unix:计数每个基因具有两个变体或一个纯合子变体的样本

Question

除了我有限的unix power工具技能之外，我还有一个编码问题。我希望通过以下两种方法来计算样本数量: i)基因中的纯合子变体(如下所示)；或者ii)基因中的两个变体(2xAB)。例如，来自：

Variant Gene    Sample1 Sample2 Sample3
   1    TP53    AA  BB  AB
   2    TP53    AB  AA  AB
   3    TP53    AB  AA  AA
   4    KRAS    AA  AB  AA
   5    KRAS    AB  AB  BB

我要找的是：

Gene Two_variants Homozygous Either
TP53     2            1        3
KRAS     1            1        2 

任何帮助都将不胜感激。谢谢。

R_G

Answer 1

在GNU awk中

awk '/\<AB\>.+\<AB\>/ { arr[$2,"AB"] += 1 }
             /\<BB\>/ { arr[$2,"BB"] += 1 }
                  END { for ( elt in arr ) {
                          split ( elt, index_parts, SUBSEP )
                          genes[index_parts[1]] = 0
                        }
                        printf "%4s%13s%11s%7s\n", "Gene", "Two_variants", "Homozygous", "Either"
                        for ( gene in genes ) {
                          printf "%4s%6d%13d%9d\n", gene, arr[gene,"AB"], arr[gene,"BB"], arr[gene,"AB"] + arr[gene,"BB"]
                        }
                      }' input.txt

Answer 2

use warnings;
use strict;
my (@header, %data);
open(my $file, "<", "input") or die("$?");
while (<$file>) {
    @header = split, next if not @header;
    my @v = split;
    $data{$v[1]}->{$_}++ for (@v[2..$#v]);
}
close $file;
print "Gene Two_variants Homozygous Either\n";
for my $k (keys %data) {
    my ($var2, $homoz) = (int($data{$k}{AB}/2), $data{$k}{BB});
    my $sum = $var2 + $homoz;
    printf("%4s %8d %9d %8d\n", $k, $var2, $homoz, $sum) if $sum;
}