perl: pattern matching for comparision pair of string - regex

Question

I have a problem to count the majority selectedresult for each pair of string. my code: it seems just count if user choose either sysA, sysB or both without considering the pair of string. I'm also have a problem to make multiple comparision and deal with 7 users for each pair.
( $file = <INFILE> ) {
#field = parse_csv($file);
chomp(#field);
#query = $field[1];
for($i=0;$i<#query;++$i) {
if ( ($field[2] eq $method) || ($field[3] eq $method)){
if ( $field[4] eq $field[2]) {
print "$query[$i]: $field[2], $field[3], $field[4]\n";
$counta++;
}
if ( $field[4] eq $field[3]) {
print "$query[$i]: $field[2], $field[3]: $field[4]\n";
$countb++;
}
if ( $field[4] eq ($field[2] && $field[3])) {
#print "$query[$i]: $field[2]$field[3]\n";
$countc++;
}
data: for each query, i have 3 different combination of string comparision.
comparison("lucene-std-rel","lucene-noLen-rr");
comparison("lucene-noLen-rr","lucene-std-rel");
comparison("lucene-noLen-rr","random");
comparison( "random", "lucene-noLen-rr");
comparison("lucene-noLen-rr","lucene-nolen-rel");
Comparison("lucene-nolen-rel","lucene-noLen-rr");
example data for one pair (7 users evaluate for each pair):
user1,male,lucene-std-rel,random,lucene-std-relrandom
user2,male,lucene-std-rel,random,lucene-std-rel
user3,male,lucene-std-rel,random,lucene-std-rel
user4,male,lucene-std-rel,random,lucene-std-rel
user5,male,lucene-std-rel,random,lucene-std-relrandom
user6,male,lucene-std-rel,random,lucene-std-rel
user7,male,lucene-std-rel,random,lucene-std-rel
example output required: query 1:male fitness models
lucene-std-rel:5, random:0, both:2 ---> majority:lucene-std-rel
any help is very much appreciated.

Answer 1

Well, without making this more complex than you requested, here is what I came up with as a possible approach.
#!/usr/bin/perl
use strict;
my %counter = ( "A" => 0, "B" => 0, "AB" => 0, "majority" => 0);
while(<DATA>){
chomp;
next unless $_;
my ($workerId,$query,$sys1,$sys2,$resultSelected) = split(',');
$counter{$resultSelected}++;
}
$counter{'majority'} = (sort {$counter{$b} <=> $counter{$a}} keys %counter)[0];
print "A: $counter{'A'} B: $counter{'B'} both(AB): $counter{'AB'} majority: $counter{'majority'}\n";
__END__
user1,male,A,B,A
user2,male,A,B,AB
user3,male,A,B,B
user4,male,A,B,A
user5,male,A,B,A
The output of this is:
A: 3 B: 1 both(AB): 1 majority: A
I don't feel like my example to you fully addresses the idea of there being more than one type with the "majority". For instance, if both A and B are 9, I'd expect them both to be listed there. I didn't bother to do that since you didn't ask, but hopefully this will get you along the right path.

Answer 2

open( INFILE, "compare.csv" ) or die("Can not open input file: $!");
while ( $file = <INFILE> ) {
#field = parse_csv($file);
chomp(#field);
#query = $field[1];
for($i=0;$i<#query;++$i) {
if ( ($field[2] eq $method) || ($field[3] eq $method)){
if ( $field[4] eq $field[2]) {
print "$query[$i]: $field[2], $field[3], $field[4]\n";
$counta++;
}
if ( $field[4] eq $field[3]) {
print "$query[$i]: $field[2], $field[3]: $field[4]\n";
$countb++;
}
if ( $field[4] eq ($field[2] && $field[3])) {
#print "$query[$i]: $field[2]$field[3]\n";
$countc++;
}
}
}
sub parse_csv {
my $text = shift;
my #new = ();
push( #new, $+ ) while $text =~ m{
"([^\"\](?:\.[^\"\])*)",?
| ([^,]+),?
| ,
}gx;
push( #new, undef ) if substr( $text, -1, 1 ) eq ',';
return #new;
}`