How to remove last word from a sentence (a string)

Question

How to go about removing the last element? For example, I have strings like these:

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

I need a neat way of showing the output:

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

This is what I could do:

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

I was wondering if there's any neat way of doing this maybe in one line, like: str.split(" ", 2).last => "testing string", which should show "my testing" instead.

EDIT

Thank you guys for the multiple and interesting answers. I appreciate your effort and time. But, I had to be fair, so I decided to benchmark everybody's answer. Here is the report with benchmark:

#!/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)

Which makes it clear that answer p str2[0...str2.rindex(' ')] is performing well. Though I liked other approaches as well, very thoughtful and interesting. Thanks.

Second Edition

As per @theTineMan comment here is the updated benchmark:

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

And if you are not a fan on regexp:

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

Answer 2

With split you can slice the last array element like this:

str1.split(' ')[0...-1].join(' ')