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I have some array deep_array that's deep in a hash and takes time to access, I also have a variable (my_local_variable) pointing to deep_array and some other local array new_array. I need to set deep_array to new_array through my_local_variable.

Something equivalent to one of these:

my_local_variable.map!.with_index {|_, i| new_array[i]}

my_local_variable.each_with_index {|_, i| b[i] = new_array[i]}

but much faster

Edit: speed

This is a rough idea of the situation I'm dealing with:

(in reality it's deeper but i'm doing fewer writes)

require 'benchmark'
H = {[1,2,3]=>[2,3,4],[3,4,5]=>[4,5,6],[5,6,7]=>[6,7,8]}
h = H[[1,2,3]]

Benchmark.bmbm(15) do |i|
  i.report('local reference') {1_000_000.times {|i| h[0] = i}}
  i.report('          index') {1_000_000.times {H[[1,2,3]][0] = i}}
end

Gives:

Rehearsal ---------------------------------------------------
local reference   0.230000   0.010000   0.240000 (  0.234168)
          index   5.780000   0.040000   5.820000 (  5.851909)
------------------------------------------ total: 6.060000sec

                      user     system      total        real
local reference   0.220000   0.000000   0.220000 (  0.226742)
          index   5.770000   0.030000   5.800000 (  5.830011)
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1 回答 1

0

告诉我我是否错了，但Array#replace似乎比任何当前答案都快，因为它不需要.duplicated 新数组并且不需要创建另一个引用：

new_arrays = 1000000.times.map {4.times.map {rand 10}}

require 'benchmark'

Benchmark.bmbm(18) do |i|
  i.report('     find, replace') do
    hash = {0=>{1=>{2=>{3=>[]}}}}
    1000000.times do |j|
      hash[0][1][2][3].replace new_arrays[j]
    end
  end

  i.report('         find, dup') do
    hash = {0=>{1=>{2=>{3=>[]}}}}
    1000000.times do |j|
      hash[0][1][2][3] = new_arrays[j].dup
    end
  end

  i.report('reference, replace') do
    hash = {0=>{1=>{2=>{3=>[]}}}}
    reference = hash[0][1][2][3]
    1000000.times do |j|
      reference.replace new_arrays[j]
    end
  end

  i.report('    above ref, dup') do
    hash = {0=>{1=>{2=>{3=>[]}}}}
    above = hash[0][1][2]
    1000000.times do |j|
      above[3] = new_arrays[j].dup
    end
  end
end

给出：

Rehearsal ------------------------------------------------------
     find, replace   1.670000   0.090000   1.760000 (  1.791100)
         find, dup   2.010000   0.110000   2.120000 (  2.193884)
reference, replace   0.460000   0.000000   0.460000 (  0.470288)
    above ref, dup   1.390000   0.020000   1.410000 (  1.421624)
--------------------------------------------- total: 5.750000sec

                         user     system      total        real
     find, replace   0.960000   0.020000   0.980000 (  1.039140)
         find, dup   1.720000   0.010000   1.730000 (  1.753176)
reference, replace   0.470000   0.000000   0.470000 (  0.472988)
    above ref, dup   1.360000   0.020000   1.380000 (  1.384833)

如果我用数组索引，差异会更大

于 2017-06-21T17:21:23.157 回答