Update: Oops, i actually meant to say: Is there anyway to do this same thing on the values of a hash?

Do you mean just the values themselves or the keys based on the values? Here is some code for both cases:

my %hash = (
    4 => 20,    # second highest key
    2 => 40,    # second highest value
    3 => 30,
    1 => 50,
    5 => 10,
);

# second highest value
print ((sort values %hash)[-2]);

# key based on the second highest value
print ((sort { $hash{$b} <=> $hash{$a} } keys %hash)[1]);
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L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
B--B--B--B--B--B--B--B--
H---H---H---H---H---H---
(the triplet paradiddle with high-hat)

use warnings;
use strict;

my @list = (4, 7, 1, 3, 2, 9);
my $second = (sort @list)[-2];
print $second;
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Prints:

7
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Of course if @list has only one element you may get an unexpected result . . . :)

my @arr = qw/ 1 9 8 7 6 2 3 4 5/;
print ( (sort @arr)[-2] );
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If the list (or hash) is large, first doing a sort is a Really Bad Idea^TM given that this is an O(n log n) operation. As choedebeck's code above illustates it can be done in time linear in the size of the list/hash, i.e., O(n). This is a significant difference for large values of n.

Hope this helps, -gjb-

sub getSecondLargest
{
  my (@list) = @_;

  my ($largest, $secondlargest);

  for(my $i = 0;$i<=$#list;$i++)
  {
    if($i == 0)
    {
      if($list[0] >= $list[1])
      {
        ($largest, $secondlargest) = ($list[0], $list[1]);
      }
      else
      {
        ($largest, $secondlargest) = ($list[1], $list[0]);
      }
    }
    if($i >=3)
    {
      if($list[$i] >= $largest)
      {
        $secondlargest = $largest;
        $largest = $list[$i];
      }
      elsif($list[$i] >= $secondlargest)
      {
        $secondlargest = $list[$i];
      }  
    }
  }
  return $secondlargest;
}
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Even if it was in C, I'd still call qsort to get the answer, rather than write a lot of code {grin}.

-- Randal L. Schwartz, Perl hacker
Be sure to read my standard disclaimer if this is a reply.

Non-sorted hash values or array compatible version:

use warnings;
use strict;

my %hash = (4 => '1', 5 => '5', 7 => '2', 1 => '7', 10 => '9', 3 => '0
+', 2 => '3', 9 => '8');

print GetSecondLargest(values %hash);

sub GetSecondLargest
{
my @largest = (shift);

while (@_)
  {
  my $new = shift;
  if ($new >= $largest [-1])
    {
    unshift @largest, $new;
    next;
    }
    
  $largest[1] = $new if ! $#largest or $new > $largest[1];
  }
  
return $#largest ? $largest[1] : undef;
}
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Prints:

# code to get second highest number without sort.
#!/usr/bin/perl

use strict ;

my @arr = qw(4 4 0 -1 -2 2 0 1) ;

my $f = $arr[0] ;
my $s = $arr[0] ;

for (my $i=1 ; $i<@arr; $i++) {
    if ($arr[$i] > $f) {
        $s = $f ;
        $f = $arr[$i] ;
    } elsif (($arr[$i] > $s || (($f == $s))) && $arr[$i]!=$f) {
        $s = $arr[$i] ;
    }
}
if ($f == $s) {
    print "\n there is no second greatest number" ;
} else {
    print "\n second highest number : $s" ;
}
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Hi,

A simplest way u can try...

@arr = (1..10);
print sort($arr[$#arr]-1);
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Regards
Franklin

Well the code above prints value-1 of the last element not the value of the next to last element. In an array of sequential numbers the result may be the same ie element 9 contains value 9.

So the print statement becomes



print sort($arr[$#arr-1]);
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?