从multidemsional数组中提取列

Question

我有一个多维数组

my @multi = ( [ 1, "first", "a" ], [ 2, "second", b ], [ 3, "third", c] ... );

我想提取一个单维数组：

[ "first", "second", "third" ... ]

这些是对@multi [0] [1]，@ multi [1] [1]，@ multi [2] [1] ...

的引用的组合

我该怎么做？

Answer 1

[ map { $_->[1] } @multi ]

Answer 2

例如，您可以这样做：

use strict;

my @multi = ( [ 1, "first", "a" ], [ 2, "second", "b" ], [ 3, "third", "c" ] );
my $res;

push @$res, $_->[1] for @multi;

use Data::Dump;
dd $res;

输出：

["first", "second", "third"]

Answer 3

数组数组

perldoc perllol，man perllol或perldoc.perl.org/perllol.html

对此进行了详细说明

您可以取消引用您的数组变量：

map { ${$_}[1] } @multi

同一事物的另一种语法：

map { $_->[1] } @multi

尝试：

my @multi = ( [ 1, "first", "a" ], [ 2, "second", b ], [ 3, "third", c] );
print join(", ", map { ${ $_ }[1] } @multi)."\n";
print join(", ", map { $_->[1] } @multi)."\n";

和

use Data::Dumper;
my @multi = ( [ 1, "first", "a" ], [ 2, "second", b ], [ 3, "third", c] );
my @other=( [ 1 , undef , "A" ] , [ 2 , [ map { ${$_}[1] } @multi ], "B" ] );
$Data::Dumper::Indent= 0;
print Data::Dumper->Dump([\@other],[qw|other|])."\n";

将呈现：

$other = [[1,undef,'A'],[2,['first','second','third'],'B']];

哪里...

$other = [[1,undef,"A"],[2,["first","second","third"],"B"]];
print $other->[0]->[0]."\n";
print $other->[1]->[0]."\n";
print $other->[1]->[1]->[0]."\n";
print $other->[1]->[1]->[2]."\n";
print $other->[1]->[2]."\n";

可以给出：

1
2
first
third
B

Answer 4

您需要使用de-reference：

my @multi = ( [ 1, "first", "a" ], [ 2, "second", b ], [ 3, "third", c]);
my $index = 0;
my @single;
foreach (@multi) {
    my $aref = $multi[$index];
    $single[$index] = $aref->[1];
    $index++;
}
print "\n @single \n";

输出： first second third