"We do have some duplicate directories so finding dupes is the next problem."

If you change this line in ++Athanasius' code:

$dirs{$File::Find::dir} = 1;
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to

++$dirs{$File::Find::dir};
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The code will run the same but now you'll have a count. You can then find duplicates like this (untested):

my @dup_dirs = grep { $dirs{$_} > 1 } keys %dirs;
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Hi Ken I tried your example but it wouldn't output anything. If I removed the pattern match string it does work but I'm not sure what it is printing out - the directory listing seems random.

#!/usr/bin/perl
# dirpathdupes

use strict;
use warnings;
use File::Find;
use Fcntl;
#*****************Path Variables**********************
our $wellpath = 'N:\\repos\\open\\Wells\\Regulated\\';
our $surveypath = 'N:\\repos\\open\\Surveys\\Regulated\\';
our $testpath = 'C:\\Temp\\';
#******************************************************* 


my %dirs;
find(\&dir_names, $testpath);
my @dup_dirs = grep { $dirs{$_} > 1 } keys %dirs;
#print "$_\n" for sort keys %dirs;
foreach my $l (@dup_dirs) {
    print "$l\n";
    }

sub dir_names
{
    # skip over everything that is not a directory
    return unless -d $File::Find::name;

    # skip over directories that don't match required pattern
    return unless $File::Find::dir =~ /[IPD]\d{8}$/;
    ++$dirs{$File::Find::dir};
}
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"Hi Ken I tried your example but it wouldn't output anything."

The technique I showed should work fine. In his response below, Athanasius has highlighted the issue with your original premise (i.e. parent vs. current directory). You can still use the technique I provided, you'll just need to work it into the code fix he's shown.

"If I removed the pattern match string it does work but I'm not sure what it is printing out - the directory listing seems random."

Hashes are unordered: keys %hash_name will return a list of keys in an apparently random order. If you're interested, see the "Hash Algorithm" section of "perlsec: Algorithmic Complexity Attacks" for more details.

sort may provide the ordering you want. If not, you may want to consider an array, or perhaps a more complex data structure, instead of a hash, to store your data. See "perldsc - Perl Data Structures Cookbook".

-- Ken