比较两个数组并删除唯一值[重复]

This question already has an answer here:

这个问题在这里已有答案:

How to sort an array in Ruby to a particular order? 3 answers

如何将Ruby中的数组排序为特定的顺序? 3个答案

I have two arrays

我有两个数组

ordered = [1, 2, 3, 4, 5]

some_list = [2, 6, 4]

I would like to compare the two arrays, then find the duplicates, and form it into a new array. The trick is to keep the array in the order provided in the ordered array.

我想比较两个数组,然后找到重复项,并将其形成一个新数组。诀窍是按照有序数组中提供的顺序保持数组。

new_array = [2, 4] # Result should be this

I've thought of one way to do it, however I think the performance can be improved.

我想过一种方法,但我认为性能可以提高。

ordered.each do |value1|
  some_list.include? value1 
    new_array << value1
  end
end

Is there any way I can improve this?

有什么办法可以改善吗?

Benchmark results

                   user     system      total        real
  using &        0.210000   0.000000   0.210000 (  0.212070)
  using select   0.220000   0.000000   0.220000 (  0.218889)

3 个解决方案

#1

Try this new_arry = ordered & some_list

试试这个new_arry = ordered&some_list

#2

It can also be done in following way:

它也可以通过以下方式完成:

uniques = ordered - some_list
duplicates = ordered - uniques

Order will be preserved.

订单将被保留。

Refer: http://ruby-doc.org/core-2.0.0/Array.html#method-i-2D

#3

ordered.select{|i| some_list.include?(i)}

Edit:

Not really sure if select is optimised for performance, but it is provided as shorter and clear alternative to the code provided by the OP.

不确定select是否针对性能进行了优化,但它是作为OP提供的代码的更短且更清晰的替代方式提供的。

A quick benchmark gave this results: Edit: adding the accepted answer alternative.

快速基准测试给出了以下结果:编辑:添加接受的答案替代方案。

           user     system      total        real
  each  0.000000   0.000000   0.000000 (  0.000005)
select  0.000000   0.000000   0.000000 (  0.000004)
     &  0.000000   0.000000   0.000000 (  0.000005)

#1