From: Eric Hodel <drbrain@...7.net>
Date: 2007-01-25T19:27:51+09:00
Subject: Re: can it be shorter?

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On Jan 24, 2007, at 15:20, Vincent Fourmond wrote:
> Vincent Fourmond wrote:
>> h = {}
>> for k,v in hash
>>   [k].flatten.each {|l| h[l] = v}
>> end
>> hash = h
>
>   Actually, I've got shorter ;-):
>
> h = {}
> for k,v in hash
>   [*k].each {|l| h[l] = v}
> end
> hash = h

but fastest (I assume all entries worked correctly):

$ ruby bm.rb
Rehearsal --------------------------------------------------------
original              10.440000   0.010000  10.450000 ( 10.492738)
original clean        11.480000   0.040000  11.520000 ( 11.632237)
vince                  5.210000   0.010000   5.220000 (  5.264394)
william               12.760000   0.020000  12.780000 ( 12.814230)
daniel                 6.900000   0.010000   6.910000 (  6.924297)
robert inject          6.560000   0.000000   6.560000 (  6.588160)
robert inject rescue   6.080000   0.010000   6.090000 (  6.094140)
---------------------------------------------- total: 59.530000sec

                            user     system      total        real
original              10.450000   0.010000  10.460000 ( 10.467971)
original clean        11.490000   0.030000  11.520000 ( 11.600672)
vince                  5.210000   0.000000   5.210000 (  5.219863)
william               12.820000   0.010000  12.830000 ( 12.839722)
daniel                 6.930000   0.010000   6.940000 (  6.945329)
robert inject          6.580000   0.010000   6.590000 (  6.591095)
robert inject rescue   6.080000   0.000000   6.080000 (  6.117226)


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require 'benchmark'

N = 500_000

hash = {
  "index" => "list",
  ["edit", "update"] => "manage_one",
  ["new", "create", "destroy"] => "manage_all"
}

Benchmark.bmbm do |bm|

  bm.report 'original' do
    N.times do
      Hash[*hash.to_a.collect{|x|
        Array === x[0] ? x[0].zip([x[1]]*x[0].size) : x
      }.flatten]
    end
  end

  bm.report 'original clean' do
    N.times do
      Hash[*hash.inject([]){|arr, (k, v)|
        arr += Array === k ? k.zip([v]* k.size) : [k, v]
      }.flatten]
    end
  end

  bm.report 'vince' do
    N.times do
      h = {}
      for k,v in hash
          [*k].each {|l| h[l] = v}
      end
    end
  end

  bm.report 'william' do
    N.times do
      Hash[ *hash.map{|k,v| k=[*k]; k.zip([v]*k.size) }.flatten ]
    end
  end

  bm.report 'daniel' do
    N.times do
      h={};hash.map{|k,v|[*k].map{|t|h[t]=v}}
    end
  end

  bm.report 'robert inject' do
    N.times do
      hash.inject({}){|h,(k,v)| k.to_a.each {|x| h[x]=v};h}
    end
  end

  bm.report 'robert inject rescue' do
    N.times do
      hash.inject({}){|h,(k,v)| k.each {|x| h[x]=v} rescue h[k]=v;h}
    end
  end

end


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-- 
Eric Hodel - drbrain@segment7.net - http://blog.segment7.net

I LIT YOUR GEM ON FIRE!


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