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将两个哈希与共享根合并

hash ruby

0

simo · 技术社区 · 7 年前

我有:

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

那份 {"base" => {"floor1" => .

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}
}}}

我合并了 apartment1 和 apartment2 进入 apartments ,得到这个:

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

4 回复 | 直到 7 年前

1

Benissimo 7 年前

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 :

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}}}}

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

另见类似问题: Ruby: merge nested hash

2

1

iGian 7 年前

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

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

3

1

Cary Swoveland 7 年前

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}}}}

最后一个例子 arr (如果可能发生),可能不会产生预期结果。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,这就是为什么我们现在可以编写

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

要支持Ruby的早期版本,请编写

arr.reduce(&:merge)

这是

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

看见 Enumerable#reduce (又名 inject

为了充分理解递归的工作原理,可能需要在方法中添加盐 puts

4

0

sawa 7 年前

您不希望在根级别合并,而是希望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}
# }}}