0000
76 -> 81         0.36917
76 ->101         0.11911
...
...

• RS='Excited State' set record selector to Excited State so awk works in block mode
• /Triplet/ test if line contains Triplet if so:
• • if (n++<2) test if counter is less then two starting by zero to get two block only, then:
• • • print RS$0 print record selector and block

PS this will work even if blank line is missing between blocks

Answer 3

This might work for you (GNU sed):

sed -E '/Triplet/{x;s/^/x/;/^x{1,2}$/{x;:a;n;/\S/ba;p;x};x};d' file

Focus on a line containing Triplet and after incrementing a counter in the hold space, determine if to print that line upto and including an empty one.

Answer 4

如果所有记录之间都有空行，则可以轻松执行以下操作：

$ awk 'BEGIN{RS="";FS=OFS="\n";n=2}($1~/Triplet/ && n-->0);(n==0){exit}' file