如何在perl中一次创建多个数组

Question

我正在尝试创建23个数组而不键入@array1,@array2依此类推,@r如果$chrid匹配数组编号(如果$chrid=1应该放入@array1),则使用数组中的变量加载它们.我怎样才能做到这一点？

这是我到目前为止:

#!/usr/bin/perl
use warnings;
use strict;

my @chr;
my $input;
open ($input, "$ARGV[0]") || die;
while (<$input>) {
    my @r = split(/\t/);
    my $snps = $r[0];
    my $pval = $r[1];
    my $pmid = $r[2];
    my $chrpos = $r[3];
    my $chrid = $r[4];
    for ($chrid) {
        push (@chr, $chrid);
    }
}

close $input;

Answer 1

您可以使用数组数组,其中每个子数组都存储在数组数组中的顺序递增索引处.这可能是什么样的,但我仍然不清楚你想要存储的数据:

use warnings;
use strict;

my @chr;
open my $input_fh, '<', $ARGV[0]
   or die "Unable to open $ARGV[0] for reading: $!";
while (< $input_fh> ) {
    # you can unpack your data in a single statement
    my ($snps, $pval, $pmid, $chrpos, $chrid) = split /\t/;

    # unclear what you actually want to store
    push @{ $chr[$chrid] }, ( $snps, $pval, $pmid, $chrpos, $chrid );

}
close $input_fh;