在Ruby中构建字符串的最快方式是什么？

在 Ternary operator年,一个想加入[“foo”,“bar”,“baz”]的人用逗号和“和”引用 The Ruby Cookbook说 @H_404_2@

If efficiency is important to you,
don’t build a new string when you can
append items onto an existing string.
[And so on]… Use str << var1 << ‘ ‘
<< var2 instead.

但是这本书是在2006年写的.

在Ruby的所有主要实现中,使用附加(即<<)仍然是构建一个较小字符串数组的大字符串的最快方式.

解决方法

可以使用Array#join,而String#<<当你不能. 使用String#的问题是它必须创建一个中间(不需要的)字符串对象,而String#<<突变原始字符串.以下是通过Array#join,String#和String#连接1000个字符串的时间结果(以秒为单位),“,”1,000次.

Ruby 1.9.2p180      user     system      total        real
Array#join      0.320000   0.000000   0.320000 (  0.330224)
String#+ 1      7.730000   0.200000   7.930000 (  8.373900)
String#+ 2      4.670000   0.600000   5.270000 (  5.546633)
String#<< 1     1.260000   0.010000   1.270000 (  1.315991)
String#<< 2     1.600000   0.020000   1.620000 (  1.793415)

JRuby 1.6.1         user     system      total        real
Array#join      0.185000   0.000000   0.185000 (  0.185000)
String#+ 1      9.118000   0.000000   9.118000 (  9.118000)
String#+ 2      4.544000   0.000000   4.544000 (  4.544000)
String#<< 1     0.865000   0.000000   0.865000 (  0.866000)
String#<< 2     0.852000   0.000000   0.852000 (  0.852000)

Ruby 1.8.7p334      user     system      total        real
Array#join      0.290000   0.010000   0.300000 (  0.305367)
String#+ 1      7.620000   0.060000   7.680000 (  7.682265)
String#+ 2      4.820000   0.130000   4.950000 (  4.957258)
String#<< 1     1.290000   0.010000   1.300000 (  1.304764)
String#<< 2     1.350000   0.010000   1.360000 (  1.347226)

Rubinius (head)     user     system      total        real
Array#join      0.864054   0.008001   0.872055 (  0.870757)
String#+ 1      9.636602   0.076005   9.712607 (  9.714820)
String#+ 2      6.456403   0.064004   6.520407 (  6.521633)
String#<< 1     2.196138   0.016001   2.212139 (  2.212564)
String#<< 2     2.176136   0.012001   2.188137 (  2.186298)

以下是基准代码：

WORDS = (1..1000).map{ rand(10000).to_s }
N = 1000

require 'benchmark'
Benchmark.bmbm do |x|
  x.report("Array#join"){
    N.times{ s = WORDS.join(',') }
  }
  x.report("String#+ 1"){
    N.times{
      s = WORDS.first
      WORDS[1..-1].each{ |w| s += ","; s += w }
    }
  }
  x.report("String#+ 2"){
    N.times{
      s = WORDS.first
      WORDS[1..-1].each{ |w| s += "," + w }
    }
  }
  x.report("String#<< 1"){
    N.times{
      s = WORDS.first.dup
      WORDS[1..-1].each{ |w| s << ","; s << w }
    }
  }
  x.report("String#<< 2"){
    N.times{
      s = WORDS.first.dup
      WORDS[1..-1].each{ |w| s << "," << w }
    }
  }
end

Ubuntu在RVM下获得的结果. Ruby上的RubyInstaller的Ruby 1.9.2p180的结果与上述1.9.2类似.

在Ruby中构建字符串的最快方式是什么？

解决方法

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