Perl, convert hash to array

Question

If I have a hash in Perl that contains complete and sequential integer mappings (ie, all keys from from 0 to n are mapped to something, no keys outside of this), is there a mean

Answer 1

OK, this is not very "built in" but works. It's also IMHO preferrable to any solution involving "sort" as it's faster.

map { $array[$_] = $hash{$_} } keys %hash; # Or use foreach instead of map

Otherwise, less efficient:

my @array = map { $hash{$_} } sort { $a<=>$b } keys %hash;

Answer 2

$Hash_value = 
{
'54' => 'abc',
'55' => 'def',
'56' => 'test',
};
while (my ($key,$value) = each %{$Hash_value})
{
 print "\n $key > $value";
}

Answer 3

This will leave keys not defined in %hashed_keys as undef:

# if we're being nitpicky about when and how much memory
# is allocated for the array (for run-time optimization):
my @keys_arr = (undef) x scalar %hashed_keys;

@keys_arr[(keys %hashed_keys)] =
    @hashed_keys{(keys %hashed_keys)};

And, if you're using references:

@{$keys_arr}[(keys %{$hashed_keys})] = 
    @{$hashed_keys}{(keys %{$hashed_keys})};

Or, more dangerously, as it assumes what you said is true (it may not always be true … Just sayin'!):

@keys_arr = @hashed_keys{(sort {$a <=> $b} keys %hashed_keys)};

But this is sort of beside the point. If they were integer-indexed to begin with, why are they in a hash now?

Answer 4

As DVK said, there is no built in way, but this will do the trick:

my @array = map {$hash{$_}} sort {$a <=> $b} keys %hash;

or this:

my @array;

keys %hash;

while (my ($k, $v) = each %hash) {
    $array[$k] = $v
}

benchmark to see which is faster, my guess would be the second.

Answer 5

Perl does not provide a built-in to solve your problem.

If you know that the keys cover a particular range 0..N, you can leverage that fact:

my $n = keys(%hash) - 1;
my @keys_and_values = map { $_ => $hash{$_} } 0 .. $n;
my @just_values     = @hash{0 .. $n};