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I have an input file with several thousand lines. I am interested in one section of it, which contains the first instance of pattern /m.o./ in the file. I need to search for this pattern, run my code, then stop the code before any other patterns of m.o. or other lines.

It looks something like this:

 >>>>> -0.2834320000 -0.9672660000 0.0000000000 6.0 C
 m.o. irrep orbital orbital orbital
 energy (a.u.) energy (e.v.) occupancy
========================================================
 1 1 -20.63710689 -561.5697 2.0000
 2 1 -20.58909944 -560.2634 2.0000
 3 1 -11.45645851 -311.7491 2.0000
 4 1 -11.29965696 -307.4823 2.0000
 5 1 -11.29203148 -307.2748 2.0000
 6 1 -1.44555716 -39.3360 2.0000
 7 1 -1.35738379 -36.9367 2.0000
 8 1 -1.07586111 -29.2760 2.0000
 9 1 -0.91591305 -24.9235 2.0000
 10 1 -0.75492584 -20.5428 2.0000
 11 1 -0.71126523 -19.3547 2.0000
 12 1 -0.70828880 -19.2737 2.0000
 13 2 -0.62802299 -17.0895 2.0000
 14 1 -0.61775719 -16.8102 2.0000
 15 2 -0.50208166 -13.6625 2.0000
 16 1 -0.49193707 -13.3864 2.0000
 17 1 -0.43731872 -11.9002 2.0000
 18 2 -0.43546575 -11.8497 2.0000
 19 2 0.07335689 1.9962 0.0000

Goals

1. Begin 3 lines below the pattern /m.o./ (where $1=1).


2. Count how many times $2 is not equal to "1" (in other files, $2 can also be 3 or 4, so I need to count by $2 != 1).


3. This counting must be in the range of lines where $3 is negative, ie. up until the second line from the bottom.

The pattern /====/ is not possible to use as it appears earlier in the document.

• The output should be 3. On the range of lines where $3 is negative, there are 3 lines where $2 is not equal to 1.

Attempt

I've searched other answers on the web, which have provided me parts of the code to use. Examples:

• 
  

  define my starting line as pattern plus 3 (source):

  
  
  

  awk '/m.o./n=NR+3n
  

  
  


• 
  

  in between the starting and final line, count how many times $2 != "1" (source)

  
  
  

  awk '$2!="1"++count
  

  
  


• 
  

  define my final line as follows:

  
  
  

  awk '{if ($3 > 0)print count; exit
  

  
  

but I don't know how to put all of this together. Importantly, I must somehow avoid counting the extra 2 in $2 on the bottom line.

I'm of course open to rewriting the above code. I just wanted to supply examples for clarity.

Thank you.

There's many ways to do this but the simplest to understand might be as follows:

You can either create a complex conditional to select rows to count:

awk 'BEGIN total=0 NR > 3 && $2 != 1 && $3 < 0 total++ END print total ' 

Or you can put the conditional inside the code block:

awk 'BEGIN total=0 NR > 3 if ( $2 != 1 && $3 < 0 ) total++ END print total ' 

You can try this awk :

awk '$1=="m.o."if(l)exit;l++;nextl&&l<3l++;nextlif($3<0&&$2!=1)c++ENDprint c' infile

whew I finally figured it out with the following line:

 awk '$1 ~ /m.o./ n=NR+3n && $3+0 > 0 n=0 if ( n != 0 && $2 != "1" && $3+0 < 0) count++; END print count ' input

The problem before is that each statement seemed to be acting on the entire document independently, so I couldn't enforce the condition to work only within a range of lines, which led it to counting many other lines that I didn't want to be counted. I kept getting values greater than the correct answer of 3.

For example, using flags -- which seemed to be a common solution on the web for this problem -- the flags didn't seem to activate at the appropriate lines, or the counting was happening outside of the line range permitted by the flags. It was counting lines that weren't even part of my pattern. Inian coded to exclude lines with the >>>> pattern (which were returning a count match for whatever reason), but there were other patterns being mismatched, and it wasn't reasonable to find them all with 20k lines in the document.

This is what finally worked for me.

 $1 ~ /m.o./ n=NR+3n

this set the script to begin at the first instance where $1 contained "m.o.". I needed to specify $1 in order to avoid the second pattern occurrence of m.o. in the script. Fortunately the second instance was in $2, so I avoided it by matching only for $1. I don't know how to avoid it if both were in the same column.

At the point of match, n is defined as the line number (NR) plus 3 in the brackets, then recorded somehow by adding it again outside the bracket. In this way, I seem to be able to use awk to start at a pattern plus an arbitrary number of lines.

 && $3+0 > 0 n=0

This allows me to end the range of lines according to a variable condition rather than matching a pattern (many other solutions on the web match a defined string pattern using /pattern/ to define the end of the line range, which I could not figure out how to adapt here).

The && I believe maintains the pattern match from previously to bind the starting point, then for any point afterward in the document where $3 > 0 (my condition), n becomes zero.

Finally, I have a way to bind the starting and ending lines.

I can now apply my desired function within that range, which is to count lines according to a condition.

 if ( n != 0 && $2 != "1" && $3+0 < 0) count++; 

I remain within my line range by invoking the first term: If n is not zero, which is only the case between my pattern match and the conditions I set. Within this line range, the script pulls lines where $2 is not 1 and $3 is negative. It increases my count variable by 1 for each instance.

 END print count ' input

At the end of the script, it prints the summed up variable count for my input file.

Hope that helps someone later eventually. Thanks to @Inian and the linked sources for the help getting here.