nefertiti has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks, I was looking for a solution to my problem of getting the top x matches for each error category. I already have the entire table stored in an array - @dataarray. I used a hash to get the count for each category. Colum 3 of the array holds the error code. The below is a simple code showing how I get the count for each category. 
my %error = (
    'disk' => 0,
    'cpu' => 0,
    'hdd' => 0,
    'graphics' => 0,
    'keyboard' => 0,
    '' => 0 #no error code
    );

for my $i (0 .. $#dataarray) {
    $error{$dataarray[$i][3]}++;
}

[download]
Column 0 and column 1 holds the computer location and name (respectively). They are unique only as a pair. Can any experts advice how I can get the top x matches for each category? In addition, I also need the error occurance count for each of the top x computer location/name. Thanks. Nefertiti

Replies are listed 'Best First'.
Re: Complex sorting and counting question.
by Corion (Patriarch) on Sep 18, 2008 at 14:40 UTC
[reply]
      Hi Monks, Below is the code provided by someone. What I do'nt get is the %code = map onwards. I'm not sure what's he's trying to get. I have difficulty understanding this part. Thanks for the help. 
      my $query = $db->prepare("select code, username, userid from transacti
+ons");
  $query->execute();
  my %code = map { $_ => [] } 0 .. 13;
  while (my $r = $qeury->fetchrow_hashref()) {
    my $c = $$r{code};
    push @{$code{$c}}, $r;
  }
  # For each code $c, $code{$c} is now an arrayref which
  # contains the relevant records (as hashrefs).  The
  # number of elements in the array equals the number
  # of occurances.  Now, for the second part...
  my %topfive;
  for my $c (keys %code) {
      my %user;
      for my $row (@{$code{$c}}) {
         # Note: this assumes that the userid field
         # is unique to each user.
         $user{$$row{userid}}{count}++;
         $user{$$row{userid}}{name} = $$row{username};
      }
      $topfive{$c} = [(sort {
        $$b[2] <=> $$a[2]
      } map {
        [$_, $user{$_}{name}, $user{$_}{count}]
      } keys %user)[0 .. 4]];     
  }

      [download]
[reply]
[d/l]
[select]

        Maybe you should ask the monk who wrote it directly then? That's why we have threaded replies here. Anyway:

        This prepares and executes an SQL statement, see DBI:

        my $query = $db->prepare("select code, username, userid from transacti
+ons");
  $query->execute();

        [download]

        This initializes the %code hash with the keys from 0 to 13. See perlop about the range operator, map and then Data::Dumper for how to inspect data structures.

          my %code = map { $_ => [] } 0 .. 13;

        [download]

        This line contains a typo that use strict; will catch for you. See strict and likely then Coping with Scoping.

          while (my $r = $qeury->fetchrow_hashref()) {

        [download]

        This loop overall fetches each retrieved hash and puts it into the corresponding array in %code:

            my $c = $$r{code};
    push @{$code{$c}}, $r;
  }
  # For each code $c, $code{$c} is now an arrayref which
  # contains the relevant records (as hashrefs).  The
  # number of elements in the array equals the number
  # of occurances.  Now, for the second part...

        [download]

        This code determines the top five userids.

          my %topfive;
  for my $c (keys %code) {
      my %user;
      for my $row (@{$code{$c}}) {
         # Note: this assumes that the userid field
         # is unique to each user.
         $user{$$row{userid}}{count}++;
         $user{$$row{userid}}{name} = $$row{username};
      }
      $topfive{$c} = [(sort {
        $$b[2] <=> $$a[2]
      } map {
        [$_, $user{$_}{name}, $user{$_}{count}]
      } keys %user)[0 .. 4]];     
  }

        [download]

        I already recommended pushing more work back into the database instead of manually sorting and counting.

[reply]
[d/l]
[select]

Back to Seekers of Perl Wisdom