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}
]
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}}
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}]
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