[英]combine keys in array of hashes
我映射查询结果以创建按organisation_id
分组的哈希数组,如下所示:
results.map do |i|
{
i['organisation_id'] => {
name: capability.name,
tags: capability.tag_list,
organisation_id: i['organisation_id'],
scores: {i['location_id'] => i['score']}
}
}
在地图外部定义capability
。
结果看起来像:
[{1=>{:name=>"cap1", :tags=>["tag A"], :scores=>{26=>4}}}, {1=>{:name=>"cap1", :tags=>["tag A"], :scores=>{12=>5}}}, {2 => {...}}...]
对于每个organisation_id
,数组中都有一个单独的条目。 我想合并这些散列并按如下方式结合scores
键:
[{1=>{:name=>"cap1", :tags=>["tag A"], :scores=>{26=>4, 12=>5}}}, {2=>{...}}... ]
要创建results
我使用以下AR:
Valuation.joins(:membership)
.where(capability: capability)
.select("valuations.id, valuations.score, valuations.capability_id, valuations.membership_id, memberships.location_id, memberships.organisation_id")
.map(&:serializable_hash)
Valuation
模型:
class Valuation < ApplicationRecord
belongs_to :membership
belongs_to :capability
end
Membership
模式:
class Membership < ApplicationRecord
belongs_to :organisation
belongs_to :location
has_many :valuations
end
results
片段:
[{"id"=>1, "score"=>4, "capability_id"=>1, "membership_id"=>1, "location_id"=>26, "organisation_id"=>1}, {"id"=>16, "score"=>3, "capability_id"=>1, "membership_id"=>2, "location_id"=>36, "organisation_id"=>1}, {"id"=>31, "score"=>3, "capability_id"=>1, "membership_id"=>3, "location_id"=>26, "organisation_id"=>2}, {"id"=>46, "score"=>6, "capability_id"=>1, "membership_id"=>4, "location_id"=>16, "organisation_id"=>2}...
我假设每个组织:名称,标签列表和organization_id保持不变。
your_hash = results.reduce({}) do |h, i|
org_id = i['organisation_id']
h[org_id] ||= {
name: capability.name,
tags: capability.taglist,
organisation_id: org_id,
scores: {}
}
h[org_id][:scores][i['location_id']] = i['score']
# If the location scores are not strictly exclusive, you can also just +=
h
end
我相信这可行,但是需要数据来对其进行测试。
results.each_with_object({}) do |i,h|
h.update(i['organisation_id'] => {
name: capability.name,
tags: capability.tag_list,
organisation_id: i['organisation_id'],
scores: {i['location_id'] => i['score']}) { |_,o,n|
o[:scores].update(n[:score]); o }
}
end.values
这使用Hash#update的形式(又名merge!
),该形式使用一个块来确定要合并的两个哈希中存在的键的值。 请查阅文档以获取每个块变量_
, o
和n
。
假设, result
就是您的最终哈希数组:
result.each_with_object({}) do |e, obj|
k, v = e.flatten
if obj[k]
obj[k][:scores] = obj[k][:scores].merge(v[:scores])
else
obj[k] = v
end
end
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