How can I use awk to print a line only if its right half of line _doesn't_ match the previous line's right half?

Answer 1

You could use associative arrays to maintain a counter for each key on the right side.

This is a proof of a concept one liner that you can use as a starting point

$ echo "[100 ps] bar\n[139 ps] foo\n[140 ps] foo" |
  awk '{count[$3]++; if (count[$3] == 1) print;}'
[100 ps] bar
[139 ps] foo

This would have to be tweaked if the right side string can contain spaces.

Answer 2

Running on ideone:

 BEGIN {prev=""}

 $3==prev {next}

{ prev = $3;
 print;}

Answer 3

what separates the right half from the left half? Is it a tab or multiple spaces? If it's a tab then:

awk -F '\t' '
    $2 in seen {next} 
    { print; seen[$2]=1 }
'

Otherwise, I'd write something like

perl -ane '
    $right_half = join " ", @F[2..-1];
    if (not $seen{$right_half}) {
        print;
        $seen{$right_half} = 1;
    }
'

Answer 4

$ awk -F"][ \t]+" '!a[$2]++' file
[100 ps]  bar
[139 ps]  foo de fa fa
[149 ps]  le pamplemouse
[177 ps]  le pomme de terre