如何有效地清除Ruby的负零浮点数?

内容来源于 Stack Overflow,并遵循CC BY-SA 3.0许可协议进行翻译与使用

  • 回答 (2)
  • 关注 (0)
  • 查看 (26)

在Ruby中,0.0 * -1 == -0.0

我有一个应用程序,我将一堆Float对象相乘-1,但是我不喜欢-0.0输出中的对象,因为它很混乱。

有没有一种智能的方式来制作Float#to_s输出,0.0而不是-0.0

我完全可以Float通过某种类型的scrubber / helper方法来运行每个对象,但以下情况只会让我更加困惑:

def clean_output(amount)
  if amount.zero?
    0.0
  else
    amount
  end
end

提问于
用户回答回答于

如果你写的代码让你感到困惑,那么这应该真的让你感到头痛:

def clean_output(amount)
  amount.zero? && 0.0 || amount
end

有了一些证据:

irb(main):005:0> f = 0.0
=> 0.0
irb(main):006:0> f.zero? && 0.0 || f
=> 0.0
irb(main):007:0> f = -0.0
=> -0.0
irb(main):008:0> f.zero? && 0.0 || f
=> 0.0
irb(main):009:0> f=1.0
=> 1.0
irb(main):010:0> f.zero? && 0.0 || f
=> 1.0

我不喜欢使用nonzero?,因为它的用例有点混乱。它是Numeric的一部分,但文档显示它用作<=>操作符Comparable的一部分。

但这是另一种过早优化没有成功的情况:

require 'benchmark'

def clean_output(amount)
  if amount.zero?
    0.0
  else
    amount
  end
end

def clean_output2(amount)
  amount.zero? && 0.0 || amount
end

def clean_output3(value)
  value + 0
end

class Numeric
  def clean_to_s
    (nonzero? || abs).to_s
  end
end


n = 5_000_000
Benchmark.bm(14) do |x|
  x.report( "clean_output:"  ) { n.times { a = clean_output(-0.0)  } }
  x.report( "clean_output2:" ) { n.times { a = clean_output2(-0.0) } }
  x.report( "clean_output3:" ) { n.times { a = clean_output3(-0.0) } }
  x.report( "clean_to_s:"    ) { n.times { a = 0.0.clean_to_s      } }
end

结果是:

ruby test.rb 
                    user     system      total        real
clean_output:   2.120000   0.000000   2.120000 (  2.127556)
clean_output2:  2.230000   0.000000   2.230000 (  2.222796)
clean_output3:  2.530000   0.000000   2.530000 (  2.534189)
clean_to_s:     7.200000   0.010000   7.210000 (  7.200648)

ruby test.rb 
                    user     system      total        real
clean_output:   2.120000   0.000000   2.120000 (  2.122890)
clean_output2:  2.200000   0.000000   2.200000 (  2.203456)
clean_output3:  2.540000   0.000000   2.540000 (  2.533085)
clean_to_s:     7.200000   0.010000   7.210000 (  7.204332)

我添加了一个没有to_s的:

require 'benchmark'

def clean_output(amount)
  if amount.zero?
    0.0
  else
    amount
  end
end

def clean_output2(amount)
  amount.zero? && 0.0 || amount
end

def clean_output3(value)
  value + 0
end

class Numeric
  def clean_to_s
    (nonzero? || abs).to_s
  end

  def clean_no_to_s
    nonzero? || abs
  end

end


n = 5_000_000
Benchmark.bm(14) do |x|
  x.report( "clean_output:"  ) { n.times { a = clean_output(-0.0)  } }
  x.report( "clean_output2:" ) { n.times { a = clean_output2(-0.0) } }
  x.report( "clean_output3:" ) { n.times { a = clean_output3(-0.0) } }
  x.report( "clean_to_s:"    ) { n.times { a = -0.0.clean_to_s     } }
  x.report( "clean_no_to_s:" ) { n.times { a = -0.0.clean_no_to_s  } }
end

结果是:

ruby test.rb 
                    user     system      total        real
clean_output:   3.030000   0.000000   3.030000 (  3.028541)
clean_output2:  2.990000   0.010000   3.000000 (  2.992095)
clean_output3:  3.610000   0.000000   3.610000 (  3.610988)
clean_to_s:     8.710000   0.010000   8.720000 (  8.718266)
clean_no_to_s:  5.170000   0.000000   5.170000 (  5.170987)

ruby test.rb 
                    user     system      total        real
clean_output:   3.050000   0.000000   3.050000 (  3.050175)
clean_output2:  3.010000   0.010000   3.020000 (  3.004055)
clean_output3:  3.520000   0.000000   3.520000 (  3.525969)
clean_to_s:     8.710000   0.000000   8.710000 (  8.710635)
clean_no_to_s:  5.140000   0.010000   5.150000 (  5.142462)

require 'benchmark'

n = 5_000_000
Benchmark.bm(9) do |x|
  x.report( "nonzero?:" ) { n.times { -0.0.nonzero? } }
  x.report( "abs:"      ) { n.times { -0.0.abs      } }
  x.report( "to_s:"     ) { n.times { -0.0.to_s     } }
end

结果如下:

ruby test.rb 
               user     system      total        real
nonzero?:  2.750000   0.000000   2.750000 (  2.754931)
abs:       2.570000   0.010000   2.580000 (  2.569420)
to_s:      4.690000   0.000000   4.690000 (  4.687808)

ruby test.rb 
               user     system      total        real
nonzero?:  2.770000   0.000000   2.770000 (  2.767523)
abs:       2.570000   0.010000   2.580000 (  2.569757)
to_s:      4.670000   0.000000   4.670000 (  4.678333)
用户回答回答于

实际上有一个不需要条件的解决方案。

def clean_output(value)
  value + 0
end

输出:

> clean_output(3.0)
=> 3.0 
> clean_output(-3.0)
=> -3.0 
> clean_output(-0.0)
=> 0.0

扫码关注云+社区