perl - 生成具有取代率的合成 DNA 序列

Question

鉴于这些输入:

my $init_seq = "AAAAAAAAAA" #length 10 bp 
my $sub_rate = 0.003;
my $nof_tags = 1000;
my @dna = qw( A C G T );

我要生成:

1. 一千个长度为 10 的标签

2. 标签中每个位置的替换率为 0.003

产生如下输出:

AAAAAAAAAA
AATAACAAAA
.....
AAGGAAAAGA # 1000th tags

在 Perl 中是否有一种紧凑的方法来做到这一点？

我坚持这个脚本的核心逻辑:

#!/usr/bin/perl

my $init_seq = "AAAAAAAAAA" #length 10 bp 
my $sub_rate = 0.003;
my $nof_tags = 1000;
my @dna = qw( A C G T );

    $i = 0;
    while ($i < length($init_seq)) {
        $roll = int(rand 4) + 1;       # $roll is now an integer between 1 and 4

        if ($roll == 1) {$base = A;}
        elsif ($roll == 2) {$base = T;}
        elsif ($roll == 3) {$base = C;}
        elsif ($roll == 4) {$base = G;};

        print $base;
    }
    continue {
        $i++;
    }

Answer 1

作为一个小的优化，替换:

    $roll = int(rand 4) + 1;       # $roll is now an integer between 1 and 4

    if ($roll == 1) {$base = A;}
    elsif ($roll == 2) {$base = T;}
    elsif ($roll == 3) {$base = C;}
    elsif ($roll == 4) {$base = G;};

与

    $base = $dna[int(rand 4)];