If you want averages over the hours, you have to use the hour as a hash key: 
#!/usr/bin/perl
use strict;
use warnings;
my %count;
my %sum;
while (<DATA>) {
    chomp;
    next if /^itemid/;
    my @F = split;
    my @J = split(/\:/,$F[2]);
    my $hour = $J[0];
    if ( $J[0] =~ /[0-9][0-9]/ ){
        $count{$hour}++;
        $sum{$hour} += $F[-1];
    };
}

for (sort keys %count ) {
    print "$_ ", $sum{$_} / $count{$_}, "\n";
}


__DATA__
itemid  FROM_UNIXTIME(clock)       value
26603   2010-10-09 00:00:23     38784360.8136
26603   2010-10-09 00:01:23     36529742.6667
26603   2010-10-09 00:02:23     36966880.8000
26603   2010-10-09 00:03:23     35666405.8667
26603   2010-10-09 01:05:23     35600190.6667
26603   2010-10-09 02:06:23     39175029.2000
26603   2010-10-09 03:07:23     35386478.4000
26603   2010-10-09 04:08:27     35773482.7500
26603   2010-10-09 05:09:23     39073367.5714
26603   2010-10-09 06:10:23     34379049.8667
26603   2010-10-09 07:11:24     33984045.7705
26603   2010-10-09 09:12:23     36276301.8305
26603   2010-10-09 10:13:23     37248366.0000

[download]

If the logfile entries are always sorted by hour, you don't even need the hash: 

use strict;
use warnings;
my $count = 0;
my $sum = 0;
my $previous_hour = -1;
while (<DATA>) {
    chomp;
    next if /^itemid/;
    my @F = split;
    my @J = split(/\:/,$F[2]);
    my $hour = $J[0];
    if ($hour != $previous_hour && $count) {
        print $previous_hour, " ", $sum / $count, "\n";
        $sum = 0;
        $count = 0;
    }
    $sum += $F[-1];
    $count++;
    $previous_hour = $hour;
}

[download]
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In reply to Re: Calculate hourly averages by moritz
in thread Calculate hourly averages by addyrocker

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