ruby - map through hash returning nil?

user_9394106 0
Viewed 70
{"key_a" => "value_a", "key_b" => "value_b", "key_c" => { "key_c_a" => nil, "key_c_b" => nil, "key_c_c" => "value_c_c"}, "key_d" => nil, "key_e" => nil }

how do i map the hash such that i get the result as "key_c_a", "key_c_b", "key_d" and key_e"?

i keep getting my answer as an empty array and here is the method in the module that is used in the main class

    def self.get_null_keys(container)
      container.map do |key, value|
        if value.is_a?(Array)
          value.map { |field| render_null_fields(field) }
        elsif value.is_a?(Hash)
          render_null_fields(value)
        elsif value.nil?
          key
        end
      end
    end
ruby
2 Answers

Do it recursively:

h={"key_a" => "value_a", "key_b" => "value_b", "key_c" => { "key_c_a" => nil, "key_c_b" => nil, "key_c_c" => "value_c_c"}, "key_d" => nil, "key_e" => nil }

def find_nils(x, memo=[]) 
    x.inject({}){ |m, (k, v)|
        v = find_nils(v, memo) if v.is_a? Hash  # arbitrarily recursive
        memo << k if v.nil? 
    }
    memo
end

Instead of inject as a skeleton you can also use each:

def find_nils(x, memo=[]) 
    x.each{ |k, v|
        v = find_nils(v, memo) if v.is_a? Hash  # arbitrarily recursive
        memo << k if v.nil? 
    }
    memo
end

Either:

p find_nils(h)
["key_c_a", "key_c_b", "key_d", "key_e"]
user_2986070

Naive recursive way to do it:

hash = {"key_a" => "value_a", "key_b" => "value_b", "key_c" => { "key_c_a" => nil, "key_c_b" => nil, "key_c_c" => "value_c_c"}, "key_d" => nil, "key_e" => nil }

def get_null_values(hash, array = [])
  hash.each do |key, value|
    array << key if value.nil?
    get_null_values(value, array) if value.is_a?(Hash)
  end
  array
end

get_null_values(hash) # ["key_c_a", "key_c_b", "key_d", "key_e"]
user_39562050
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