Have you consider an SQL answer to your problem is place of a Perl

Yes, I have. You can use order by e_id, cus, ta, compute sum(col4), sum(col5) by e_id, cus, ta. I want to know if you can do it in Perl. I know there is a way to achieve the shown data structure because I did it once before with a stock file. Unfortunately, I can't figure it out this time and I can't find the code.

I know there is a way to achieve the shown data structure because I did it once before with a stock file.

Perhaps How can I visualize my complex data structure? will help you with your data structures; it's what I use to help me with mine.

Unfortunately, I can't figure it out this time and I can't find the code.

Oh, I really do recommend some form or fashion of a Version Control System. I am constantly looking back at things (even very broken things) to find out how I accomplished a particular task, or how not to try to solve a given problem. Recently, I've even begun putting various notes (usually OpenOffice docs) in because, in my case, TortoiseSVN makes it so very easy to keep things in synch between my main work machine and the netbook I carry whenever I leave my home office.

HTH,

planetscape

Show us some code. Ideally a trivial database using something like the following as a starting point:

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

unlink 'results';
open my $resFile, '>', 'results';
print $resFile <<DATA;
Joe,A,1,85,90
Joe,A,1,80,99
Joe,A,2,50,70
Joe,A,2,60,65
Joe,A,2,87,89
Joe,B,1,82,92
Joe,B,3,30,51
Rob,A,1,85,90
Rob,A,1,80,99
Rob,A,2,50,70
Rob,A,2,60,65
Rob,A,2,87,89
Rob,B,1,82,92
Rob,B,3,30,51
DATA
close $resFile;

my $dbh = DBI->connect("DBI:CSV:") or die $DBI::errstr;
my $sth = $dbh->prepare("SELECT * FROM results");

$sth->execute();

my $arrayRef  = $sth->fetchall_arrayref;

print +(join ', ', @$_), "\n" for @$arrayRef;

exit;
[download]

Prints:

Joe, A, 1, 80, 99
Joe, A, 2, 50, 70
Joe, A, 2, 60, 65
Joe, A, 2, 87, 89
Joe, B, 1, 82, 92
Joe, B, 3, 30, 51
Rob, A, 1, 85, 90
Rob, A, 1, 80, 99
Rob, A, 2, 50, 70
Rob, A, 2, 60, 65
Rob, A, 2, 87, 89
Rob, B, 1, 82, 92
Rob, B, 3, 30, 51
[download]

I have updated the question with all of the code that I have. I am able to get the data results (5 columns) as shown but I cannot figure out a way to calculate the 4th and 5th column once ordered by the first three columns.

Something like:

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

my $tableName = 'hist';

unlink $tableName;
open my $resFile, '>', $tableName;
print $resFile <<DATA;
e_id,cus,ta,gd1,gd2
Joe,A,1,85,90
Joe,A,1,80,99
Joe,A,2,50,70
Joe,A,2,60,65
Joe,A,2,87,89
Joe,B,1,82,92
Joe,B,3,30,51
Rob,A,1,64,77
Rob,B,2,20,32
DATA
close $resFile;

my $dbh = DBI->connect("DBI:CSV:") or die $DBI::errstr;
my $sth = $dbh->prepare("SELECT * FROM $tableName");

$sth->execute();

my $query = $dbh->selectall_arrayref(
    "select e_id, cus, ta, gd1, gd2 from $tableName order by e_id, cus
+, ta",
    {Slice => {}});
my $gd1Total;
my $gd2Total;
my $currGroup;

for my $row (@$query, {map {$_, 0} qw(e_id cus ta gd1 gd2)}) {
    my $group = join ',', @{$row}{qw(e_id cus ta)};

    $currGroup = $group if !defined $currGroup;

    if ($group ne $currGroup) {
        print "  Total $gd1Total, $gd2Total\n";
        $gd1Total = $gd2Total = 0;
        $currGroup = $group;
        last if $group eq '0,0,0';
    }

    print +(join ', ', @{$row}{qw(e_id cus ta gd1 gd2)}), "\n";
    $gd1Total += $row->{gd1};
    $gd2Total += $row->{gd2};
}

exit;
[download]

Prints:

Joe, A, 1, 85, 90
Joe, A, 1, 80, 99
  Total 165, 189
Joe, A, 2, 50, 70
Joe, A, 2, 60, 65
Joe, A, 2, 87, 89
  Total 197, 224
Joe, B, 1, 82, 92
  Total 82, 92
Joe, B, 3, 30, 51
  Total 30, 51
Rob, A, 1, 64, 77
  Total 64, 77
Rob, B, 2, 20, 32
  Total 20, 32
[download]

You'll have to fix the 'total' calculation to get the results you show however. My sample code simply sums the values.

True laziness is hard work

use warnings;
use strict;
use Data::Dump qw(pp);

my %hash;
while (<DATA>) {
    chomp;
    next if /\A\s*\z/;
    my @pieces = split ',' => $_;
    for my $idx (3..4) {
        push @{
            $hash
            {$pieces[0]}
            {$pieces[1]}
            {$pieces[2]}
            [$idx-3]
        }, $pieces[$idx];
    }    
}
pp %hash;

__DATA__
Joe,A,1,85,90
Joe,A,1,80,99
Joe,A,2,50,70
Joe,A,2,60,65
Joe,A,2,87,89
Joe,B,1,82,92
Joe,B,3,30,51
Rob,A,1,64,77
Rob,B,2,20,32
[download]

(
  "Joe",
  {
    A => { 1 => [[85, 80], [90, 99]], 2 => [[50, 60, 87], [70, 65, 89]
+] },
    B => { 1 => [[82], [92]], 3 => [[30], [51]] },
  },
  "Rob",
  { A => { 1 => [[64], [77]] }, B => { 2 => [[20], [32]] } },
)
[download]

If your only reason is finding the sum of the 4th & 5th column ... then a simple sql query might do it for you.

Here's an example:

select e_id,cus, sum(gd1) as g1, sum(gd2) as g2 from test group by e_i
+d,cus,ta;
[download]

SELECT null, e_id, Cus, ta, gd1, gd2 FROM tabla1 t
union all
select 'Total', e_id, Cus, ta, sum(gd1), SUM(gd2) from tabla1 group by
+ e_id,Cus, ta
order by e_id,Cus, ta;
[download]

Thanks to all for the different answers. It works perfect.