在Ruby中的數組嵌套迭代過程中跳過值

Question

我讀過一個輸入文件，並將每列解壓縮到單個數組中，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（輸入文件）

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

我應該如何遍歷以嵌套的方式的陣列以這樣的方式即，陣列的第一個元素 - 的「id」（ id_a_po87y）應該在文件Jack_details.txt（存儲數組的第一個元素）中檢查它的屬性，並且應該從行名稱中刪除 - leader（同樣是組數組的第一個數組元素）

同樣，id \ ruio66應該檢查它的可用性2dn文件中的ilty - 行ID中的John_Marsha - leader_1，如果存在，請將其刪除。

等..

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

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

and so on 

我的代碼，但值的數組元素

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 

我在做什麼錯跳過？

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 

輸出

For ID: id\a_po87y 
Not 
group row leader 

Not 
group row leader_1 

For ID: id\ruio66 

Answer 1

如果我理解正確的話，也許你可以採取slightl Ÿ不同的方法：

考慮到這

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

那麼你並不需要嵌套的循環，你可以嘗試循環只有一次，使用索引來獲取每個值形成每個陣列，東西像這樣：

(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 

此代碼假定id，file_name和group將始終具有相同的尺寸。