需要一些帮助来创建hashrefs的hashrefs,最后一个键作为对数组的引用。
use Data::Dumper;
my $foo = "a:b:c:d:a";
my $bar = "a:b:c:d:z";
my $hoh = {};
sub createHash {
my ($hoh,$orig,$rest,$last) = @_;
$rest = $rest || $orig;
$_ = $rest;
if (/^(.*?):(.*)$/) {
$hoh->{$1} = $hoh->{$1} || {};
createHash($hoh->{$1},$orig,$2,$1);
}
elsif (defined($last)) {
push (@{$hoh->{value}} , [$rest,$orig]);
}
return $hoh;
}
$hoh = createHash($hoh,$foo,undef);
$hoh = createHash($hoh,$bar,undef);
print Dumper($hoh);
$VAR1 = {
'a' => {
'b' => {
'c' => {
'd' => [
[
'a',
'a:b:c:d:a'
],
[
'z',
'a:b:c:d:z'
]
]
}
}
}
};
您可以将其与键盘输出进行比较。注意细微差别;而不是'd'是具有arrayref value
的hashref,'d'是arrayref,没有value
。
答案 0 :(得分:2)
我建议Data::Diver,虽然它有点尴尬,因为它总是想在最后创建标量引用,而这不是我们想要的。因此,我作弊。
这里的主要内容是我们可以通过一次解密所有密钥并使用while循环(在Data::Diver内)而不是递归来节省工作量(主要是维护),这本质上是递归解密更有趣:-)将它与即使它是递归的事实相结合,它隐藏在一个漂亮,整洁的函数调用中,它是一个双赢: - )
use Data::Dumper;
use Data::Diver qw(DiveRef);
my $foo = "a:b:c:d:a";
my $bar = "a:b:c:d:z";
my $hoh = {};
sub add_item
{
my $href = shift;
my $str = shift;
my @keys = split /:/, $str;
# force an array to be autovivified if it isn't already there.
# (this is kinda cheating)
my $cheat = DiveRef($href, @keys[0..$#keys-1], 0);
my $ref = DiveRef($href, @keys[0..$#keys-1]);
# if we cheated (thus $$cheat will be undef), we need to pop that
# off.
pop @$$ref unless $$cheat;
# store this at the end.
push @{$$ref}, [ $keys[-1], $str ];
return;
}
add_item($hoh, $foo);
add_item($hoh, $bar);
print Dumper($hoh);
希望有所帮助,
更新:与tye交谈后,他提供了一种更简洁的方法来执行此操作。它仍然使用Data::Diver,但嵌入了一个更简单的解决方法。 (他声称perl在这里有一个错误:lvalue subs和push - 我不知道更好,所以我会接受他的话。)
use Data::Dumper;
use Data::Diver qw(DiveRef DiveVal);
my $foo = "a:b:c:d:a";
my $bar = "a:b:c:d:z";
my $hoh = {};
sub add_item
{
my $href = shift;
my $str = shift;
my @keys= split /:/, $str;
my $last= pop @keys;
push @{ DiveVal( $href, \( @keys ) ) ||= []}, [ $last, $str ];
return;
}
add_item($hoh, $foo);
add_item($hoh, $bar);
print Dumper($hoh);
答案 1 :(得分:1)
perl -MData::Dumper -F: -anle'($p,$l)=splice@F,-2,2;$x=\$h;$x=\($$x->{$_}||={})for@F;push@{$$x->{$p}||=[]},[$l=>$_]}{print Dumper($h)' <<EOI
a:b:c:d:a
a:b:c:d:z
a:b:c:d:f
EOI
答案 2 :(得分:1)
改变
push (@{$hoh->{value}} , [$rest,$orig]);
到
push (@{$hoh->{$last}} , [$rest,$orig]);
use Data::Dumper;
my $foo = "a:b:c:d:a";
my $bar = "a:b:c:d:z";
my $hoh = {};
sub createHash {
my ($hoh,$orig,$rest,$last) = @_;
$rest = $rest || $orig;
$_ = $rest;
if (/^(.?):(.+)$/) {
$_ = $1;
$rest = $2;
if ($rest =~ /:/) {
$hoh->{$_} = $hoh->{$_} || {};
createHash($hoh->{$_},$orig,$rest,$_);
} else {
push(@{$hoh->{$_}}, [$rest, $orig]);
}
}
return $hoh;
}
$hoh = createHash($hoh,$foo,undef);
$hoh = createHash($hoh,$bar,undef);
print Dumper($hoh);
答案 3 :(得分:0)
use Data::Dumper;
my $hoh = {};
foreach my $str ('a:b:c:d:a','a:b:c:d:z'){
my @vals = split /:/,$str;
my $hr = $hoh;
my $lastkey = @vals[-2];
for (0..$#vals-2){
$hr->{$vals[$_]}= $hr->{$vals[$_]} || {};
$hr=$hr->{$vals[$_]};
}
if (defined $lastkey){
push @{$hr->{$lastkey}}, [@vals[-1], $str];
}
}
print Dumper($hoh);
在回顾Hynek之后,我认为我们正在使用类似的方法
use Data::Dumper;
my $foo = "a:b:c:d:a";
my $bar = "a:b:c:d:z";
my $hoh = {};
sub createHash {
my ($hoh,$str_orig,$str_rest,$lastkey,$parent) = @_;
$str_rest = $str_rest || $str_orig || "";
$_ = $str_rest;
if (/^(.*?):(.*)$/)
{
$parent = $hoh;
$hoh->{$1} = $hoh->{$1} || {};
createHash($hoh->{$1},$str_orig,$2,$1,$parent);
}
elsif (defined($lastkey))
{
delete($parent->{$lastkey}) if ref $parent->{$lastkey} ne "ARRAY";
push (@{$parent->{$lastkey}} , [$str_rest,$str_orig]);
}
return $hoh;
}
$hoh = createHash($hoh,$foo);
$hoh = createHash($hoh,$bar);
print Dumper($hoh);