Re^2: Looping Over Hash Skips an Element?

[mmartin]

in reply to Re: Looping Over Hash Skips an Element?
in thread Looping Over Hash Skips an Element?

Hey toolic, thanks for the reply.

Duhh haha... I guess I had a brain fart on that one..! lol thanks.

I was just using the printing as an example. What I'm reallying trying to do is to add 8 consecutive values from the hash and place that summed value in an array.

Using the same hash, would this be what I should do?

my @sum_values;
my $x = 0;

foreach my $key ( keys %ifSpeeds )
{
    if ($counter <= 8) {
        $sum_values[$x] += $ifSpeeds{ $key };
        $counter++;
    } else { 
        $x++;
        $sum_values[$x] = $ifSpeeds{ $key };
        $counter = 1;
    }
}

for (my $x = 0; $x <= $#sum_values; $x++)
{
    print "$x) $sum_values\n";
}

The final result (given the hash with 32 elements) would be that the array would contains 4 values, which include the following:

   [0] --> (sum of Gi3/1 to Gi3/8)
   [1] --> (sum of Gi3/9 to Gi3/16)
   [2] --> (sum of Gi3/17 to Gi3/24)
   [3] --> (sum of Gi3/25 to Gi3/32)

Does all that look correct to you?


Thanks Again,
Matt

[toolic]

How about:

use List::MoreUtils qw(natatime);
use List::Util      qw(sum);
use Data::Dumper;

my @sums;
my $it = natatime(8, (values %ifSpeeds));
while (my @vals = $it->()) {
    push @sums, sum(@vals);
}

print Dumper(\@sums);

__END__

$VAR1 = [
          '6200000000',
          '6200000000',
          '6200000000',
          '2600000000'
        ];

[Eliya]

    if ($counter <= 8) {
        $sum_values[$x] += $ifSpeeds{ $key };
        $counter++;
    } else { 
        $x++;
        $sum_values[$x] = $ifSpeeds{ $key };
        $counter = 1;
    }

To avoid the almost-duplication of the key statement, you could also write

    $sum_values[$x] += $ifSpeeds{ $key };
    $x++ unless ++$counter % 8;

or even simply

    $sum_values[$counter++/8] += $ifSpeeds{ $key };     # the [] impli
+es int(...)

(Perl does not generate a warning when you do something like $sum += ..., with $sum being undef initially — it treats it as 0)