Question

我有一个哈希数组

[{:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>10},
 {:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>10},
 {:id=>2, :book=>{:title=>"title2", :desc=>"title2", :author=>"title2"}, :pages=>30}]

如何求和pages值，仅保留唯一键？例如：

[{:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>20},
 {:id=>2, :book=>{:title=>"title2", :desc=>"title2", :author=>"title2"}, :pages=>30}]

Answer 1

让我们按ID分组，并将每个结果映射到一个结构，其中页面是该ID的所有页面的总和。

array.
  group_by { |item| item[:id] }.
  map do |id, items| 
    page_sum = items.sum { |i| i[:pages] }
    Hash[:id, id, :book, items.first[:book], :pages, page_sum] 
  end

Answer 2

arr = [
  {:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>10},
  {:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>10},
  {:id=>2, :book=>{:title=>"title2", :desc=>"title2", :author=>"title2"}, :pages=>30}
]

arr.each_with_object({}) do |g,h|
  h.update(g[:id]=>g) { |_,o,n| o.merge(pages: o[:pages] + n[:pages]) }
end.values
  #=> [{:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>30},
  #    {:id=>2, :book=>{:title=>"title2", :desc=>"title2", :author=>"title2"}, :pages=>30}]

请注意values的接收者等于

{1=>{:id=>1, :book=>{:title=>"title1", :desc=>"title1", :author=>"title1"}, :pages=>30},
 2=>{:id=>2, :book=>{:title=>"title2", :desc=>"title2", :author=>"title2"}, :pages=>30}}

这使用Hash#update（也称为merge!）形式，并使用了该块

{ |_,o,n| o.merge(pages: o[:pages] + n[:pages]) }

确定合并的两个哈希中存在的键（块变量_）的值。有关块变量o和n的说明，请参阅文档。

Hash#update和Enumerable#group_by是处理此类问题时通常采用的两种方法。都可以使用。它们的效率大致相等，因此选择很大程度上是个人喜好之一。

如何合并按值求和的哈希数组？

2 个答案: