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in reply to Re^4: Concerning hash operations (appending, concatenating)
in thread Concerning hash operations (appending, concatenating)

Here's my benchmark:
use Benchmark 'cmpthese'; use strict; my %hash = 1 .. 10_000; my %add = 9_001 .. 11_000; cmpthese(-5, { BASE => sub { my %copy = %hash; }, JAPHY => sub { my %copy = %hash; my @new = grep !exists $copy{$_}, keys %add; @copy{@new} = @add{@new}; }, ZAXO => sub { my %copy = %hash; %copy = (%add, %copy); }, NULL => sub { my %copy = %hash; %copy = (%copy); }, });
Here are the results I get:
Rate ZAXO NULL JAPHY BASE ZAXO 15.1/s -- -22% -61% -76% NULL 19.3/s 28% -- -50% -69% JAPHY 38.9/s 157% 101% -- -38% BASE 62.4/s 312% 223% 60% --
The two cases 'BASE' and 'NULL' show that Perl does not optimize %hash = %hash at all, since 'NULL' is three times slower than 'BASE'. 'JAPHY' is 2.5 times faster than 'ZAXO' in this run, with 500 old keys and 500 new keys. When I change %add to be (5_001 .. 12_000), the run times are:
Rate ZAXO NULL JAPHY BASE ZAXO 11.4/s -- -39% -58% -81% NULL 18.6/s 63% -- -31% -69% JAPHY 26.9/s 136% 45% -- -56% BASE 60.5/s 431% 225% 125% --
The point is that %a = (%b, %a) is slow because it is O(N+M) where N is the size of %a and M is the size of %b. My code is O(M) -- specifically, O(M(1+x)) where x is the fraction of keys in %b that aren't in %a -- because it doesn't iterate over %a at all, only %b.
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Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart