Ruby multiple group_by of array

Question

I have an array like this:

[{:a=>"2017-01-01", :b=>"2", :c=>"1"}, {:a=>"2017-01-01", :b=>"2", :c=>"2"}, {:a=>"2017-01-02", :b=>"5", :c=>"1"}]

If, for example, I group the array with the key ":a", the result is:

p = y.group_by { |g| g[:a]}

 => {"2017-01-01"=>[{:a=>"2017-01-01", :b=>"2", :c=>"1"}, {:a=>"2017-01-01", :b=>"2", :c=>"2"}], "2017-01-02"=>[{:a=>"2017-01-02", :b=>"5", :c=>"1"}]}

Now I want to group for each ":a" the key ":b" like:

 => {"2017-01-01"=>["2" => [{:a => "2017-01-01", :b => "2", :c => "1" }], ... }

How can I do? Thanks in advance.

Edit:

multiple group by ... 1° group_by a, 2° group_by b the result of 1°, etc..

y.group_by {|g| g[:a]}.map do |k,v|
 [
  k, v.group_by { |d| d[:b] }.map do |p,q|
   [p, q.group_by { |f| f[:c] }]
  end.to_h
 ]
end.to_h

1. [:a=>... is not a valid Ruby object. Please edit to correct. 2. It's not obvious what ... stands for. Please replace all ... with the values you desire as output for the example. 3. When you give an example (generally) assign a variable to each input object (e.g., arr = [{:a=>...) so that readers can refer to those variables (here just one) in answers and comments without having to define them. — Cary Swoveland
– Cary Swoveland, Commented Feb 9, 2017 at 19:12

Tom Lord · Accepted Answer · 2017-02-09 12:09:51Z

7

Quick answer, using your code above:

y = [{:a=>"2017-01-01", :b=>"2", :c=>"1"}, {:a=>"2017-01-01", :b=>"2", :c=>"2"}, {:a=>"2017-01-02", :b=>"5", :c=>"1"}]
p = y.group_by { |g| g[:a]}
q = p.map {|date, list| [date, list.group_by {|g| g[:b]}]}.to_h

This gives the desired result of:

q == {
  "2017-01-01" => {
    "2"=> [
      {:a=>"2017-01-01", :b=>"2", :c=>"1"},
      {:a=>"2017-01-01", :b=>"2", :c=>"2"}
    ]
  },
  "2017-01-02" => {
    "5"=> [
      {:a=>"2017-01-02", :b=>"5", :c=>"1"}
    ]
  }
}

This slightly odd pattern of mapping the hash to a list of arrays, then converting it back to a hash (using to_h) can be simplified slightly if you are working with a Rails v4.2+ project, by use of Hash#transform_values:

p.transform_values {|list| list.group_by {|g| g[:b]}}

However with that said, it is generally considered bad practice to be manipulating complex, nested hash objects like this. If your code gets too complicated, you may wish to consider redesigning how it works, in a more Object-Oriented manner.

edited Feb 9, 2017 at 12:09

answered Feb 9, 2017 at 11:58

Tom Lord

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9 Comments

sirion1987 Over a year ago

Thank you very much for the answer.

sirion1987 Over a year ago

I would submit to you another question. If I want to add an additional group_by the second filter, how do I proceed?

Tom Lord Over a year ago

An additional group_by? Do you mean for a final result like: {"2017-01-01" => { "2"=> {"1" => {:a=>"2017-01-01", :b=>"2", :c=>"1"} ....??

sirion1987 Over a year ago

Yes, I would an additional grop_by... I found this solution

y.group_by {|g| g[:a]}.map do |k,v|     [      k, v.group_by { |d| d[:b] }.map do |p,q|       [p, q.group_by { |f| f[:c] }]      end.to_h     ]    end.to_h

Tom Lord Over a year ago

Erm... Yes, something like that would work. But please, do note my above comment: "...bad practice to be manipulating complex, nested hash objects... consider redesigning how it works...". There are better ways of handling complex data structures in ruby than big nested hashes!

|

Oleksandr Holubenko · Accepted Answer · 2017-02-09 12:12:38Z

I think, a few from possible solutions is:

 y = [
  {
    :a=>"2017-01-01", 
    :b=>"2", 
    :c=>"1"
  }, 
  {
    :a=>"2017-01-01", 
    :b=>"2", 
    :c=>"2"
  }, 
  {
    :a=>"2017-01-02", 
    :b=>"5", 
    :c=>"1"
  }
]

first:

y.group_by { |g| [g[:a], g[:b]] }

but result would seems:

#=> {["2017-01-01", "2"]=>[{:a=>"2017-01-01", :b=>"2", :c=>"1"}, {:a=>"2017-01-01", :b=>"2", :c=>"2"}], ["2017-01-02", "5"]=>[{:a=>"2017-01-02", :b=>"5", :c=>"1"}]}

and second one:

y.group_by { |g| g[:a] }.map do |key, value| 
   [key, value.group_by { |g| g[:b] }] 
end.to_h

#=> {"2017-01-01"=>{"2"=>[{:a=>"2017-01-01", :b=>"2", :c=>"1"}, {:a=>"2017-01-01", :b=>"2", :c=>"2"}]}, "2017-01-02"=>{"5"=>[{:a=>"2017-01-02", :b=>"5", :c=>"1"}]}}

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