Group array of hashes by multiple keys to find average in Ruby

Question

I have found solution for my problem done in JS but i need it done in Ruby (RoR). Here is link to problem and solution: Find average value for array of hashes using multiple group by

So i have array of hashes that needs to be grouped by keys (first subject_id then element_id) and then find average values for them. Number of hashes in array is not fixed.

Below is input array:

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}
]

Result:

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}
]

MaicolBen · Accepted Answer · 2017-03-23 01:23:08Z

The result shown in the question isn't the average, it's the sum, so the result will be different:

  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

And the result from this is:

  [{: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}]

Cary Swoveland · Accepted Answer · 2017-03-23 03:30:45Z

I suggest that a counting hash be used to obtain the subtotals for the key :value and then construct the required of array of hashes from that hash. This uses the form of Hash#new that takes an argument that is the hash's default value. That means that if a hash h does not have a key k, h[k] returns the default value.

Computing totals

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, as a first step, unpacks the expression

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

changing it to

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

If h does not have a key [g[:subject_id], g[:element_id]], h[[g[:subject_id], g[:element_id]]] on the right side of the equality returns the default value, 0.

Note that

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}

Computing averages

Only a small change is needed to compute averages.

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}]

Note

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}}

Thank you for help, i made a mistake in example where i showed sum as result instead of average.

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