从Perl中少于15个字符的字符串中获取15个字符

Question

我有一个序列和一个表示残差(字符)位置的数字。我想从残留物的两边各取7个残留物。下面是实现这一点的代码：

my $seq = substr($sequence, $location-8, 14);

这从残留物的两边各抓取7个。然而，有些序列两边的残基都少于7个。因此，当发生这种情况时，我会收到一个错误消息：

substr outside of string at test9.pl line 52 (#1) (W substr)(F) You tried to reference a substr() that pointed outside of a string.  That is, the absolute value of the offset was larger than the length of the string.

如何更改空白处，并用另一个字母(例如X)替换它们？

例如，如果有一个序列ABCDEFGH，并且$location指向D，那么我需要在两边各有7个，因此结果将是: XXXXABCDEFGHXXX

Answer 1

对我上面的评论进行扩展。我将创建一个my_substr函数来封装填充和位置移位。

my $sequence = "ABCDEFGH";
my $location = 3;

sub my_substr {
 my ($seq, $location, $pad_length) = @_;
 my $pad = "X"x$pad_length;
 return substr("$pad$seq$pad", $location, (2*$pad_length+1));
}

print my_substr($sequence, $location, 7) . "\n";

收益率

XXXXABCDEFGHXXX

Answer 2

这是一个非常冗长的答案，但或多或少会得到你想要的：

use strict;
use warnings;

my $sequence = 'ABCDEFGH';
my $wings = 7;

my $location = index $sequence, 'D';
die "D not found" if $location == -1;

my $start = $location - $wings;
my $length = 1 + 2 * $wings;

my $leftpad = 0;
if ($start < 0) {
    $leftpad = -1 * $start;
    $start = 0;
}
my $seq = substr($sequence, $start, $length);
$seq = ('X' x $leftpad) . $seq if $leftpad;

my $rightpad = $length - length ($seq);
$seq .= 'X' x $rightpad if $rightpad > 0;

print $seq;

或者为了避免所有额外的工作，可以只创建一个包含padding的新$sequence变量：

my $sequence = 'ABCDEFGH';
my $wings = 7;

my $location = index $sequence, 'D';
die "D not found" if $location == -1;

my $paddedseq = ('X' x $wings) . $sequence . ('X' x $wings);
my $seq = substr($paddedseq, $location, 1 + 2 * $wings);

print $seq;