Update, should take my own advice, misread your intention entirely. I didn't realise the "y co-ordinate" needed to be summed

Look at the structure of your data. You want to sort a list of numbers where each list is associated with a unique key What do you think is the data structure you need to put the data into?
Something like:
%records{$number_before_colon => @list_of_numbers_seperated_by_commas}
So read each file splitting the result into an index and an array and add the array to a hash on the index.

while(<INFILE>){
   chomp;
   ($index,@record)=split/[:,]/;
   push @{$records{$index}},@record;
}
[download]

When you need to do data conversion, half the battle is examining the relationship between the data structures. Now you can access the data in a sorted order:

for $index (sort{$a<=>$b}(keys (%records))){
        print $OUTFILE "$index: ",join(",",(sort{$a<=>$b}(@{$records{$
+index}}))),"\n";
}
[download]

Wrapping it up

#!/usr/bin/perl
use strict;
use warnings;
my @files=qw(temp.data temp1.data temp2.data);
my ($file,$index, %records,$INFILE,$OUTFILE);
for $file (@files){
        open($INFILE,"<","$file")|| die "Failed to open $file: $!";
        while(<$INFILE>){
                my @record;
                chomp;
                ($index,@record)=split/[:,]/;
                push @{$records{$index}},@record;
        }
        close $INFILE;
}

open ($OUTFILE,">","newfile.data")|| die "Failed to open newfile.data:
+ $!";
for $index (sort{$a<=>$b}(keys (%records))){
        print $OUTFILE "$index: ",join(",",(sort{$a<=>$b}(@{$records{$
+index}}))),"\n";
}
close $OUTFILE
[download]

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