Requirement: To get the count of directories under the input directory that matches the following criteria
use strict;
use File::Find;
my ($inputdir) = @ARGV;
my (@branches, $branch, $directory, @directories);
my $count = 0;
find(\&wanted, $inputdir);
while ( defined($directory = shift @directories) ) {
if (-d $directory){
next if ($directory =~ "DIR1" || $directory =~ "DIR2" || $directory =~ "DIR3");
push @branches, $directory;
$count++;
}
}
print "Total number of directories: $count \n";
sub wanted{
push @directories, $File::Find::name;
return @directories;
}
This piece of code is giving the required output but it's taking quite a lot of time.
Please suggest ways to reduce the time taken by improving this code.
The File::Find::Rule can skip whole branches altogether
use warnings;
use strict;
use File::Find::Rule;
my $start_dir = shift || '.';
my $re_skip = qr/DIR(?:1|2|3)/;
my $ok = File::Find::Rule->directory; # add selection rules as needed
my $skip = File::Find::Rule->directory
->name(qr/$re_skip/)
->prune
->discard;
my @dirs = File::Find::Rule -> any($skip, $ok) -> in($start_dir);
print "Total: ", scalar @dirs, "\n";
This still has to take some time with a large filesystem but it will be much better.
In a one-liner, if all you need from this is just a quick count
perl -MFile::Find::Rule -wE'
$ffr = File::Find::Rule;
$skip = $ffr->directory->name(qr/DIR(?:1|2|3)/)->prune->discard;
say scalar $ffr->any($skip, $ffr->directory)->in(".")'
where I've consolidated some of the code from the script.
The next step would be to use multi-threaded execution (I'd use fork
here). Group subdirectories so that they are roughly balanced in their sub-counts and run something like the above in parallel over those groups. The gain will depend on your hardware but there should be a good speedup factor.