Re: Concerning hash operations (appending, concatenating)

[inq123]

Note that %combined = (%hash1, %hash2); suffers from performance penalty. If performance's a concern, it's always better to do map { $hash1{$_} = $hash2{$_} } keys %hash2; (unless you want to keep %hash1 and %hash2 intact, which doesn't seem to be the case in your question)

use Benchmark;

my $size = 100000;
my %hash1 = map { $_ => 1 } (0..$size);
my %hash2 = map { $_ => 1 } ($size..(2*$size));
my $t0 = new Benchmark;
my %combined = (%hash1, %hash2);
my $t1 = new Benchmark;
print "Combining took:",timestr(timediff($t1, $t0)),"\n";
$t0 = new Benchmark;
map { $hash1{$_} = $hash2{$_} } keys %hash2;
my $t1 = new Benchmark;
print "Map took:",timestr(timediff($t1, $t0)),"\n";

When size = 100000 and 1000000 respectively, the results:
inq123@perlmonks$ perl test.pl
Combining took: 0 wallclock secs ( 0.28 usr +  0.02 sys =  0.30 CPU)
Map took: 0 wallclock secs ( 0.18 usr +  0.00 sys =  0.18 CPU)
inq123@perlmonks$ perl test.pl
Combining took:42 wallclock secs (41.53 usr +  0.21 sys = 41.74 CPU)
Map took: 2 wallclock secs ( 1.89 usr +  0.05 sys =  1.94 CPU)

So performance penalty is manifested when the hashs contain tens of thousands of elements, which is not too rare.

[tilly]

General advice. Just because there are many ways to do it in Perl is not a reason for picking a less readable one. If you are not returning data from map, then you should write it like this:

$hash1{$_} = $hash2{$_} for keys %hash2;

That is easier to read, and much more clearly signals intent. It is also at least as fast as the map version. (It used to be a lot faster, but in Perl 5.8 there is an optimization that causes map to shortcircuit to become a for if it is in null context.)

Furthermore performance is far less likely to matter than most people think, and when it does having micro-optimized as you went is generally a bad strategy for getting it. (You want to keep code clean and then look for a better algorithm, or move a small section into C.) Therefore I would generally use the following strategy because it is even clearer, even though it is marginally slower on my machine (about 10% so):

@hash1{keys %hash2} = values %hash2;

And, of course, in the rare case that performance really mattered and I really wanted to work in Perl, it is fastest to avoid having to do 2 sets of hash lookups on %hash2:

$combined{$k} = $v while my ($k, $v) = each %hash2;