【perl脚本】把01数据转化为基因型数据

use strict;
use warnings;

open DATA, "<data.txt" or die ("cannot open file:$!");
open OUT, ">out.txt";

my @num = (5,6,6,5,5,5);
my @type = ("A","B","C","D","E","F","G");

foreach my $tmp1 (<DATA>){
    chomp $tmp1;
    my @arr;
    @arr = split(/    /,$tmp1);
    print OUT $arr[0]."\t";
    my $order;
    my $o = 0;
    foreach my $tmp2 (@num){
        my $gen = "";
        foreach my $tmp3 (1...$tmp2){
            if ($arr[$o+$tmp3]){
                $gen = $gen.$type[$tmp3-1]
            }
        }
        if (length($gen) == 1){
            print OUT $gen."    ".$gen;
        }elsif(length($gen) == 2){
            print OUT substr($gen,0,1)."    ".substr($gen,1,1);
        }else{
        print OUT "error";
        }
        print OUT "\t";
        $o += $tmp2;
    }
    print OUT "\n";

}

close(DATA);
close(OUT);

原始数据，第一列是样本id，之后是各位点的有无，M1-M6的等位基因分别是5，6，6，5，5，5