Ruby中，哪种方法执行速度最快，能给出具体比较吗？

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要确定哪种方法运行得更快，可以在Ruby中使用Benchmark模块来测量代码的执行时间。以下是一个简化的例子，展示了如何使用`Benchmark`模块来比较两种不同实现`count_between`方法的性能。

rubyrequire 'benchmark'

def count_between_naive(list_of_integers, lower_bound, upper_bound) count=0 list_of_integers.each do |x| count +=1 if lower_bound <=x && x <=upper_bound end countend

def count_between_optimized(list_of_integers, lower_bound, upper_bound) list_of_integers.count { |x| lower_bound <=x && x <=upper_bound }end

Benchmark.bm do |x| list_of_integers=Array.new(1000000) { rand(1..1000000) } lower_bound=500000 upper_bound=600000

x.report(Naive:) { count_between_naive(list_of_integers, lower_bound, upper_bound) } x.report(Optimized:) { count_between_optimized(list_of_integers, lower_bound, upper_bound) }end

这段代码首先定义了两种不同的`count_between`方法：一种使用循环和条件语句（naive方法），另一种使用Ruby的内置`count`方法和块（optimized方法）。然后，使用`Benchmark`模块的`bm`方法来比较这两种方法的执行时间。注意，为了避免数组的初始化过于耗时，这里使用了随机数生成器来创建一个包含100万个元素的数组。

如何确定哪种方法运行得更快？很难在 Ruby文档中阅读Benchmark并实际实现它.谢谢

def count_between(list_of_integers, lower_bound, upper_bound) count = 0 list_of_integers.each do |x| (x >= lower_bound && x <= upper_bound) ? count += 1 : next end count end

要么

def count_between(list_of_integers, lower_bound, upper_bound) count = 0 list_of_integers.each do |x| count += 1 if x.between?(lower_bound, upper_bound) end count end

require 'benchmark' def count_between_1(list_of_integers, lower_bound, upper_bound) count = 0 list_of_integers.each do |x| (x >= lower_bound && x <= upper_bound) ? count += 1 : next end count end def count_between_2(list_of_integers, lower_bound, upper_bound) count = 0 list_of_integers.each do |x| count += 1 if x.between?(lower_bound, upper_bound) end count end list_of_integers = (1..100_000).to_a lower_bound = 5 upper_bound = 80_000 Benchmark.bm do |x| x.report do count_between_1(list_of_integers, lower_bound, upper_bound) end x.report do count_between_2(list_of_integers, lower_bound, upper_bound) end end

结果如下：

user system total real 0.010000 0.000000 0.010000 ( 0.008910) 0.010000 0.000000 0.010000 ( 0.018098)

所以第一个变种要快得多.