Ruby - Group an array of hashes by key

Question

I have an array of hashes (edited):

data = [
    {id: 1, name: "Amy", win: 1, defeat: 0},
    {id: 1, name: "Amy", win: 1, defeat: 3},
    {id: 2, name: "Carl", win: 0, defeat: 1},
    {id: 2, name: "Carl", win: 2, defeat: 1}
]

How can I group or merge into something like this using the key "name" as reference:

data = [
    {id: 1, name: "Amy", win: 2, defeat: 3},
    {id: 2, name: "Carl", win: 2, defeat: 2}
]

edited I forgot to mention that I have an ID too that can't be added.

Does this (stackoverflow.com/questions/5191578/…) help you? — Henrik Andersson
– Henrik Andersson, Commented Jun 30, 2014 at 18:42
The rule for merger is not clear. What gives id: 1 for name: "Amy" while giving it win: 2? — sawa
– sawa, Commented Jun 30, 2014 at 19:09
@sawa The merger rule is crystal clear: sum wins and defeats grouped by name. — mu is too short
– mu is too short, Commented Jun 30, 2014 at 19:12

Arup Rakshit · Accepted Answer · 2014-06-30 19:07:08Z

7

Here is my try

#!/usr/bin/env ruby

data = [
    {"name"=> "Amy", "win" => 1, "defeat" => 0},
    {"name"=> "Amy", "win" => 1, "defeat" => 3},
    {"name"=> "Carl", "win" => 0, "defeat" => 1},
    {"name"=> "Carl", "win" => 2, "defeat" => 1}
]

merged_hash = data.group_by { |h| h['name'] }.map do |_,val| 
  val.inject do |h1,h2| 
    h1.merge(h2) do |k,o,n|
      k == 'name' ? o : o + n 
    end
  end
end

merged_hash
# => [{"name"=>"Amy", "win"=>2, "defeat"=>3},
#     {"name"=>"Carl", "win"=>2, "defeat"=>2}]

Answer to the edited post :-

#!/usr/bin/env ruby

data = [
    {id: 1, name: "Amy", win: 1, defeat: 0},
    {id: 1, name: "Amy", win: 1, defeat: 3},
    {id: 2, name: "Carl", win: 0, defeat: 1},
    {id: 2, name: "Carl", win: 2, defeat: 1}
]

merged_hash = data.group_by { |h| h.values_at(:name, :id) }.map do |_,val| 
  val.inject do |h1,h2| 
    h1.merge(h2) do |k,o,n|
      %i(id name).include?(k) ? o : o + n 
    end
  end
end

merged_hash
# => [{:id=>1, :name=>"Amy", :win=>2, :defeat=>3},
#     {:id=>2, :name=>"Carl", :win=>2, :defeat=>2}]

edited Jun 30, 2014 at 19:07

answered Jun 30, 2014 at 18:46

Arup Rakshit

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2 Comments

6ft Dan Over a year ago

Each of his data lines are JSON. You can do a JSON.parse before each line like this

data = []; data << JSON.parse('    {"name": "Amy",  "win": 1, "defeat": 0} '); data << JSON.parse('    {"name": "Amy",  "win": 1, "defeat": 3} '); data << JSON.parse('    {"name": "Carl", "win": 0, "defeat": 1} '); data << JSON.parse('    {"name": "Carl", "win": 2, "defeat": 1} ');

. You can use this before following this answer.

Seralto Over a year ago

@ArupRakshit, your solution is OK, but I forgot to mention that I have an Id too, and I wrote the hash correctly this time.

mu is too short · Accepted Answer · 2019-03-23 17:46:10Z

1

You can do it in one pass with each_with_object and a Hash-memo with an appropriate default. For example:

data.each_with_object(Hash.new { |h, k| h[k] = { :id => k.first, :name => k.last, :win => 0, :defeat => 0 } }) do |h, m|
  k = h.values_at(:id, :name)
  m[k][:win   ] += h[:win   ]
  m[k][:defeat] += h[:defeat]
end.values

The basic trick is to cache the results indexed by an appropriate key ([ h[:id], h[:name] ] in this case) and use the values to store what you're after. The default proc on the m Hash autovivifies cached values and then you can apply simple summations during iteration. And a final values call to unwrap the cache.

edited Mar 23, 2019 at 17:46

answered Jun 30, 2014 at 19:11

mu is too short

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Comments

reto · Accepted Answer · 2014-07-01 11:13:20Z

0

Good place where you can use group_by

result = []
data.group_by{|d| d[:id]}.each do {|name, records|
  win = 0
  defeat = 0

  records.each do |r|
    win += r[:win]
    defeat += r[:defeat]
  end
  f = records.first
  results << {:id => f[:id], :name => f[:name], :win => win, :defeat => defeat}
end

answered Jul 1, 2014 at 11:13

reto

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