foreach my $i ($#$arr)
{
    $avg = ($i + ($avg ? $avg : $i)) / 2;
}
[download]

This only runs once, with $i taking the value of the last index of $arr. You probably meant @$arr.

The next problem is that this isn't an average in the normal sense, it's a weighted average where the last element gets much more weight than the first, more like:

$avg = (...(($arr->[0]/2 + $arr->[1])/2 + $arr->[2])/2 + ... $arr->[$#
+$arr])/2;
[download]

So perhaps you meant a decaying average, but it's not clear from the context. Here's something equivalent, with output:

my @x = (10, 7, 5, 10, 50, 70, 75, 72, 79, 80);

sub avg {
    my @arr = @_; 
    my $avg; 
    foreach my $i (@arr) {
        print "$i: $avg\n"; 
        $avg = ($i + ($avg ? $avg : $i))/2;
    }
    return $avg;
}

print avg(@x),"\n";

10:       # undef
7: 10
5: 8.5
10: 6.75
50: 8.375
70: 29.1875
75: 49.59375
72: 62.296875
79: 67.1484375
80: 73.07421875
76.537109375

print avg(reverse @x),"\n";

80:       # undef
79: 80
72: 79.5
75: 75.75
70: 75.375
50: 72.6875
10: 61.34375
5: 35.671875
7: 20.3359375
10: 13.66796875
11.833984375
[download]

Making a change to produce the uniformly weighted average:

my @x = (10, 7, 5, 10, 50, 70, 75, 72, 79, 80);

sub avg2 {
    my @arr = @_; 
    my $sum; 
    foreach my $i (@arr) {
        print "$i: $sum\n"; 
        $sum += $i;
    }
    return $sum/@arr;
}

print avg(@x),"\n";

10:       # undef
7: 10
5: 17
10: 22
50: 32
70: 82
75: 152
72: 227
79: 299
80: 378
45.8

print avg(reverse @x),"\n";

80:       # undef
79: 80
72: 159
75: 231
70: 306
50: 376
10: 426
5: 436
7: 441
10: 448
45.8
[download]

where the forward and reverse lists produce the same average.

A lot of the syntax would be less cumbersome if you started with my @arr = (...).

Serious Error: "can't even get this to work" is not recognized as a valid error description.

Seriously, you'll need to tell us more about what you expected or want; why the output deviates from your heart's desire; and -- tho none appear when I test or execute your code -- error messages or warnings received from code you haven't shown.

Seriously? The code runs fine for me.

The expected output of the code or what I'd like. Since I'm not sure if I'm on the right track with the code, what I'd like are three sets with 5,7,10,10, 50, and 70,72,75,79,80.

I'm not sure how else to explain this - between the code I'm stuck on, the use case, and abstract....?

What is significant about those groups of numbers?

Perl can't mind read.

Michael

Not sure if this is any use but is does give the expected output. It loops through the sorted list working out the effect of adding each to 'current group average'. If change is greater than the value 'diff' it starts a new group

#!perl
use strict;

my @data = (10,7,5,10,50,70,75,72,79,80);
my @arr = sort {$a<=>$b} @data;
my $diff = 2; # group closeness

my @avg; #[count,sum]
my @grp;
my $g=0;

for my $i (0..$#arr){
  my $val = $arr[$i];
  #print "value $val\n";
  if ($i == 0){
    push @{$grp[$g]},$val;
    $avg[$g] = [1,$val];
  } else {
    #work out new average with this element
    my ($n,$sum) = @{$avg[$g]};
    #print "count $n sum $sum\n";
    my $avg = $sum/$n;
    my $new_avg = ($sum+$val)/($n+1);
    if (abs($new_avg - $avg) < $diff){
      # join group
      push @{$grp[$g]},$val;
      $avg[$g] = [$n+1,$sum+$val];
    } else {
      # start new group
      ++$g;
      push @{$grp[$g]},$val;
      $avg[$g] = [1,$val];
    }
  }
}

for (@grp) {
  print join ',',@$_,"\n";
}
[download]

POJ

-Scott

I want to see if someone sends a ton of packets that I DROP in a short time vs someone that, over time, ends up sending the same number (or more) DROPped packets (or DENY, etc). Technically, I'm looking at a time stamp and converting it to epoch, but the general theory holds up to what I presented.

Agreed... More info please!