Count the occurence of values in Ruby -

i'm trying count numbers of different values in mysql table columns. range of possible values integer , 0-10. following code working, i'm wondering if there more elegant way this?

# example data mysql result = [{ :column1 => "2", :column2 => "3", :column3 => "1"},{ :column1 => "2", :column2 => "3", :column3 => "1"},{ :column1 => "1", :column2 => "2", :column3 => "3"}]  # init hash final_result = hash.new { |h, k| h[k] = {  } }  # loop result columns result.each |single_column|      # loop single items inside columns     single_column.each |single_result|              # create column if not exist             if final_result[single_result[0]][single_result[1]].nil?                 final_result[single_result[0]][single_result[1]] = 1             else                 final_result[single_result[0]][single_result[1]] += 1             end     end end  puts final_result # => {:column1=>{"2"=>2, "1"=>1}, :column2=>{"3"=>2, "2"=>1}, :column3=>{"1"=>2, "3"=>1}} 

there's room cleaning here. obvious part long, clunky if statement. test vs nil? pointless, remember in ruby only things logically false false , nil, since false never going show here, test vs. nil can removed.

more that, though, you're on right track custom hash.new call, don't go far enough. why not initialize second tier zero?

that results in code looks like:

result = [   { :column1 => "2", :column2 => "3", :column3 => "1"},   { :column1 => "2", :column2 => "3", :column3 => "1"},   { :column1 => "1", :column2 => "2", :column3 => "3"} ]  # init hash final_result = hash.new { |h, k| h[k] = hash.new(0) }  # loop result columns result.each |single_column|   single_column.each |r|     final_result[r[0]][r[1]] += 1   end end  puts final_result.inspect