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• perldoc -q array (http://perldoc.perl.org/perlfaq4.html#Data:-Arrays)
• Write some code.
• Post again when you have specific questions about your code

Welcome skalman,

Maybe a hash is what you want. Build the hash so that the names from 'array 2' are the keys and the key value is the counter, i.e. (untested):

my %hash = ();
for my $no ( 0..$#array2 )
{
      if ( $array2[$no] ) { $hash{ $array2[$no] }++; }
}
[download]

foreach my $key ( sort keys %hash )
{    print "$key:$hash{$key}\n"; }
[download]

UPDATE:Fixed as 'perl' Laurent_R :-)

Regards...Ed

This should be reading array2 (where there are multiple occurrences of the users) rather than array1 for populating the counter hash, shouldn't it?
Je suis Charlie.

Laurent_R,

You are correct, as usual!

Regards...Ed

"Well done is better than well said." - Benjamin Franklin

I would probably do it like this:

use strict;
use warnings;

my @array1 = qw/ jed jethro zed zug /;
my @array2 = qw/ jed jed jed zack zed zug zug /;

print join(':', @$_), "\n" for count_occurrences( \@array1, \@array2 )
+;

sub count_occurrences {
    my( $wanted, $have ) = @_;
    my %counter;
    @counter{@$wanted} = (0)x@$wanted;
    $counter{$_}++ for @$have;
    return map { [ $_ => $counter{$_} ] } @$wanted;
}
[download]

The output is:

jed:3
jethro:0
zed:1
zug:2
[download]

Notice how this method first sets counters to zero for all elements in @array1. Then it adds a count of those names in @array2 that appeared in @array1. Then a data structure is returned, and we print it.

The nuance here is that if a name appears in @array2 but not in @array1, it won't show up in the final count. After reading your question a couple of times, I got the impression that this behavior would be desirable. Another nuance is that the output will appear in the original order of names from @array1.