在Ruby中的数组嵌套迭代过程中跳过值

Question

我读过一个输入文件，并将每列解压缩到单个数组中，如下所示。

id = ["id\a_po87y", "id\ruio66", "id\rzd766", "id\ruio66", etc..] 

store = ["Jack", "John_Marsha", "123_Smart_option", "John_Marsha", etc...] 

group = ["leader", "leader_1", "fresher", "automation_dev", etc...] 

id_details.txt（输入文件）

id\a_po87y Jack leader 
id\ruio66 John_Marsha leader_1 
id\rzd766 123_Smart_option fresher 
id\ruio66 John_Marsha automation_dev 
.... 
etc 

我应该如何遍历以嵌套的方式的阵列以这样的方式即，阵列的第一个元素 - 的“id”（ id_a_po87y）应该在文件Jack_details.txt（存储数组的第一个元素）中检查它的属性，并且应该从行名称中删除 - leader（同样是组数组的第一个数组元素）

同样，id \ ruio66应该检查它的可用性2dn文件中的ilty - 行ID中的John_Marsha - leader_1，如果存在，请将其删除。

等..

id[0]--> store[0]-->group[0] 

id[1]--> store[1]-->group[1] 

and so on 

我的代码，但值的数组元素

file_dir = 'E:/ruby_work' 

file = File.open("E:/ruby_work/id_details.txt", "r") 
contents = file.each_line.map { |line| line.split("\t") }.transpose 
id, file_name, group = contents 

id.each do |ids| 
    puts "For ID: #{ids}" 
    file_name.each do |name| 
    value = File.open("#{file_dir}/#{name}_details.txt") 
    text = File.read(value) 
    #puts text 
    text.each_line do |el| 
     group.each do |gr| 
     if el.match(/#{gr}/) then 
      print "group row #{gr}\n" 
      print "Its matching\n" 
      replace = text.gsub(/#{Regexp.escape(ids)}\,\s/, '').gsub(/#{Regexp.escape(ids)}/, '').gsub /,\s*$/, '' 
     else print "Not\n" 
      print "group row #{gr}\n" 
     end 
     group.shift 
     end 
    end 
    file_name.shift 
    end 
end 
id.shift 

我在做什么错跳过？

Jack_details.txt

Joined on Feb 7, 2016 
Created by: Solomon (ruio66) 

[groups] 
leader_1 = id\rty67, id\mztrt, id\ruio66, ncr\025kc, id\a_po87y 
automation = id\bzo0l4, ccr\poxz7j 
automation_dev = id\ruio66 

John_Marsha_details.txt

Joined on Jan 7, 2016 
Created by: Jack Rondon 

[groups] 
leader_1 = id\sop0r2, id\34_dev, id\mz4d5, id\ruio66 
fresher = id\kzpo98, id\gz8sl7, id\cp0jxr, id\fzxlol, 
automation_dev = id\ruio66 

输出

For ID: id\a_po87y 
Not 
group row leader 

Not 
group row leader_1 

For ID: id\ruio66 

Answer 1

如果我理解正确的话，也许你可以采取slightl Ÿ不同的方法：

考虑到这

id[0]--> store[0]-->group[0] 
id[1]--> store[1]-->group[1] 

那么你并不需要嵌套的循环，你可以尝试循环只有一次，使用索引来获取每个值形成每个阵列，东西像这样：

(0..id.size - 1).each do |i| 
    puts "For ID: #{id[i]}" 
    value = File.open("#{file_dir}/#{file_name[i]}_details.txt") 
    text = File.read(value) 

    text.each_line do |line| 
    if line.match(/#{group[i]}/) 
     print "group row #{group[i]}\n" 
     print "Its matching\n" 
     replace = text.gsub(/#{Regexp.escape(id[i])}\,\s/, '') 
        .gsub(/#{Regexp.escape(id[i])}/, '') 
        .gsub(/,\s*$/, '') 
    else 
     print "Not\n" 
     print "group row #{group[i]}\n" 
    end 
    end 
end 

此代码假定id，file_name和group将始终具有相同的尺寸。