Transforming deep nested hash into multidimensional array in ruby

Question

I want to transform nested hashes into multidimensional arrays recursivley.

This is my current code:

def deep_to_a(hash)
  return  hash.to_a.each {|k, v|
    if k.is_a?(Hash) then
      hash[k.to_a] = hash.delete(k)
      deep_to_a(k)
    elsif v.is_a?(Hash) then
      hash[k] = hash[k].to_a
      if hash[k].any?{|k, v| k.is_a?(Hash) || v.is_a?(Hash)}
        deep_to_a(v)
      end
    end
  }
end

I want to get:

h = {11=>{12=>13, 14=>15}, 16=>17}
p deep_to_a(h) # => [[11, [[12, 13], [14, 15]]], [16, 17]]

But I get

[[11, {12=>13, 14=>15}], [16, 17]]

How can I make it work?

k.is_a?(Hash) – none of your keys is a hash.

Stefan
– Stefan

2018-05-17 14:54:38 +00:00
Commented May 17, 2018 at 14:54 — Stefan
– Stefan, Commented May 17, 2018 at 14:54

set0gut1 · Accepted Answer · 2018-05-19 07:44:31Z

2

A destructive function is hard to debug. In this case, map is better than each + destructive assignment.

The result of hash.to_a is an array, so your iteration |k, v| is incorrect.

def deep_to_a(hash)
  hash.map do |v|
    if v.is_a?(Hash) or v.is_a?(Array) then
      deep_to_a(v)
    else
      v
    end
  end
end

h = {11=>{12=>13, 14=>15}, 16=>17}
p deep_to_a(h)
# [[11, [[12, 13], [14, 15]]], [16, 17]]

edited May 19, 2018 at 7:44

answered May 17, 2018 at 15:13

set0gut1

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3 Comments

iGian Over a year ago

I was on the track, but this is great.

Cary Swoveland Over a year ago

You don't need return either, as the return value from the block is the last value computed, so it is returned by the method. Note that you could instead (but not better) write, case v; when Hash, Array; deep_to_a(v); else v; end. Note then is optional with an if and generally not used.

set0gut1 Over a year ago

@CarySwoveland Certainly... I updated my post. Thankyou again!

Cary Swoveland · Accepted Answer · 2018-05-17 20:41:33Z

1

def deep_to_a(h)
  h.map { |k,v| [k, Hash === v ? deep_to_a(v) : v] }
end

deep_to_a({11=>{12=>13, 14=>15}, 16=>17})
  #=> [[11, [[12, 13], [14, 15]]], [16, 17]]

answered May 17, 2018 at 20:41

Cary Swoveland

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