如何检查条件，然后使用awk在带有条件的行之前的行中打印特定字段？

Question

I have a file with the following output:

58.752391 0.000  1 1  6.152565 2.757839 14.558406 0.000000 2.156979  0.000000 0.000000  0 0 0  1
16.089417316313 0.000000000000 6.171292860915 2.757949885550  -150168 0
6.953218e-310 0.000000e+00 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0 0 0 0 0.000000   0.000000 0.000000  0
0.000000 0.000000 0 0 0 0 0.000000   0.000000 0.000000  0
-1.000000 -1.000000
0 

14034.172996 0.000  13 13  1.107936 1.107936 -1.000000 -1.000000 -1.000000  23.670258 34.172995  0 0 0  0
3085.963203076240 0.667625281751 10.905159250868 8.915904022910  -150168 639
6.953218e-310 0.000000e+00 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
3.120454 8.844320 8 1 127.895 0 23.670258   1.107936 0.000000  0
1.107936 8.390152 13 5 1e+50 0 34.172995   1.107936 0.000000  0
-1.000000 -1.000000
3 CE1(5-1;8-1) SN1 SN2 

and so on...

I would like to check whether the two numbers in the 6th line (and then every 8th line) are different than -1.000000. If so, depending on which (the first, second or both) one is different, I'd like to produce the following output:

1) The first number is different than -1.000000:

this different number, 4 lines before this number - the third column (in this example 6.171292860915 if the number was different than -1.000000), 2 lines before this number - 8th column (in this example 0.000000), the same line - 7th column, the line after this number

2) The second number is different than -1.000000:

this different number, 4 lines before this number - the fourth column, 1 line before this number - 8th column, the same line - 7th column, the line after this number

3) The first and second number are different than -1.000000:

Output from 1) Output from 2)

I know how to use sed to extract these lines where -1.000000 in my example are. Now I think I should include some awk's ifs in my sed command to check whether the line contains -1.000000 or other numbers and then print what I need. I have no idea, however, how to refer to lines before that which is being examined by sed. I'd be grateful for any help or clues. Thank you!

Answer 1

Perl to the rescue:

#!/usr/bin/perl
use warnings;
use strict;

# Read in "paragraph mode".
$/ = q();

# Auto add newlines.
$\ = "\n";

while (<>) {
    my @lines = map [ split ], split /\n/;  # Create an array of arrays.
    if (-1 != $lines[5][0]) {
        print join ' ', $lines[5][0], $lines[1][2], $lines[3][7],
                        $lines[3][6], @{ $lines[6] };
    }
    if (-1 != $lines[5][1]) {
        print join ' ', $lines[5][1], $lines[1][3], $lines[4][7],
                        $lines[4][6], @{ $lines[6] };
    }
}

Answer 2

In awk, the variable NR holds the line number, so the expression

NR % 8 == 6 { ..... }

will select lines 6, 14, 22 and so on. You only need a counter, but variables are automatically initialized to zero, so you get the sequence number (1 for line 6, 2 for line 14 and so on) with an expression like this

++seqno

Hope that helps....

Answer 3

one more solution in awk

awk '{if(NR%8==2){a=$3;b=$4};if(NR%8==4){a=$8","$7","a};if(NR%8==5){b=$8","$7","b};if(NR%8==6){c=$1;d=$2};if(NR%8==7){if(c!=-1.000000)print a,$0;if(d!=-1.000000)print b,$0 }}' inputfilename

if I break the above command as below

awk '{
     if(NR%8==2){a=$3;b=$4};
     if(NR%8==4){a=$8","$7","a};
     if(NR%8==5){b=$8","$7","b};
     if(NR%8==6){c=$1;d=$2};
     if(NR%8==7){
               if(c!=-1.000000)print a,$0;
               if(d!=-1.000000)print b,$0 
                }
     }' inputfilename

I am storing the output details in a and b from the beginning while reading file line by line like NR%8==2 is line 2 , NR%8==4 is line 4 of file and so on. At the 7th line (NR%8==7), I am checking for values c and d stored from line 6 (NR%8==6), if c and d has mismatch values we will print the output with 7th line content.