ruby - 比较除元素 x、y、z 之外的两个数组(ruby)

Question

还有其他更简单、更好的方法吗？

require 'pp'

a1 = ["02/28/10","Webinars","131","0","26 Feb 2010","0","3d, 8h, 49m, 18s"]
a2 = ["02/20/10","Webinars","131","9","26 Feb 2010","0","3d, 8h, 49m, 18s"]

def compare(array1,array2,ignore)

  tmp1 = Array.new
  tmp2 = Array.new
  0.upto(array1.length-1) {|index|
    if !ignore.include?(index)
      tmp1 << array1[index]
      tmp2 << array2[index]
    end
  }
  if tmp1 == tmp2 
    return true
  else
    return false    
  end
end

pp a1
pp a2
puts
puts compare(a1,a2,[0,3])

输出是

["02/28/10", "Webinars", "131", "0", "26 Feb 2010", "0", "3d, 8h, 49m, 18s"]
["02/20/10", "Webinars", "131", "9", "26 Feb 2010", "0", "3d, 8h, 49m, 18s"]

true

Answer 1

最简单的代码(需要 Ruby 1.8.7 或更高版本):

def compare(array_a, array_b, ignore_list)
  array_a.zip(array_b).each_with_index.all? do |a, b, idx|
    a == b or ignore_list.include? idx
  end
end

我怀疑它也会更快(因为它使用单个 zip 而不是为每个项目单独询问数组)——尽管这可能不是速度非常重要的情况。

顺便说一句，几乎任何时候我都直接在 Ruby 中索引数组(如 some_array[i])而不是使用高阶方法如 map 或 each，我认为这表明我可能在标准库中遗漏了一些东西，并且该算法的效率可能低于高度优化的库函数。