Question

我已经找到了解决我在JS中完成的问题的解决方案，但我需要在Ruby（RoR）中完成。以下是问题和解决方案的链接：Find average value for array of hashes using multiple group by

所以我有一系列哈希需要按键分组（先subject_id然后element_id），然后找到它们的平均值。数组中的哈希数不固定。

下面是输入数组：

a=[
{:subject_id=>1, :element_id=>2, :value=>55},
{:subject_id=>1, :element_id=>4, :value=>33},
{:subject_id=>1, :element_id=>2, :value=>33},
{:subject_id=>1, :element_id=>4, :value=>1},
{:subject_id=>1, :element_id=>2, :value=>7},
{:subject_id=>1, :element_id=>4, :value=>4},
{:subject_id=>2, :element_id=>2, :value=>3},
{:subject_id=>2, :element_id=>2, :value=>5},
{:subject_id=>2, :element_id=>4, :value=>9}
]

结果：

b=[
{:subject_id=>1, :element_id=>2, :value=>95},
{:subject_id=>1, :element_id=>4, :value=>38},
{:subject_id=>2, :element_id=>2, :value=>8},
{:subject_id=>2, :element_id=>4, :value=>9}
]

Answer 1

我建议使用计算哈希来获取密钥:value的小计，然后从该哈希构造所需的哈希数组。这使用Hash#new的形式，它接受一个参数，即哈希的默认值。这意味着如果哈希h没有密钥k，h[k]将返回默认值。

计算总计

a.each_with_object(Hash.new(0)) {|g,h| h[[g[:subject_id], g[:element_id]]] += g[:value]}.
  map {|(sub, el), tot| { subject_id: sub, element_id: el, value: tot}}
  #=> [{:subject_id=>1, :element_id=>2, :value=>95},
  #    {:subject_id=>1, :element_id=>4, :value=>38},
  #    {:subject_id=>2, :element_id=>2, :value=>8},
  #    {:subject_id=>2, :element_id=>4, :value=>9}]

Ruby，作为第一步，解压缩表达式

h[[g[:subject_id], g[:element_id]]] += g[:value]

将其更改为

h[[g[:subject_id], g[:element_id]]] = h[[g[:subject_id], g[:element_id]]] + g[:value]

如果h没有密钥[g[:subject_id], g[:element_id]]，则等式右侧的h[[g[:subject_id], g[:element_id]]]会返回默认值0。

请注意

a.each_with_object(Hash.new(0)) {|g,h| h[[g[:subject_id], g[:element_id]]] += g[:value]}
  #=> {[1, 2]=>95, [1, 4]=>38, [2, 2]=>8, [2, 4]=>9}

计算平均值

计算平均值只需要很小的改动。

a.each_with_object({}) do |g,h|
  pair = [g[:element_id], g[:subject_id]] 
  h[pair] = {tot: 0, count: 0} unless h.key?(pair)
  h[pair] = {tot: h[pair][:tot] + g[:value], count: h[pair][:count]+1}
end.map {|(sub, el),h| {subject_id: sub, element_id: el,
                        average: (h[:tot].to_f/h[:count]).round(1)}} 
  #=> [{:subject_id=>2, :element_id=>1, :average=>31.7},
  #    {:subject_id=>4, :element_id=>1, :average=>12.7},
  #    {:subject_id=>2, :element_id=>2, :average=>4.0},
  #    {:subject_id=>4, :element_id=>2, :average=>9.0}]

请注意

a.each_with_object({}) do |g,h|
  pair = [g[:element_id], g[:subject_id]] 
  h[pair] = {tot: 0, count: 0} unless h.key?(pair)
  h[pair] = {tot: h[pair][:tot] + g[:value], count: h[pair][:count]+1}
end
  #=> {[2, 1]=>{:tot=>95, :count=>3}, [4, 1]=>{:tot=>38, :count=>3},
  #    [2, 2]=>{:tot=> 8, :count=>2}, [4, 2]=>{:tot=> 9, :count=>1}}

Answer 2

问题中显示的结果不是平均值，而是总和，因此结果会有所不同：

  def groupByAndAverage(a)
    b = []

    a.each_with_index do |element, key|
      index = b.index do |x| 
        x != element && 
          x[:subject_id] == element[:subject_id] && 
            x[:element_id] == element[:element_id] 
      end
      if index
        b[index][:value] += element[:value]
        b[index][:amount] += 1
      else
        b.push a[key].merge(amount: 1)
      end

      true
    end
    b.map do |element| 
      element[:value] = element[:value] / element[:amount]
      element.delete(:amount)
      element
    end
    b
  end

结果是：

  [{:subject_id=>1, :element_id=>2, :value=>31}, 
    {:subject_id=>1, :element_id=>4, :value=>12}, 
    {:subject_id=>2, :element_id=>2, :value=>4}, 
    {:subject_id=>2, :element_id=>4, :value=>9}]

通过多个键分组哈希值，以便在Ruby中找到平均值

2 个答案: