Hash with array in value

Question

Want to make such a conversion, but do not get right result

 old = {
      1 => %w(A E I O U L N R S T),
      2 => %w(D G),
      3 => %w(B C M P),
      4 => %w(F H V W Y),
      5 => %w(K),
      8 => %w(J X),
      10 => %w(Q Z)
    }

expected = {
  'a' => 1, 'b' => 3, 'c' => 3, 'd' => 2, 'e' => 1,
  'f' => 4, 'g' => 2, 'h' => 4, 'i' => 1, 'j' => 8,
  'k' => 5, 'l' => 1, 'm' => 3, 'n' => 1, 'o' => 1,
  'p' => 3, 'q' => 10, 'r' => 1, 's' => 1, 't' => 1,
  'u' => 1, 'v' => 4, 'w' => 4, 'x' => 8, 'y' => 4,
  'z' => 10
}

Implementing my conversion with nested loop. I thought that while inside loop is looping the outer loop will not increment. Here is my code:

class Etl

    def self.transform(old)

        result = {}

        old.each do |key, value|
            value.each_with_index do |v, i|
                result[v[i]] = key
            end
        end

        result

    end


end

The inverse hash part is interesting, but as a whole this is a bad question because downcasing has nothing to do with it. You should separate them into individual questions. — sawa
– sawa, Commented Feb 28, 2017 at 14:01

falsetru · Accepted Answer · 2017-02-27 16:47:06Z

5

You don't need to use index, just use v:

def self.transform(old)
    result = {}
    old.each do |key, value|
        value.each do |v|
            result[v.downcase] = key
        end
    end
    result
end

NOTE: used String#downcase to match the expected hash's key case.

Alternative using Array#map, Enumerable#flat_map, Enumerable#to_h:

def self.transform(old)
    old.flat_map { |key, value|
      value.map { |v|
        [v.downcase, key]
      }
    }.sort.to_h
end

edited Feb 27, 2017 at 16:47

answered Feb 27, 2017 at 15:55

falsetru

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2 Comments

Stefan Over a year ago

You can incorporate sort before to_h.

falsetru Over a year ago

@Stefan, Thank you for the comment. I added .sort before .to_h accordingly.

Cary Swoveland · Accepted Answer · 2017-02-27 21:50:00Z

2

old.flat_map { |k, v| v.map(&:downcase).product([k]) }.sort.to_h
  #=> {"a"=>1, "b"=>3, "c"=>3, "d"=>2, "e"=>1, "f"=>4, "g"=>2, "h"=>4, "i"=>1,
  #    "j"=>8, "k"=>5, "l"=>1, "m"=>3, "n"=>1, "o"=>1, "p"=>3, "q"=>10, "r"=>1,
  #    "s"=>1, "t"=>1, "u"=>1, "v"=>4, "w"=>4, "x"=>8, "y"=>4, "z"=>10}

answered Feb 27, 2017 at 21:50

Cary Swoveland

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