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Posted by harryfmudd [AT] comcast [DOT] on December 9, 2006, 2:42 pm
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John W. Krahn wrote:
> DJ Stunks wrote:
>
>>Hi all, topical question here regarding Sort::Maker
>>
>>I have a poorly designed db table over which I have no control. The
>>data isn't normalized but for each row I'd like to return the column
>>numbers which have the highest number of counts, and the highest column
>>number with any counts at all. I refer to these values as mode and max
>>respectively, but please refer to my code and it's output below for a
>>better understanding. (I have provided sample input data and mimicked
>>the fetchrow_hashref function provided by DBI)
>>
>>Both sorts seem klunky, and after the discussion about the efficiency
>>gains given by Sort::Maker in another thread I'd like to take advantage
>>if I can (the sorts must be performed for each row returned by the
>>database).
>>
>>I'm not sure how to pass in the information in the $hashref to
>>make_sorter. Advice would be greatly appreciated.
>>
>>TIA,
>>-jp
>>
>>Code:
>> #!/usr/bin/perl
>>
>> use strict;
>> use warnings;
>>
>> print "id\tmode\tmax\n";
>> while ( my $hashref = fetchrow_hashref() ) {
>> my $id = $hashref->;
>>
>> # mode = col number with greatest count
>> my $mode = ( map { $_->[0] }
>> sort { $b->[1] <=> $a->[1] }
>> map { [ $_, $hashref-> ] } 1..5 )[0];
>>
>> # max = highest col number with any count
>> my $max = ( sort { $b <=> $a }
>> map { $_ if $hashref-> > 0 } 1..5 )[0];
>>
>> $max = 1 if $max eq ''; #all zeros
>
>
>
> If you want efficiency then you should just use loops instead of sorting:
>
>
> my ( $mode, $curr ) = ( 1, 0 );
> for ( 1 .. 5 ) {
> if ( $hashref->{ "col$_" } > $curr ) {
> $curr = $hashref->{ "col$_" };
> $mode = $_;
> }
> }
>
> my $max = 1;
> for ( reverse 1 .. 5 ) {
> if ( $hashref->{ "col$_" } > 0 ) {
> $max = $_;
> last;
> }
> }
>
>
>
> John
Have you looked at List::Util? It's a standard module in Perl 5.8, and
contains a max function. So the first loop becomes something like
use List::Util qw;
my $mode = max (map } 1..5);
Assuming the only zeros are trailing, the second loop could be
my $max = grep {$_ > 0} map } 1..5
since the value of a list in scalar context is the number of elements in
the list. If you have zeros embedded, this won't give the right answer,
since what it really does is find the number of values > 0, not the
highest index with a value > 0.
Of course, for either of these you might want to make the hashref into a
real list, to avoid mapping twice.
Tom Wyant
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