For a lot of stuff, perl's list munging capabilities are much better/straightforward than dealing with array indices.

The problem at hand is one of the few instances to NOT take advantage of perl's list-centric features and work with the indices directly.

-Greg

Can I use the same concept on a hash value? Let's say I want to perform the same operation on values of the hash and return the sorted keys with new data. For instance:

#!/usr/local/bin/perl
use strict;

        my %pvalues = ( 1=> 0.5453980,
                        2=> 0.4902384,
                        3=> 0.8167950,
                        4=> 0.2821822,
                        5=> 0.4693030,
                        6=> 0.6491767,
                        7=> 0.9802138,
                        8=> 0.1155778,
                        9=> 0.9585124,
                        10=> 0.4069490
                        );
[download]

They keys of course are unique but the values may not. I have done the following.

        my $numberPvalues = scalar keys %pvalues;
        my $numberPvaluesCounter = scalar keys %pvalues;
        my $numberPos = 0;
        
        # now loop through all p values and calculate FDR value
        # store FDR values in hash, key = pos, value is FDRq
        my @reversPvalues =();
        my %sortedTransfPvalues =();
        my @inversPvalue =();
        my @cumulminPvalue =();
        

 while ($numberPvaluesCounter >= 1 ){

             #Sort the original pvalue hash based on the values descen
+ding manner
             foreach my $posCurrentPvalue (sort {$pvalues{$b} <=> $pva
+lues{$a}} keys %pvalues)
                 {
                  #calculate the transformed pvalues based on its rank
+ in the sorted value in the hash
                  my $transformedPvalue = $pvalues{$posCurrentPvalue} 
+* $numberPvalues / $numberPvaluesCounter;
                        $numberPvaluesCounter--;
                        #new indcies
                        $numberPos++;
                        push (@inversPvalue, $transformedPvalue);
                         #$sortedTransfPvalues{$numberPos}{$posCurrent
+Pvalue} = $transformedPvalue;
                         #$sortedTransfPvalues{$posCurrentPvalue}{$num
+berPos} = $transformedPvalue;
                         print "$posCurrentPvalue\t$transformedPvalue\
+n";

                 }#end foreach and sorting values
        }#end while for count down
[download]

Can I use the same concept on a hash value?

Yes. In exactly analogous fashion. Order the keys by value and then process your values in that order:

#!/usr/local/bin/perl
use strict;
use Data::Dump qw[ pp ];

my %pvalues = (
    1=> 0.5453980,
    2=> 0.4902384,
    3=> 0.8167950,
    4=> 0.2821822,
    5=> 0.4693030,
    6=> 0.6491767,
    7=> 0.9802138,
    8=> 0.1155778,
    9=> 0.9585124,
    10=> 0.4069490
);

my @orderedKeys = sort {
    $pvalues{ $b } <=> $pvalues{ $a }
} keys %pvalues;

my $d = my $n = values %pvalues;
$pvalues{ $_ } *= $n / $d-- for @orderedKeys;

pp \%pvalues;

__END__
c:\test>junk68
{
  1  => "0.908996666666667",
  2  => "0.9804768",
  3  => "1.02099375",
  4  => "1.410911",
  5  => "1.1732575",
  6  => "0.927395285714286",
  7  => "0.9802138",
  8  => "1.155778",
  9  => "1.06501377777778",
  10 => "1.35649666666667",
}
[download]

Of course, hash keys aren't intrinsically ordered, so you'll have to order them as needed for output, but one of the structure dumpers will do that for you.

---

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

"I'd rather go naked than blow up my ass"

Thanks for helping to solve this problem. For my final implementation of p.adjust function from R in Perl I need to provide you the entire algorithm. Basically you have solved a lot of it. I do not have any problem with section (d), but based on what I was thinking and you implemented the order that I get at (c) is different from what it should be. </code>

(a) sort these values in an decreasing order

0.9802138 0.9585124 0.8167950 0.6491767 0.5453980 0.4902384 0.4693030 0.4069490 0.2821822 0.1155778

(b) assign order to each above observation in a decreasing order, i.e.,

10 9 8 7 6 5 4 3 2 1

(c) then calculate

0.9802138*10/10 0.9585124*10/9 0.8167950*10/8 ..... 0 .1155778*10/1 (note that: the 10 in the numerator corresponds to the total number of tests conducted in this toy example, you should change this to the corresponding actual number of tests when you do real data analysis. Also the order assignment will start with the total number of tests and decrease to 1.)

You will obtain the following

0.9802138 1.0650138 1.0209938 0.9273953 0.9089967 0.9804768 1.1732575 1.3564967 1.4109110 1.1557780

(d) then do the the cumulative minimum adjustment, i.e.

since 0.9802138<1.0650138, replace 1.0650138 with 0.9802138,

then among the first three values 0.9802138, 0.9802138, 1.0209938, 0.9802138 is smallest, replace 1.0209938 with 0.9802138

then among the first four values 0.9802138, 0.9802138, 0.9802138, 0.9273953, 0.9273953 is the smallest, and the fourth value will not be changed

similarly, the fifth value will not be changed and all the rest of values will be changed to 0.9089967

Finally you should have 0.9802138, 0.9802138, 0.9802138, 0.9273953 0.9089967 0.9089967 0.9089967 0.9089967 0.9089967 0.9089967

rearrange these value according to the original order of p-value,

0.5453980 0.4902384 0.8167950 0.2821822 0.4693030 0.6491767 0.9802138 0.1155778 0.9585124 0.4069490

[6]      [5]        [8]     [2]      [4]     [7]        [10]     [1]  
+        [9]     [3]
[download]

then you will obtain the following values

0.9089967 0.9089967 0.9802138 0.9089967 0.9089967 0.9273953 0.9802138 0.9089967 0.9802138 0.9089967

then you compare each of these values (in red) with 1, if it is greater than 1, then replace that value with 1. Since in this particular example, all the above values (in red) are less than 1, your final BH adjusted q-values are

0.9089967 0.9089967 0.9802138 0.9089967 0.9089967 0.9273953 0.9802138 0.9089967 0.9802138 0.9089967