From: Kent Dahl <kentda@...>
Date: 2002-06-10T00:18:23+09:00
Subject: Re: Hash#===()

Han Holl wrote:
> Yes,  like a test for set membership. It seems
> intuitive to me.

I'd be more inclined to alias it to the Enumerable#include? myself. That
is, unless someone does use Hash#=== for something as is. (Doesn't it
just go up to Object#=== and degenerate into a == as it is now?)

But as for your benchmarks:
  1) hash key lookups are supposed to be fast, with constant lookup
time. This is not surprising. I feel that you are comparing apples to
oranges.
  2) your tests aren't completely fair, as the hashobject is
pregenerated, while the others are literals in the case statements
themselves. Doesn't yield much difference, but still:

<CODE>
require 'benchmark'
include Benchmark

class Hash
  alias === has_key?
end

array1  = "abc", "def", "kli", "lop", "tli", "ppp"
regexp1 = /^(abc|def|kli|lop|tli|ppp)$/
regexp2 = /^(?:abc|def|kli|lop|tli|ppp)$/

n = 100000
hash = { "abc"=> 1, "def"=> 1, "kli"=> 1, "lop"=> 1, "tli"=> 1, "ppp"=>
1 }
[ "ppp", "abc", "for" ].each do |a|
 bm(7) do |x|
    # first the explicit array version
   x.report(a+' a'){for i in 1..n;
    case a
    when "abc", "def", "kli", "lop", "tli", "ppp"
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when *array1 #"abc", "def", "kli", "lop", "tli", "ppp"
    end
   end}

   x.report(a+' r1'){for i in 1..n;
    case a
    when /^(abc|def|kli|lop|tli|ppp)$/
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when regexp1 #/^(abc|def|kli|lop|tli|ppp)$/
    end
   end}

   x.report(a+' r2'){for i in 1..n;
    case a
    when /^(?:abc|def|kli|lop|tli|ppp)$/
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when regexp2 #/^(?:abc|def|kli|lop|tli|ppp)$/
    end
   end}

   x.report(a+' h'){for i in 1..n;
    case a
    when  { "abc"=> 1, "def"=> 1, "kli"=> 1, "lop"=> 1, "tli"=> 1,
"ppp"=> 1 }
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when hash
    end
   end}

 end
end
</CODE>

<OUTPUT>
ppp a    0.640000   0.000000   0.640000 (  0.645890)
ppp      0.290000   0.000000   0.290000 (  0.299588)
ppp r1   0.420000   0.010000   0.430000 (  0.424303)
ppp      0.420000   0.000000   0.420000 (  0.418180)
ppp r2   0.380000   0.000000   0.380000 (  0.382281)
ppp      0.380000   0.000000   0.380000 (  0.382734)
ppp h    1.800000   0.000000   1.800000 (  1.799063)
ppp      0.130000   0.000000   0.130000 (  0.128566)
             user     system      total        real
abc a    0.170000   0.000000   0.170000 (  0.167413)
abc      0.110000   0.000000   0.110000 (  0.109137)
abc r1   0.340000   0.000000   0.340000 (  0.335180)
abc      0.330000   0.000000   0.330000 (  0.331492)
abc r2   0.310000   0.000000   0.310000 (  0.315915)
abc      0.320000   0.000000   0.320000 (  0.318973)
abc h    1.820000   0.000000   1.820000 (  1.815141)
abc      0.130000   0.000000   0.130000 (  0.128977)
             user     system      total        real
for a    0.660000   0.000000   0.660000 (  0.659464)
for      0.300000   0.000000   0.300000 (  0.296972)
for r1   0.270000   0.000000   0.270000 (  0.276839)
for      0.280000   0.000000   0.280000 (  0.277514)
for r2   0.260000   0.000000   0.260000 (  0.260307)
for      0.260000   0.000000   0.260000 (  0.261449)
for h    1.790000   0.000000   1.790000 (  1.795176)
for      0.120000   0.000000   0.120000 (  0.115688)
</OUTPUT>  

Note that pregenerating the regexps doesn't yield much, but expanding a
pregenerated array seems to have very much effect, even going below the
hash version for 'abc'. But then that is because the first element of
the array is 'abc'.

-- 
(\[ Kent Dahl ]/)_    _~_    __[ http://www.stud.ntnu.no/~kentda/ ]___/~
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