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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
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)
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