ruby - 如何从句子(字符串)中删除最后一个词

Question

如何删除最后一个元素？例如，我有这样的字符串:

str1 = "My testing String"
str2 = "My another testing string"

我需要一种简洁的方式来显示输出:

str1 = "My testing"
str2 = "My another testing"

这是我能做的:

str1 = str1.split(" ")
str1.delete(str1.last)
str1.join(" ")
# => "My testing"

我想知道是否有任何简洁的方法可以在一行中执行此操作，例如:str.split("", 2).last => "testing string"，它应该显示 “我的测试” 代替。

编辑

感谢你们提供的多个有趣的答案。我感谢你的努力和时间。但是，我必须公平，所以我决定以每个人的答案为基准。这是带有基准的报告:

#!/usr/bin/ruby
require 'benchmark'
str2 = "My another testing string"
n = 500
Benchmark.bm(20) do |x|
  x.report("str2[/(.*)\s/,1]                  "){ n.times { str2[/(.*)\s/,1] } }
  x.report("str2[0...str2.rindex(' ')]        "){ n.times { str2[0...str2.rindex(' ')] } }
  x.report("str2.split(' ')[0...-1].join(' ') "){ n.times { str2.split(' ')[0...-1].join(' ') } }
  x.report("str2[/.*(?=\s)/]                  "){ n.times { str2[/.*(?=\s)/] } }
end


                                     user     system      total        real
str2[/(.*) /,1]                    0.000000   0.000000   0.000000 (  0.001394)
str2[0...str2.rindex(' ')]         0.000000   0.000000   0.000000 (  0.000956)
str2.split(' ')[0...-1].join(' ')  0.010000   0.000000   0.010000 (  0.002569)
str2[/.*(?= )/]                    0.000000   0.000000   0.000000 (  0.001351)

这清楚地表明答案 p str2[0...str2.rindex(' ')] 表现良好。尽管我也喜欢其他方法，但非常周到且有趣。谢谢。

第二版

根据@theTineMan 的评论，这里是更新的基准:

require 'benchmark'
str2 = "My another testing string"
n = 999999
Benchmark.bmbm(20) do |x|
  x.report("str2[/(.*)\s/,1]                  "){ n.times { str2[/(.*)\s/,1] } }
  x.report("str2[0...str2.rindex(' ')]        "){ n.times { str2[0...str2.rindex(' ')] } }
  x.report("str2.split(' ')[0...-1].join(' ') "){ n.times { str2.split(' ')[0...-1].join(' ') } }
  x.report("str2[/.*(?=\s)/]                  "){ n.times { str2[/.*(?=\s)/] } }
end

Rehearsal ----------------------------------------------------------------------
str2[/(.*) /,1]                      1.030000   0.000000   1.030000 (  1.033787)
str2[0...str2.rindex(' ')]           0.850000   0.000000   0.850000 (  0.853124)
str2.split(' ')[0...-1].join(' ')    4.740000   0.000000   4.740000 (  4.750215)
str2[/.*(?= )/]                      0.990000   0.000000   0.990000 (  0.990726)
------------------------------------------------------------- total: 7.610000sec

                                         user     system      total        real
str2[/(.*) /,1]                      1.020000   0.000000   1.020000 (  1.014772)
str2[0...str2.rindex(' ')]           0.830000   0.000000   0.830000 (  0.839385)
str2.split(' ')[0...-1].join(' ')    4.620000   0.010000   4.630000 (  4.629874)
str2[/.*(?= )/]                      0.990000   0.000000   0.990000 (  0.988224)

Answer 1

str2 = "My another testing string"     
p str2[/(.*)\s/,1] #=> My another testing

如果您不喜欢正则表达式:

str2 = "My another testing string"
p str2[0...str2.rindex(' ')] #=> My another testing