[lug] Output every xth line?

[Brad Doctor]

Not a CS grad student? :)

Cut and paste this into a file, then use it like so:

cat <filename> | ./<script_name>.pl <start> <seq>

Remove the ($cnt) if you don't want the line number printed out.
---------------

#!/usr/bin/perl

my $start = shift||die "need starting line\n";
my $mult  = shift||die "need sequential line (every <nth>)\n";

my $cnt   = 0;
my $i_cnt = 0;

while (<STDIN>) {

        $cnt++;
        $i_cnt++;

        if ($cnt == $start) {
                print "($cnt) $_";
                $i_cnt = 0;
                next;
        } # if cnt == start

        if ($i_cnt == $mult) {
                print "($cnt) $_";
                $i_cnt = 0;
        } # if i_cnt == mult
} # while

---------------


> Folks, I need your Unix knowledge.  I have an input file that contains
> data that isn't marked nice enough that I can use grep to parse it.
> 
> Rather, I find I need a script that says "starting with line x, output
> that line and every yth line after it".  That is, start at line 5 and
> output line 5, 5+y, 5+2y, &c.  Can you think of a quick shell script
> that can do this?  My attempts have been rather laughable mainly because
> I am neither a perl hacker or a shell script hacker.  (I could make one
> in Fortran 95, but it wouldn't be very portable.)
> 
> Any help for the poor grad student?
> 
> Matt
> -- 
> I am a theoretical chemist.  Fear me!
>  Matt Thompson -- http://ucsub.colorado.edu/~thompsma/
>  440 UCB, Boulder, CO  80309-0440
>  JILA A510, 303-492-4662

-- 
Brad Doctor, CISSP