Question

我有一些数组deep_array深入哈希并需要时间访问，我还有一个变量（my_local_variable）指向deep_array和其他一些本地数组new_array。我需要将deep_array设置为new_array到my_local_variable。

相当于其中一项的东西：

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

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

但更快

编辑：速度

这是我处理的情况的粗略概念：

（实际上它更深，但我写的更少）

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

给出：

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)

Answer 1

告诉我，如果我错了，Array#replace似乎比任何当前答案都要快，因为它不需要新的数组.dup被许可，并且不需要另一个引用创建：

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)

如果我使用数组索引

，差异会更大

ruby通过指向原始

1 个答案: