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Ruby deep nested hash from array



2019 Community Moderator ElectionCalling shell commands from RubyA concise explanation of nil v. empty v. blank in Ruby on RailsHow to write a switch statement in RubyHow to convert a string to lower or upper case in RubyCheck if a value exists in an array in Rubyclass << self idiom in RubyWhat is attr_accessor in Ruby?How to remove a key from Hash and get the remaining hash in Ruby/Rails?Why is it bad style to `rescue Exception => e` in Ruby?Access / process (nested) objects, arrays or JSON










1















I need to convert this array:



[:a, :b, :c, :d, :e]


Into this hash:



:a=>:b=>:c=>:d=>:e


Unfortunately, I'm missing the Recursion Lobe of the brain. I found BottomlessHash



# http://firedev.com/posts/2015/bottomless-ruby-hash/
class BottomlessHash < Hash
def initialize
super &-> h, k h[k] = self.class.new
end
end


While I'm struggling to understand the "pretzel stab", it does the trick if you write it out explicitly,



bhash = BottomlessHash.new
bhash[:a][:b][:c][:d] = :e
bhash # => :a=>:b=>:c=>:d=>:e


However I can't figure out a way to pass arbitrary values programmatically.



store doesn't work, nor does send("[:a][:b][:c][:d]=", :e)










share|improve this question



















  • 1





    What should be the result if the array has 0 or 1 elements?

    – ndnenkov
    2 days ago












  • What is your question?

    – sawa
    yesterday















1















I need to convert this array:



[:a, :b, :c, :d, :e]


Into this hash:



:a=>:b=>:c=>:d=>:e


Unfortunately, I'm missing the Recursion Lobe of the brain. I found BottomlessHash



# http://firedev.com/posts/2015/bottomless-ruby-hash/
class BottomlessHash < Hash
def initialize
super &-> h, k h[k] = self.class.new
end
end


While I'm struggling to understand the "pretzel stab", it does the trick if you write it out explicitly,



bhash = BottomlessHash.new
bhash[:a][:b][:c][:d] = :e
bhash # => :a=>:b=>:c=>:d=>:e


However I can't figure out a way to pass arbitrary values programmatically.



store doesn't work, nor does send("[:a][:b][:c][:d]=", :e)










share|improve this question



















  • 1





    What should be the result if the array has 0 or 1 elements?

    – ndnenkov
    2 days ago












  • What is your question?

    – sawa
    yesterday













1












1








1








I need to convert this array:



[:a, :b, :c, :d, :e]


Into this hash:



:a=>:b=>:c=>:d=>:e


Unfortunately, I'm missing the Recursion Lobe of the brain. I found BottomlessHash



# http://firedev.com/posts/2015/bottomless-ruby-hash/
class BottomlessHash < Hash
def initialize
super &-> h, k h[k] = self.class.new
end
end


While I'm struggling to understand the "pretzel stab", it does the trick if you write it out explicitly,



bhash = BottomlessHash.new
bhash[:a][:b][:c][:d] = :e
bhash # => :a=>:b=>:c=>:d=>:e


However I can't figure out a way to pass arbitrary values programmatically.



store doesn't work, nor does send("[:a][:b][:c][:d]=", :e)










share|improve this question
















I need to convert this array:



[:a, :b, :c, :d, :e]


Into this hash:



:a=>:b=>:c=>:d=>:e


Unfortunately, I'm missing the Recursion Lobe of the brain. I found BottomlessHash



# http://firedev.com/posts/2015/bottomless-ruby-hash/
class BottomlessHash < Hash
def initialize
super &-> h, k h[k] = self.class.new
end
end


While I'm struggling to understand the "pretzel stab", it does the trick if you write it out explicitly,



bhash = BottomlessHash.new
bhash[:a][:b][:c][:d] = :e
bhash # => :a=>:b=>:c=>:d=>:e


However I can't figure out a way to pass arbitrary values programmatically.



store doesn't work, nor does send("[:a][:b][:c][:d]=", :e)







ruby recursion






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago









ndnenkov

28.2k74981




28.2k74981










asked 2 days ago









JP DuffyJP Duffy

8431914




8431914







  • 1





    What should be the result if the array has 0 or 1 elements?

    – ndnenkov
    2 days ago












  • What is your question?

    – sawa
    yesterday












  • 1





    What should be the result if the array has 0 or 1 elements?

    – ndnenkov
    2 days ago












  • What is your question?

    – sawa
    yesterday







1




1





What should be the result if the array has 0 or 1 elements?

– ndnenkov
2 days ago






What should be the result if the array has 0 or 1 elements?

– ndnenkov
2 days ago














What is your question?

– sawa
yesterday





What is your question?

– sawa
yesterday












4 Answers
4






active

oldest

votes


















2














[:a, :b, :c, :d, :e].reverse_each.inject k => h
# => :a=>:b=>:c=>:d=>:e





share|improve this answer























  • While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

    – DebanjanB
    yesterday


















3














What send does is call a method, just one, with zero or more arguments. It can't call multiple methods at once. Your example here:



send("[:a][:b][:c][:d]=", :e)


This is trying to call a method named, literally, [:a][:b][:b][:d]= which doesn't exist, so the send fails.



Now this bit of actual Ruby code:



x[:a][:b][:c][:d] = :e


Becomes interpreted by Ruby as:



x.send(:[], :a).send(:[], :b).send(:[], :c).send(:[]=, :d, :e)


Which is just a really long, ugly way of doing what the original code did. The key here is that each [...] part represents a method call that returns something and then the next part is evaluated against that, or chained on.



Now for the original part, this pretzel-stab:



super &-> h, k h[k] = self.class.new 


The & means "pass through this Proc as a block argument", as in for a method with the signature:



initialize(&block)


Where &block represents a block, if given, to that method, as in:



Hash.new h[k] = 


In a more basic implementation.



The -> h, k ... part is traditionally written as:



lambda ... 


Where that's probably more recognizable.






share|improve this answer






























    1














    array = [*:a..:e]
    array[0..-2].reverse.reduce(array[-1]) a => b
    # :a=>:b=>:c=>:d=>:e





    share|improve this answer






























      1














      items = [:a, :b, :c, :d, :e]

      # Iterative version
      def hashify(items)
      items = items.dup

      result = items.pop
      result = items.pop => result until items.empty?
      result
      end

      hashify(items) # => :a=>:b=>:c=>:d=>:e

      # Recursive version
      def hashify(items)
      return items.last if items.size < 2

      *head, penultimate, last = items
      hashify(head + [penultimate => last])
      end

      hashify(items) # => :a=>:b=>:c=>:d=>:e





      share|improve this answer
























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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        [:a, :b, :c, :d, :e].reverse_each.inject k => h
        # => :a=>:b=>:c=>:d=>:e





        share|improve this answer























        • While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

          – DebanjanB
          yesterday















        2














        [:a, :b, :c, :d, :e].reverse_each.inject k => h
        # => :a=>:b=>:c=>:d=>:e





        share|improve this answer























        • While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

          – DebanjanB
          yesterday













        2












        2








        2







        [:a, :b, :c, :d, :e].reverse_each.inject k => h
        # => :a=>:b=>:c=>:d=>:e





        share|improve this answer













        [:a, :b, :c, :d, :e].reverse_each.inject k => h
        # => :a=>:b=>:c=>:d=>:e






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered yesterday









        sawasawa

        132k29205304




        132k29205304












        • While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

          – DebanjanB
          yesterday

















        • While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

          – DebanjanB
          yesterday
















        While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

        – DebanjanB
        yesterday





        While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

        – DebanjanB
        yesterday













        3














        What send does is call a method, just one, with zero or more arguments. It can't call multiple methods at once. Your example here:



        send("[:a][:b][:c][:d]=", :e)


        This is trying to call a method named, literally, [:a][:b][:b][:d]= which doesn't exist, so the send fails.



        Now this bit of actual Ruby code:



        x[:a][:b][:c][:d] = :e


        Becomes interpreted by Ruby as:



        x.send(:[], :a).send(:[], :b).send(:[], :c).send(:[]=, :d, :e)


        Which is just a really long, ugly way of doing what the original code did. The key here is that each [...] part represents a method call that returns something and then the next part is evaluated against that, or chained on.



        Now for the original part, this pretzel-stab:



        super &-> h, k h[k] = self.class.new 


        The & means "pass through this Proc as a block argument", as in for a method with the signature:



        initialize(&block)


        Where &block represents a block, if given, to that method, as in:



        Hash.new h[k] = 


        In a more basic implementation.



        The -> h, k ... part is traditionally written as:



        lambda ... 


        Where that's probably more recognizable.






        share|improve this answer



























          3














          What send does is call a method, just one, with zero or more arguments. It can't call multiple methods at once. Your example here:



          send("[:a][:b][:c][:d]=", :e)


          This is trying to call a method named, literally, [:a][:b][:b][:d]= which doesn't exist, so the send fails.



          Now this bit of actual Ruby code:



          x[:a][:b][:c][:d] = :e


          Becomes interpreted by Ruby as:



          x.send(:[], :a).send(:[], :b).send(:[], :c).send(:[]=, :d, :e)


          Which is just a really long, ugly way of doing what the original code did. The key here is that each [...] part represents a method call that returns something and then the next part is evaluated against that, or chained on.



          Now for the original part, this pretzel-stab:



          super &-> h, k h[k] = self.class.new 


          The & means "pass through this Proc as a block argument", as in for a method with the signature:



          initialize(&block)


          Where &block represents a block, if given, to that method, as in:



          Hash.new h[k] = 


          In a more basic implementation.



          The -> h, k ... part is traditionally written as:



          lambda ... 


          Where that's probably more recognizable.






          share|improve this answer

























            3












            3








            3







            What send does is call a method, just one, with zero or more arguments. It can't call multiple methods at once. Your example here:



            send("[:a][:b][:c][:d]=", :e)


            This is trying to call a method named, literally, [:a][:b][:b][:d]= which doesn't exist, so the send fails.



            Now this bit of actual Ruby code:



            x[:a][:b][:c][:d] = :e


            Becomes interpreted by Ruby as:



            x.send(:[], :a).send(:[], :b).send(:[], :c).send(:[]=, :d, :e)


            Which is just a really long, ugly way of doing what the original code did. The key here is that each [...] part represents a method call that returns something and then the next part is evaluated against that, or chained on.



            Now for the original part, this pretzel-stab:



            super &-> h, k h[k] = self.class.new 


            The & means "pass through this Proc as a block argument", as in for a method with the signature:



            initialize(&block)


            Where &block represents a block, if given, to that method, as in:



            Hash.new h[k] = 


            In a more basic implementation.



            The -> h, k ... part is traditionally written as:



            lambda ... 


            Where that's probably more recognizable.






            share|improve this answer













            What send does is call a method, just one, with zero or more arguments. It can't call multiple methods at once. Your example here:



            send("[:a][:b][:c][:d]=", :e)


            This is trying to call a method named, literally, [:a][:b][:b][:d]= which doesn't exist, so the send fails.



            Now this bit of actual Ruby code:



            x[:a][:b][:c][:d] = :e


            Becomes interpreted by Ruby as:



            x.send(:[], :a).send(:[], :b).send(:[], :c).send(:[]=, :d, :e)


            Which is just a really long, ugly way of doing what the original code did. The key here is that each [...] part represents a method call that returns something and then the next part is evaluated against that, or chained on.



            Now for the original part, this pretzel-stab:



            super &-> h, k h[k] = self.class.new 


            The & means "pass through this Proc as a block argument", as in for a method with the signature:



            initialize(&block)


            Where &block represents a block, if given, to that method, as in:



            Hash.new h[k] = 


            In a more basic implementation.



            The -> h, k ... part is traditionally written as:



            lambda ... 


            Where that's probably more recognizable.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 2 days ago









            tadmantadman

            156k18178209




            156k18178209





















                1














                array = [*:a..:e]
                array[0..-2].reverse.reduce(array[-1]) a => b
                # :a=>:b=>:c=>:d=>:e





                share|improve this answer



























                  1














                  array = [*:a..:e]
                  array[0..-2].reverse.reduce(array[-1]) a => b
                  # :a=>:b=>:c=>:d=>:e





                  share|improve this answer

























                    1












                    1








                    1







                    array = [*:a..:e]
                    array[0..-2].reverse.reduce(array[-1]) a => b
                    # :a=>:b=>:c=>:d=>:e





                    share|improve this answer













                    array = [*:a..:e]
                    array[0..-2].reverse.reduce(array[-1]) a => b
                    # :a=>:b=>:c=>:d=>:e






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 2 days ago









                    Sebastian PalmaSebastian Palma

                    16k42035




                    16k42035





















                        1














                        items = [:a, :b, :c, :d, :e]

                        # Iterative version
                        def hashify(items)
                        items = items.dup

                        result = items.pop
                        result = items.pop => result until items.empty?
                        result
                        end

                        hashify(items) # => :a=>:b=>:c=>:d=>:e

                        # Recursive version
                        def hashify(items)
                        return items.last if items.size < 2

                        *head, penultimate, last = items
                        hashify(head + [penultimate => last])
                        end

                        hashify(items) # => :a=>:b=>:c=>:d=>:e





                        share|improve this answer





























                          1














                          items = [:a, :b, :c, :d, :e]

                          # Iterative version
                          def hashify(items)
                          items = items.dup

                          result = items.pop
                          result = items.pop => result until items.empty?
                          result
                          end

                          hashify(items) # => :a=>:b=>:c=>:d=>:e

                          # Recursive version
                          def hashify(items)
                          return items.last if items.size < 2

                          *head, penultimate, last = items
                          hashify(head + [penultimate => last])
                          end

                          hashify(items) # => :a=>:b=>:c=>:d=>:e





                          share|improve this answer



























                            1












                            1








                            1







                            items = [:a, :b, :c, :d, :e]

                            # Iterative version
                            def hashify(items)
                            items = items.dup

                            result = items.pop
                            result = items.pop => result until items.empty?
                            result
                            end

                            hashify(items) # => :a=>:b=>:c=>:d=>:e

                            # Recursive version
                            def hashify(items)
                            return items.last if items.size < 2

                            *head, penultimate, last = items
                            hashify(head + [penultimate => last])
                            end

                            hashify(items) # => :a=>:b=>:c=>:d=>:e





                            share|improve this answer















                            items = [:a, :b, :c, :d, :e]

                            # Iterative version
                            def hashify(items)
                            items = items.dup

                            result = items.pop
                            result = items.pop => result until items.empty?
                            result
                            end

                            hashify(items) # => :a=>:b=>:c=>:d=>:e

                            # Recursive version
                            def hashify(items)
                            return items.last if items.size < 2

                            *head, penultimate, last = items
                            hashify(head + [penultimate => last])
                            end

                            hashify(items) # => :a=>:b=>:c=>:d=>:e






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 2 days ago

























                            answered 2 days ago









                            ndnenkovndnenkov

                            28.2k74981




                            28.2k74981



























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