Question

我试图将数组中的数据重新排列为哈希格式，但是如果使用嵌套，我就会搞砸了

样本数据

[
  ["England", "London", "University College London ", "Faculty of Law"], 
  ["England", "London", "University College London ", "Faculty of Engineering"], 
  ["England", "London", "Imperial College London", "Faculty of Medicine"], 
  ["England", "Manchester", "University of Manchester", "Faculty of Engineering"]
]

预期产出

{:name=>"England", 
:level=>1, 
:children=>[{:name=>"London", 
             :level=>2, 
             :children=>[{:name=>"University College London ", 
                          :level=>3, 
                          :children=>[{:name=>"Faculty of Law", 
                                       :level=>4, 
                                       :children=>[]},
                                      {:name=>"Faculty of Engineering", 
                                       :level=>4, :children=>[]}]},
                         {:name=>"Imperial College London", 
                          :level=>3, 
                          :children=>[{:name=>"Faculty of Engineering", 
                                       :level=>4, 
                                      :children=>[]}]
                         }]
           }]
}

希望我提供明确的解释

PS。编辑以显示我尝试过的内容首先我制作哈希数组然后做这样的事情我认为不会混淆这么多

result = []
arr.each do |b|
  if result.any? {|r| r[:name] == b[:name]}
    if result.first[:children].any? {|r| r[:name] == b[:children].first[:name]}
      if result.first[:children].any?{|c| c[:children].any? {|r| r[:name] == b[:children].first[:children].first[:name] && c[:name] == b[:children].first[:name] }}
        if result.first[:children].any? {|r| r[:children].any? {|c| c[:children].any?{|k| k[:name] == b[:children].first[:children].first[:children].first[:name] && (c[:name] == b[:children].first[:children].first)}}}

        else
          result.first[:children].any?{|c| c[:children].any? {|r| r[:name] == b[:children].first[:children].first[:name] ; r[:children] << b[:children].first[:children].first[:children].first}}

        end #fourth
      else
        result.first[:children].any? {|r| r[:name] == b[:children].first[:name]; r[:children] << b[:children].first[:children].first}
      end 
    else
      result.any? {|r| r[:name] == b[:name] ; r[:children] << b[:children].first}
    end 
  else result << b 
  end
end

Answer 1

你可以像这样递归：

def map_objects(array, level = 1)
  new_obj = []
  array.group_by(&:shift).each do |key, val|
    new_obj << {:name=>key, :level=>level, :children=>map_objects(val, level + 1)} if key
  end
  new_obj
end

对于你的数组，它会像这样返回：

# [
# {:name => "England", :level => 1, :children => [
#     {:name => "London", :level => 2, :children => [
#         {:name => "University College London ", :level => 3, :children => [
#             {:name => "Faculty of Law", :level => 4, :children => []
#             },
#             {:name => "Faculty of Engineering", :level => 4, :children => []
#             }]
#         },
#         {:name => "Imperial College London", :level => 3, :children => [
#             {:name => "Faculty of Medicine", :level => 4, :children => []
#             }]
#         }]
#     },
#     {:name => "Manchester", :level => 2, :children => [
#         {:name => "University of Manchester", :level => 3, :children => [
#             {:name => "Faculty of Engineering", :level => 4, :children => []
#             }]
#         }]
#     }]
# }
# ]

Answer 2

我希望以下代码可以帮助您：

input = [
    ["England", "London", "University College London ", "Faculty of Law"], 
    ["England", "London", "University College London ", "Faculty of Engineering"], 
    ["England", "London", "Imperial College London", "Faculty of Medicine"], 
    ["England", "Manchester", "University of Manchester", "Faculty of Engineering"]
  ]

output = Array.new

input.each do |i|
    if output.select { |out| out[:name] == i[0] }.empty?
        output << { :name => i[0], :level => 1, :child => [] }
    end

    level1 = output.select { |out| out[:name] == i[0] }
    level1.each do |l1|
        if l1[:child].select { |l| l[:name] == i[1] }.empty?
            l1[:child] << { :name => i[1], :level => 2, :child => [] }
        end
    end

    level1.each do |l1|
        level2 = l1[:child].select { |l| l[:name] == i[1] }
        level2.each do |l2|
            if l2[:child].select { |l| l[:name] == i[2] }.empty?
                l2[:child] << { :name => i[2], :level => 3, :child => [] } 
            end
        end
    end

    level1.each do |l1|
        level2 = l1[:child].select { |l| l[:name] == i[1] }
        level2.each do |l2|
            level3 = l2[:child].select { |l| l[:name] == i[2] }
            level3.each do |l3|
                if l3[:child].select { |l| l[:name] == i[3] }.empty?
                    l3[:child] << { :name => i[3], :level => 4 } 
                end
            end
        end
    end
end

puts output

使用ruby从数组创建嵌套哈希

2 个答案: