将两个哈希与共享根合并答案

【问题标题】：Merging two hashes with shared root将两个哈希与共享根合并
【发布时间】：2019-02-04 18:46:18
【问题描述】：

我有：

apartment1 = {"base" => {"floor1" => {"apartment1" => {"rooms_number" => 4}}}}
apartment2 = {"base" => {"floor1" => {"apartment2" => {"rooms_number" => 6}}}}

分享{"base" => {"floor1" =>。

我将如何合并共享部分下的公寓以获得：

{"base" => {"floor1" => {
  "apartment1" => {"rooms_number" => 6},
  "apartment2" => {"rooms_number" => 6}
}}}

我将apartment1 和apartment2 合并到apartments，得到了这个：

apartments = {}
apartments.merge!(apartment1)
# => {"base" => {"floor1" => {"apartment1" => {"rooms_number" => 4}}}} 
apartments.merge!(apartment2)
# => {"base" => {"floor1" => {"apartment2" => {"rooms_number" => 6}}}}

【问题讨论】：

标签： ruby hash

【解决方案1】：

如果您正在使用 Rails（或者在任何情况下如果您愿意使用 ActiveSupport），您可以通过 deep_merge 进行此操作：

2.5.1 :001 > apartment1= {"base" => {"floor1" => {"apartment1" => {"rooms_number" => 4} } }}
 => {"base"=>{"floor1"=>{"apartment1"=>{"rooms_number"=>4}}}} 
2.5.1 :002 > apartment2= {"base" => {"floor1" => {"apartment2" => {"rooms_number" => 6} } }}
 => {"base"=>{"floor1"=>{"apartment2"=>{"rooms_number"=>6}}}} 
2.5.1 :003 > apartments = apartment1.deep_merge apartment2
 => {"base"=>{"floor1"=>{"apartment1"=>{"rooms_number"=>4}, "apartment2"=>{"rooms_number"=>6}}}}

见https://apidock.com/rails/Hash/deep_merge

另请参阅此类似问题： Ruby: merge nested hash

【讨论】：

【解决方案2】：

手动处理：

apartments = apartment1
apartments['base']['floor1'].merge!(apartment2['base']['floor1'])

p apartments
#=> {"base"=>{"floor1"=>{"apartment1"=>{"rooms_number"=>4}, "apartment2"=>{"rooms_number"=>6}}}}

或者从 Rails 中窃取（！）深度合并算法：

def deep_merge(first, second)
  first.merge(second) do |key, oldval, newval|
    oldval = oldval.to_hash if oldval.respond_to?(:to_hash)
    newval = newval.to_hash if newval.respond_to?(:to_hash)
    oldval.class.to_s == 'Hash' && newval.class.to_s == 'Hash' ? deep_merge(oldval, newval) : newval
  end
end

所以你可以在纯 Ruby 中使用它：

deep_merge(apartment1, apartment2) #=> {"base"=>{"floor1"=>{"apartment1"=>{"rooms_number"=>4}, "apartment2"=>{"rooms_number"=>6}}}}

【讨论】：

【解决方案3】：

以下递归方法应提供所需的结果。

def combine_em(arr)
  (k1, k2), (v1, v2) = arr.map(&:flatten).transpose
  (k1==k2 && v1.is_a?(Hash)) ? { k1=>combine_em([v1, v2]) } :
    {}.merge(*arr) 
end

arr = [{"base"=>{"floor1"=>{"apt1"=>{"room"=>4}}}},
       {"base"=>{"floor1"=>{"apt2"=>{"room"=>6}}}}]
combine_em arr
  #=>  {"base"=>{"floor1"=>{"apt1"=>{"room"=>4},
  #                         "apt2"=>{"room"=>6}}}}

arr = [{"base"=>{"floor1"=>{"level1"=>{"apt1"=>{"room"=>4}}}}},
       {"base"=>{"floor1"=>{"level1"=>{"apt2"=>{"room"=>6}}}}}]
combine_em arr
  #=> {"base"=>{"floor1"=>{"level1"=>{"apt1"=>{"room"=>4},
  #                                   "apt2"=>{"room"=>6}}}}}

arr = [{"base"=>{"floor1"=>{"apt1"=>{"room"=>4}}}},
       {"base"=>{"floor2"=>{"apt1"=>{"room"=>6}}}}]
combine_em arr
  #=>  {"base"=>{"floor1"=>{"apt1"=>{"room"=>4}},
  #              "floor2"=>{"apt1"=>{"room"=>6}}}}

arr = [{"base"=>{"floor1"=>{"apt1"=>{"room1"=>4}}}},
       {"base"=>{"floor1"=>{"apt1"=>{"room2"=>6}}}}]
combine_em arr
  #=>  {"base"=>{"floor1"=>{"apt1"=>{"room1"=>4,
  #                                  "room2"=>6}}}}
arr = [{"base1"=>{"floor1"=>{"apt1"=>{"room"=>4}}}},
       {"base2"=>{"floor2"=>{"apt1"=>{"room"=>6}}}}]
combine_em arr
  #=> {"base1"=>{"floor1"=>{"apt1"=>{"room"=>4}}},
  #    "base2"=>{"floor2"=>{"apt1"=>{"room"=>6}}}}

arr = [{"base"=>{"floor1"=>{"apt1"=>{"room"=>4}}}},
       {"base"=>{"floor1"=>{"apt1"=>{"room"=>6}}}}]
combine_em arr
  #=>  {"base"=>{"floor1"=>{"apt1"=>{"room"=>6}}}}

arr 的最后一个示例（如果可能发生）可能不会给出预期的结果。如果是这样，则有必要在这种情况下指定所需的返回值。

Hash#merge 在 Ruby v2.6 中被更改为允许多个参数，这就是我们现在可以编写的原因

arr = [{:a=>1}, {:b=>2}, {:c=>3}]
{}.merge(*arr)
  #=> {:a=>1, :b=>2, :c=>3}

要支持早期版本的 Ruby，请编写

arr.reduce(&:merge)

这是的简写

arr.reduce { |h,g| h.merge(g) }

见Enumerable#reduce（又名inject）。

要全面了解递归的工作原理，可能需要使用 puts 语句对方法进行加盐。

【讨论】：

【解决方案4】：

您不想在根级别合并，而是在更深的两个级别合并。

apartment1.merge(apartment2){|_, h1, h2| h1.merge(h2){|_, h1, h2| h1.merge(h2)}}
# =>
# {"base" => {"floor1" => {
#   "apartment1" => {"rooms_number" => 4},
#   "apartment2" => {"rooms_number" => 6}
# }}}

【讨论】：

这是最短的答案，但是，我们将不胜感激！