perl - 如何在perl中使用数组匹配两个序列

Question

当循环遍历两个数组时，我对如何在一个循环中移动指针但在另一个循环中保持不变感到困惑。例如:

数组 1:A T C G T C G A G C G

数组 2:A C G T C C T G T C G

所以第一个数组中的 A 与第二个数组中的 A 匹配，所以我们继续下一个元素。但是由于 T 与第二个索引中的 C 不匹配，我希望程序将该 T 与数组 2 中的下一个 G 进行比较，依此类推，直到找到匹配的 T。

my ($array1ref, $array2ref) = @_;

my @array1 = @$array1ref;
my @array2= @$array2ref;
my $count = 0; 
foreach my $element (@array1) {
 foreach my $element2 (@array2) {
 if ($element eq $element2) {
 $count++;
  }else { ???????????


}

Answer 1

您可以使用 while循环搜索匹配项。如果找到匹配项，则在两个数组中前进。如果不这样做，请推进第二个阵列。最后，您可以打印第一个数组中剩余的不匹配字符:

# [1, 2, 3] is a reference to an anonymous array (1, 2, 3)
# qw(1, 2, 3) is shorthand quoted-word for ('1', '2', '3')
my $arr1 = [qw(A T C G T C G A G C G)];
my $arr2 = [qw(A C G T C C T G T C G)];

my $idx1 = 0;
my $idx2 = 0;

# Find matched characters
# @$arr_ref is the size of the array referenced by $arr_ref
while ($idx1 < @$arr1 && $idx2 < @$arr2) {
    my $char1 = $arr1->[$idx1];
    my $char2 = $arr2->[$idx2];
    if ($char1 eq $char2) {
        # Matched character, advance arr1 and arr2
        printf("%s %s  -- arr1[%d] matches arr2[%d]\n", $char1, $char2, $idx1, $idx2);
        ++$idx1;
        ++$idx2;
    } else {
        # Unmatched character, advance arr2
        printf(". %s  -- skipping arr2[%d]\n", $char2, $idx2);
        ++$idx2;
    }
}

# Remaining unmatched characters
while ($idx1 < @$arr1) {
    my $char1 = $arr1->[$idx1];
    printf("%s .  -- arr1[%d] is beyond the end of arr2\n", $char1, $idx1);
    $idx1++;
}

脚本打印:

A A  -- arr1[0] matches arr2[0]
. C  -- skipping arr2[1]
. G  -- skipping arr2[2]
T T  -- arr1[1] matches arr2[3]
C C  -- arr1[2] matches arr2[4]
. C  -- skipping arr2[5]
. T  -- skipping arr2[6]
G G  -- arr1[3] matches arr2[7]
T T  -- arr1[4] matches arr2[8]
C C  -- arr1[5] matches arr2[9]
G G  -- arr1[6] matches arr2[10]
A .  -- arr1[7] is beyond the end of arr2
G .  -- arr1[8] is beyond the end of arr2
C .  -- arr1[9] is beyond the end of arr2
G .  -- arr1[10] is beyond the end of arr2