代码之家 › 专栏 › 技术社区 › Bharath M

我有一个数组,如描述中所述,需要在Ruby中将其转换为格式化散列

hash ruby arrays

0

Bharath M · 技术社区 · 8 年前

我有一个包含许多数组的文件,如下所示:

[["default", "'drop'"],
["rule", "49", "action", "'accept'"],
["rule", "49", "description", "'This one is for'"],
["rule", "49", "destination", "address", "'1.2.3.4/20'"],
["rule", "50", "action", "'accept'"],
["rule", "50", "description", "'Once more'"],
["rule", "50", "destination", "address", "'1.2.3.5/20'"]]

我想要一个Ruby代码,它可以使它看起来像:

{
  'default': 'drop',
  'rule': { 
    '49': { 
      'destination': { 
        'address': '1.2.3.4/20' 
      },
      'action': 'accept', 
      'description': 'This one is for' 
    },
    '50': {
      'destination': {
        'address': '1.2.3.5/20'
      },
      'action': 'accept',
      'description': 'Once more'
    }
  }
}

我尝试了不同的方法,替换和保留只有最后一把或几把钥匙丢失。请帮帮我。

3 回复 | 直到 8 年前

1

4

Cary Swoveland 8 年前

这是另一个递归解决方案。使用递归的一个优点是数组可以是任意大小的。

arr = [
  ["default", "'drop'"],
  ["rule", "49", "action", "'accept'"],
  ["rule", "49", "description", "'This one is for'"],
  ["rule", "49", "destination", "address", "'1.2.3.4/20'"],
  ["rule", "50", "action", "'accept'"],
  ["rule", "50", "description", "'Once more'"], 
  ["rule", "50", "destination", "address", "'1.2.3.5/20'"]
]

def recurse(arr)
  arr.group_by(&:first).transform_values do |a|
    a.map! { |r| r.drop(1) }
    a.size == 1 && a[0].size == 1 ? a[0][0] : recurse(a)
  end
end

recurse arr
  #=> { "default"=>"'drop'",
  #     "rule"=>{
  #       "49"=>{
  #         "action"=>"'accept'",
  #         "description"=>"'This one is for'",
  #         "destination"=>{"address"=>"'1.2.3.4/20'"}
  #       },
  #       "50"=>{
  #         "action"=>"'accept'",
  #         "description"=>"'Once more'",
  #         "destination"=>{"address"=>"'1.2.3.5/20'"}
  #       }
  #     }
  #   }

注意第一步如下。

arr.group_by(&:first)
  #=> {"default"=>[["default", "'drop'"]],
  #    "rule"=>[["rule", "49", "action", "'accept'"],
  #             ["rule", "49", "description", "'This one is for'"],
  #             ["rule", "49", "destination", "address", "'1.2.3.4/20'"],
  #             ["rule", "50", "action", "'accept'"],
  #             ["rule", "50", "description", "'Once more'"],
  #             ["rule", "50", "destination", "address", "'1.2.3.5/20'"]]}

2

Simple Lime 8 年前

这就利用了将块传递给 Hash::new 它通过 default_proc :

output = Hash.new { |hash, key| hash[key] = Hash.new(&hash.default_proc) }

现在,我们只需迭代数组,我们知道每个子数组中的最后一个值将成为最终值(即,不是子散列值),并且当我们访问一个没有值的键时,其他所有需要创建一个新的子散列(如上所述),这样我们就可以使用 Hash#dig ,我们还需要将所有键转换为符号,并且(看起来)从值的开头和结尾删除单引号:

array.each_with_object(output) do |(*nesting, key, value), hash|
  hash = nesting.empty? ? hash : hash.dig(*nesting.map(&:to_sym))

  hash[key.to_sym] = value.gsub(/(\A\'|\'\z)/, '')
end

用它来运行 array 设置为您的输入数组,我们将得到以下结果:

{
  :default => "drop",
  :rule => {
    :"49" => {
      :action => "accept",
      :description => "This one is for",
      :destination => {
        :address => "1.2.3.4/20"
      }
    },
    :"50" => {
      :action => "accept",
      :description => "Once more",
      :destination => {
        :address => "1.2.3.5/20"
      }
    }
  }
}

3

2

Lars Haugseth 8 年前

class Hash
  def deep_store(keys, value)
    if keys.size > 1
      self[keys.first] ||= {}
      self[keys.first].deep_store keys[1..-1], value
    else
      self[keys.first] = value
    end
    self
  end
end

input = [
  ["default", "'drop'"],
  ["rule", "49", "action", "'accept'"],
  ["rule", "49", "description", "'This one is for'"],
  ["rule", "49", "destination", "address", "'1.2.3.4/20'"],
  ["rule", "50", "action", "'accept'"],
  ["rule", "50", "description", "'Once more'"],
  ["rule", "50", "destination", "address", "'1.2.3.5/20'"]
]

result = input.each_with_object({}) do |(*keys, value), hash|
  hash.deep_store keys, value
end

puts result

# => {"default"=>"'drop'", "rule"=>{"49"=>{"action"=>"'accept'", "description"=>"'This one is for'", "destination"=>{"address"=>"'1.2.3.4/20'"}}, "50"=>{"action"=>"'accept'", "description"=>"'Once more'", "destination"=>{"address"=>"'1.2.3.5/20'"}}}}