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copied!<p>In <a href="http://www.ruby-doc.org/core-1.8.7/Hash.html" rel="nofollow">Ruby 1.8.7</a>, the order of elements in a hash is documented to be not under our control, so none of the above methods work. In <a href="http://www.ruby-doc.org/core-1.9.3/Hash.html" rel="nofollow">Ruby 1.9.3</a>, things work and are documented in the way that the other answers rely upon.</p> <pre> $ irb1.8 h = { "4" => "happiness", "10" => "cool", "lala" => "54", "1" => "spider" } Hash[h.to_a().reverse()] => {"lala"=>"54", "1"=>"spider", "10"=>"cool", "4"=>"happiness"} quit $ irb1.9.1 h = { "4" => "happiness", "10" => "cool", "lala" => "54", "1" => "spider" } Hash[h.to_a().reverse()] =>{"1"=>"spider", "lala"=>"54", "10"=>"cool", "4"=>"happiness"} </pre> <p>The Ruby 1.8.7 way was ingrained so firmly for me that I misunderstood the question for quite some time. I thought it requested a way to <a href="http://www.ruby-doc.org/core-1.9.3/Hash.html#method-i-invert" rel="nofollow">Hash#invert</a>: ie to transform the hash such that the range maps to the domain. That method discards duplicates. Luís Ramalho proffers <a href="http://www.luisramalho.com/how-to-invert-a-hash-in-ruby-and-not-overwriting-repeated-keys" rel="nofollow">a method</a> that doesn't, but it's a bit clunky. This is a little shorter:</p> <pre> $ irb def invertWithDuplicates(original) inverse = Hash.new() { |hash, key| hash[key] = []; } original.each_pair() { |key, value| inverse[value].push(key); } return inverse end h = { "4" => "happiness", "10" => "cool", "lala" => "54", "1" => "cool" } invertWithDuplicates(h) => {"happiness"=>["4"], "cool"=>["1", "10"], "54"=>["lala"]} </pre> <p>Sorry to drift away from the OP's intended topic, though I submit that this does fit the post's title "Reverse a hash in Ruby".</p>

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