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perl - 根据插入顺序迭代哈希?

转载 作者:行者123 更新时间:2023-12-04 02:59:58 27 4
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不想对条目进行排序。

使用它也不会保留顺序

 foreach my $val (keys %hash) {
...
}

最佳答案

在 Perl 5 中,哈希默认是无序的。你可以使用 tie Tie::IxHash 覆盖此行为。但是请注意,这会影响性能和其他考虑因素(例如订单不会保留在副本中的事实)。

#!/usr/bin/perl

use strict;
use warnings;

use Tie::IxHash;

tie my %hash, "Tie::IxHash"
or die "could not tie %hash";

$hash{one} = 1;
$hash{two} = 2;
$hash{three} = 3;

for my $k (keys %hash) {
print "$k $hash{$k}\n";
}

更好的选择可能是使用数组或哈希值:
#!/usr/bin/perl

use strict;
use warnings;

my %hash;
$hash{one} = { data => 1, order => 1 };
$hash{three} = { data => 3, order => 2 };
$hash{two} = { data => 2, order => 3 };

for my $k (sort { $hash{$a}{order} <=> $hash{$b}{order} } keys %hash) {
print "$k $hash{$k}{data}\n";
}

至于性能,以下是基准测试的结果:
IndexedOO: a, b, c, d, e, f
HashOrdered: a, b, c, d, e, f
IxHashOO: a, b, c, d, e, f
hash: f, e, a, c, b, d
hoh_pis: a, b, c, d, e, f
IxHash: a, b, c, d, e, f
hoh: a, b, c, d, e, f
Indexed: a, b, c, d, e, f
Rate IxHash hoh Indexed IxHashOO IndexedOO hoh_pis HashOrdered hash
IxHash 261/s -- -18% -26% -48% -54% -57% -66% -80%
hoh 316/s 21% -- -10% -37% -44% -48% -59% -75%
Indexed 353/s 35% 12% -- -29% -38% -42% -55% -72%
IxHashOO 499/s 91% 58% 41% -- -12% -18% -36% -61%
IndexedOO 569/s 118% 80% 61% 14% -- -7% -27% -56%
hoh_pis 611/s 134% 93% 73% 22% 7% -- -21% -52%
HashOrdered 778/s 198% 146% 120% 56% 37% 27% -- -39%
hash 1279/s 391% 305% 262% 156% 125% 109% 64% --

从这里我们可以看到,如果您不需要它表现得像普通哈希(即绑定(bind)哈希),那么使用 Hash::Ordered 是一种方法。

这是基准:
#!/usr/bin/perl

use strict;
use warnings;

use Tie::IxHash;
use Tie::Hash::Indexed;
use Hash::Ordered;
use Benchmark;

#this is O(n) instead of O(n log n) or worse
sub perfect_insert_sort {
my $h = shift;
my @k;
for my $k (keys %$h) {
$k[$h->{$k}{order}] = $k;
}
return @k;
}

my @keys = qw/a b c d e f/;
my %subs = (
hash => sub {
my $i;
my %h = map { $_ => $i++ } @keys;
return join ", ", keys %h;
},
hoh => sub {
my $i;
my $order;
my %h = map {
$_ => { data => $i++, order => $order++ }
} @keys;
return join ", ", sort { $h{$a}{order} <=> $h{$b}{order} }
keys %h;
},
hoh_pis => sub {
my $i;
my $order;
my %h = map {
$_ => { data => $i++, order => $order++ }
} @keys;
return join ", ", perfect_insert_sort \%h;
},
IxHash => sub {
my $i;
tie my %h, "Tie::IxHash";
%h = map { $_ => $i++ } @keys;
return join ", ", keys %h;
},
Indexed => sub {
my $i;
tie my %h, "Tie::Hash::Indexed";
%h = map { $_ => $i++ } @keys;
return join ", ", keys %h;
},
IxHashOO => sub {
my $i;
my $o = tie my %h, "Tie::IxHash",
map { $_ => $i++ } @keys;
return join ", ", $o->Keys;
},
IndexedOO => sub {
my $i;
my $o = tie my %h, "Tie::Hash::Indexed",
map { $_ => $i++ } @keys;
my @k = my $k = $o->FIRSTKEY;
while ($k = $o->NEXTKEY($k)) {
push @k, $k;
}
return join ", ", @k;
},
HashOrdered => sub {
my $i;
my $oh = Hash::Ordered->new( map { $_ => $i++ } @keys );
return join ", ", $oh->keys;
},
);

for my $sub (keys %subs) {
print "$sub: ", $subs{$sub}(), "\n";
}

@keys = 1 .. 1_000;
Benchmark::cmpthese -2, \%subs;

关于perl - 根据插入顺序迭代哈希?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3638690/

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