这个 Perl 子程序有什么问题? [英] What is wrong with this Perl subroutine?

问题描述

我正在尝试实现一个计算输入字符串的 d-neighbors 的子程序.这是种植主题搜索的一部分，但我的问题更笼统.代码如下:

I'm trying to implement a subroutine that calculates the d-neighbors of an input string. This is apart of an implementation of planted motif search, but my question is much more general. Here is the code:

#subroutine for generating d-neighbors
sub generate_d_neighbors{
    # $sequence is the sequence to generate d-neighbors from
    # $HD is the Hamming Distance
    my ($sequence, $HD) = @_;

    for(my $i = 0; $i=$HD; $i++){
        my @l = ['A', 'C', 'T', 'G'];
        my @t = splice(@l,$sequence[$i]);  
       #TODO
    }
}

错误发生在最后一行，说明:

The error is occurring at the last line, saying that:

全局符号@sequence"需要明确的包名(你忘记声明我的@sequence"了吗?

据我所知，Perl 不像在 Java 中那样采用 subroutine(param1, param2) 形式的参数，但是为什么 $sequence 不被识别是否已经初始化?

It was my understanding that Perl does not take parameters in the form subroutine(param1, param2) like in Java for example, but why is $sequence not being recognized as already having been initialized?

推荐答案

您的代码存在一些问题:

There are some problems with your code:

sub generate_d_neighbors{
    my ($sequence, $HD) = @_;

    for(my $i = 0; $i=$HD; $i++){
        my @l = ['A', 'C', 'T', 'G'];
        my @t = splice(@l,$sequence[$i]);  
    }
}

首先，我们来看看

    for(my $i = 0; $i=$HD; $i++){

假设 $HD 非零，这个循环永远不会终止，因为条件永远不会为假.如果您希望 $i 的范围从 0 到 $HD，请将语句写为 for my $i (0 .. $HD) 会更好.

Assuming $HD is nonzero, this loop will never terminate because the condition will never be false. If you wanted $i to range from 0 to $HD, writing the statement as for my $i (0 .. $HD) would have been better.

第二，你有<​​/p>

Second, you have

        my @t = splice(@l,$sequence[$i]);  

您似乎假设有一个数组 @sequence 并且您正试图访问它的第一个元素.但是，$sequence 是对数组的引用.因此，您应该使用

where you seem to assume there is an array @sequence and you are trying to access its first element. However, $sequence is a reference to an array. Therefore, you should use

$sequence->[$i]

第三(感谢@Ikegami)，你有

Third (thanks @Ikegami), you have

        my @l = ['A', 'C', 'T', 'G'];

在 for 循环的主体中.然后 @l 将包含单个元素，对包含元素 'A', 'C', 的匿名数组的引用'T' 和 'G'.相反，请使用:

in the body of the for-loop. Then @l will contain a single element, a reference to an anonymous array containing the elements 'A', 'C', 'T', and 'G'. Instead, use:

my @l = qw(A C T G);

我不确定你想用 splice(@l, $sequence->[$i]) 实现什么，但可以更好地写成:

I am not sure exactly what you want to achieve with splice(@l, $sequence->[$i]), but that can be better written as:

 my @t = @l[0 .. ($sequence->[$i] - 1)];  

实际上，您可以将两个分配减少为:

In fact, you could reduce the two assignments to:

 my @t = qw(A C T G)[0 .. ($sequence->[$i] - 1)];

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