Perl 如何创建多个列表的组合而不使用硬编码循环?
我有如下数据:Perl 如何创建多个列表的组合而不使用硬编码循环?,perl,algorithm,nested-loops,Perl,Algorithm,Nested Loops,我有如下数据: my @homopol = ( ["T","C","CC","G"], # part1 ["T","TT","C","G","A"], #part2 ["C","CCC","G"], #part3 ...upto part K=~50 ); my @prob = ([1.00,0.63,0.002,1.00,0
my @homopol = (
["T","C","CC","G"], # part1
["T","TT","C","G","A"], #part2
["C","CCC","G"], #part3 ...upto part K=~50
);
my @prob = ([1.00,0.63,0.002,1.00,0.83],
[0.72,0.03,1.00, 0.85,1.00],
[1.00,0.97,0.02]);
# Note also that the dimension of @homopol is always exactly the same with @prob.
# Although number of elements can differ from 'part' to 'part'.
我想做的是
part1
到partK
@prob
中找到相应元素的乘积T-T-C 1 x 0.72 x 1 = 0.720
T-T-CCC 1 x 0.72 x 0.97 = 0.698
T-T-G 1 x 0.72 x 0.02 = 0.014
...
G-G-G 1 x 0.85 x 0.02 = 0.017
G-A-C 1 x 1 x 1 = 1.000
G-A-CCC 1 x 1 x 0.97 = 0.970
G-A-G 1 x 1 x 0.02 = 0.020
问题是我下面的代码是通过硬编码实现的
循环。由于@homopol
的部件数量可以变化且较大
(例如~K=50),我们需要一种灵活紧凑的方法来获得相同的结果。有吗?
我在考虑使用,但不确定如何实现
use strict;
use Data::Dumper;
use Carp;
my @homopol = (["T","C","CC","G"],
["T","TT","C","G","A"],
["C","CCC","G"]);
my @prob = ([1.00,0.63,0.002,1.00,0.83],
[0.72,0.03,1.00, 0.85,1.00],
[1.00,0.97,0.02]);
my $i_of_part1 = -1;
foreach my $base_part1 ( @{ $homopol[0] } ) {
$i_of_part1++;
my $probpart1 = $prob[0]->[$i_of_part1];
my $i_of_part2 =-1;
foreach my $base_part2 ( @{ $homopol[1] } ) {
$i_of_part2++;
my $probpart2 = $prob[1]->[$i_of_part2];
my $i_of_part3 = -1;
foreach my $base_part3 ( @{ $homopol[2] } ) {
$i_of_part3++;
my $probpart3 = $prob[2]->[$i_of_part3];
my $nstr = $base_part1."".$base_part2."".$base_part3;
my $prob_prod = sprintf("%.3f",$probpart1 * $probpart2 *$probpart3);
print "$base_part1-$base_part2-$base_part3 \t";
print "$probpart1 x $probpart2 x $probpart3 = $prob_prod\n";
}
}
}
为什么不使用递归呢?将深度作为一个参数传递,让函数在循环内以深度+1调用自身。您可以通过创建一个与@homopol数组长度相同的标记数组(N say)来实现,以跟踪您正在查看的组合。实际上,这个数组就像一个
以N为基数的数字,元素为数字。迭代的方式与以N为基数写下连续数字的方式相同,例如(0 0…0),(0 0…1),…,(0 0…N-1),(0 0…1 0),…方法1:从指数计算 计算homopol中长度的乘积(长度1*长度2*…*长度n)。然后,将i从零迭代到乘积。现在,您需要的索引是i%length1,(i/length1)%length2,(i/length1/length2)%length3 方法2:递归 我被打败了,看尼基的回答。:-) 在不更改输入数据的情况下使用的解决方案如下所示:
use Algorithm::Loops;
# Turns ([a, b, c], [d, e], ...) into ([0, 1, 2], [0, 1], ...)
my @lists_of_indices = map { [ 0 .. @$_ ] } @homopol;
NestedLoops( [ @lists_of_indices ], sub {
my @indices = @_;
my $prob_prod = 1; # Multiplicative identity
my @base_string;
my @prob_string;
for my $n (0 .. $#indices) {
push @base_string, $hompol[$n][ $indices[$n] ];
push @prob_string, sprintf("%.3f", $prob[$n][ $indices[$n] ]);
$prob_prod *= $prob[$n][ $indices[$n] ];
}
print join "-", @base_string; print "\t";
print join "x", @prob_string; print " = ";
printf "%.3f\n", $prob_prod;
});
但我认为,通过将结构更改为类似的结构,实际上可以使代码更清晰
[
{ T => 1.00, C => 0.63, CC => 0.002, G => 0.83 },
{ T => 0.72, TT => 0.03, ... },
...
]
因为没有并行数据结构,您可以简单地迭代可用的基序列,而不是迭代索引,然后在两个不同的位置查找这些索引。我建议,这将创建一个迭代器来生成所有集合的叉积。因为它使用迭代器,所以不需要预先生成每个组合;相反,它会根据需要生成每一个
use strict;
use warnings;
use Set::CrossProduct;
my @homopol = (
[qw(T C CC G)],
[qw(T TT C G A)],
[qw(C CCC G)],
);
my @prob = (
[1.00,0.63,0.002,1.00],
[0.72,0.03,1.00, 0.85,1.00],
[1.00,0.97,0.02],
);
# Prepare by storing the data in a list of lists of pairs.
my @combined;
for my $i (0 .. $#homopol){
push @combined, [];
push @{$combined[-1]}, [$homopol[$i][$_], $prob[$i][$_]]
for 0 .. @{$homopol[$i]} - 1;
};
my $iterator = Set::CrossProduct->new([ @combined ]);
while( my $tuple = $iterator->get ){
my @h = map { $_->[0] } @$tuple;
my @p = map { $_->[1] } @$tuple;
my $product = 1;
$product *= $_ for @p;
print join('-', @h), ' ', join(' x ', @p), ' = ', $product, "\n";
}
@霍布斯:你的方法也创造了不必要的成对和单一组合(例如T-T,T-TT,T-)。我们有什么方法可以修改它吗?
NestedLoops
接受一个可选的过滤器子项,它允许您控制代码将被调用的组合。但是默认情况下,它应该做与问题中的原始代码相同的事情,所以我不确定它应该做什么。您可以使用while(my$tuple=$iterator->next)
来避免将所有这些内容都放在内存中。@FM&Brian:您的新修复程序给出了错误的结果。我得到了无限循环,每行都有“T-T-c1x0.72x1=0.720”。@傻瓜小子,对不起,又修好了。应该在编辑之前运行代码。我们需要的方法是get
而不是next
。哦,那是我的错。很抱歉next()向前看,但没有得到下一个元组。不需要时为什么要使用递归?Perl没有尾部递归,因此它在其他语言中工作的主要原因通常会扼杀Perl。递归到50的深度在任何语言中都是完全可以接受的。不要为了避免递归而不必要地使代码复杂化。使代码复杂化?使用递归更为复杂。:)而且,你不知道它只有50深。当人们在新的情况下使用程序时,程序往往会扩展其限制。既然很容易避免风险,为什么要冒险呢?:)嗯,没有什么能比得上早晨DNA编码的气味了。:)