在Ruby中使用Hash反转?

我有一个格式的哈希:

{key1 => [a, b, c], key2 => [d, e, f]} 

我想最终得到:

 { a => key1, b => key1, c => key1, d => key2 ... } 

实现这一目标的最简单方法是什么?

我正在使用Ruby on Rails。

UPDATE

好吧,我设法从服务器日志中提取真实对象,它是通过AJAX推送的。

  Parameters: {"status"=>{"1"=>["1", "14"], "2"=>["7", "12", "8", "13"]}} 

 hash = {:key1 => ["a", "b", "c"], :key2 => ["d", "e", "f"]} 

第一个变种

 hash.map{|k, v| v.map{|f| {f => k}}}.flatten #=> [{"a"=>:key1}, {"b"=>:key1}, {"c"=>:key1}, {"d"=>:key2}, {"e"=>:key2}, {"f"=>:key2}] 

要么

 hash.inject({}){|h, (k,v)| v.map{|f| h[f] = k}; h} #=> {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2} 

UPD

好的,你的哈希是:

 hash = {"status"=>{"1"=>["1", "14"], "2"=>["7", "12", "8", "13"]}} hash["status"].inject({}){|h, (k,v)| v.map{|f| h[f] = k}; h} #=> {"12"=>"2", "7"=>"2", "13"=>"2", "8"=>"2", "14"=>"1", "1"=>"1"} 

很多其他好的答案。 只是想为Ruby 2.0和1.9.3折腾这个:

 hash = {apple: [1, 14], orange: [7, 12, 8, 13]} Hash[hash.flat_map{ |k, v| v.map{ |i| [i, k] } }] # => {1=>:apple, 14=>:apple, 7=>:orange, 12=>:orange, 8=>:orange, 13=>:orange} 

这是利用: Hash::[]Enumerable#flat_map

在这些新版本中还有Enumerable::each_with_object ,它与Enumerable::inject / Enumerable::reduce非常相似:

 hash.each_with_object(Hash.new){ |(k, v), inverse| v.each{ |e| inverse[e] = k } } 

使用具有100个密钥的原始哈希执行快速基准测试 (Ruby 2.0.0p0; 2012 Macbook Air),每个密钥具有100个不同的值:

 Hash::[] w/ Enumerable#flat_map 155.7 (±9.0%) i/s - 780 in 5.066286s Enumerable#each_with_object w/ Enumerable#each 199.7 (±21.0%) i/s - 940 in 5.068926s 

显示该数据集的each_with_object变体更快。

好的,我们猜猜看。 你说你有一个数组,但我同意Benoit你可能拥有的是哈希。 function方法:

  h = {:key1 => ["a", "b", "c"], :key2 => ["d", "e", "f"]} h.map { |k, vs| Hash[vs.map { |v| [v, k] }] }.inject(:merge) #=> {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2} 

也:

  h.map { |k, vs| Hash[vs.product([k])] }.inject(:merge) #=> {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2} 

在值对应多个键的情况下,例如本例中的“c”…

 { :key1 => ["a", "b", "c"], :key2 => ["c", "d", "e"]} 

……其他一些答案不会给出预期的结果。 我们需要反向散列来将键存储在数组中,如下所示:

 { "a" => [:key1], "b" => [:key1], "c" => [:key1, :key2], "d" => [:key2], "e" => [:key2] } 

这应该是诀窍:

 reverse = {} hash.each{ |k,vs| vs.each{ |v| reverse[v] ||= [] reverse[v] << k } } 

这是我的用例,我会像OP一样定义我的问题(事实上,搜索类似的短语让我在这里),所以我怀疑这个答案可能有助于其他搜索者。

如果您想要反转这样格式化的哈希,以下内容可能对您有所帮助:

 a = {:key1 => ["a", "b", "c"], :key2 => ["d", "e", "f"]} a.inject({}) do |memo, (key, values)| values.each {|value| memo[value] = key } memo end 

这会返回:

 {"a"=>:key1, "b"=>:key1, "c"=>:key1, "d"=>:key2, "e"=>:key2, "f"=>:key2} 
 new_hash={} hash = {"key1" => ['a', 'b', 'c'], "key2" => ['d','e','f']} hash.each_pair{|key, val|val.each{|v| new_hash[v] = key }} 

这给了

 new_hash # {"a"=>"key1", "b"=>"key1", "c"=>"key1", "d"=>"key2", "e"=>"key2", "f"=>"key2"} 

如果你想正确处理重复值,那么你应该使用Ruby的Facets中的Hash#inverse

Hash#inverse保留重复值 ,例如,它确保hash.inverse.inverse == hash

之一:

  • 从这里使用Hash#inverse: http : //www.unixgods.org/Ruby/invert_hash.html

  • 使用FacetsOfRuby库’facets’中的Hash#inverse

用法是这样的:

 require 'facets' h = {:key1 => [:a, :b, :c], :key2 => [:d, :e, :f]} => {:key1=>[:a, :b, :c], :key2=>[:d, :e, :f]} h.inverse => {:a=>:key1, :b=>:key1, :c=>:key1, :d=>:key2, :e=>:key2, :f=>:key2} 

代码如下所示:

 # this doesn't looks quite as elegant as the other solutions here, # but if you call inverse twice, it will preserve the elements of the original hash # true inversion of Ruby Hash / preserves all elements in original hash # eg hash.inverse.inverse ~ h class Hash def inverse i = Hash.new self.each_pair{ |k,v| if (v.class == Array) v.each{ |x| i[x] = i.has_key?(x) ? [k,i[x]].flatten : k } else i[v] = i.has_key?(v) ? [k,i[v]].flatten : k end } return i end end h = {:key1 => [:a, :b, :c], :key2 => [:d, :e, :f]} => {:key1=>[:a, :b, :c], :key2=>[:d, :e, :f]} h.inverse => {:a=>:key1, :b=>:key1, :c=>:key1, :d=>:key2, :e=>:key2, :f=>:key2} 

实现您所需要的一种方法:

 arr = [{["k1"] => ["a", "b", "c"]}, {["k2"] => ["d", "e", "f"]}] results_arr = [] arr.each do |hsh| hsh.values.flatten.each do |val| results_arr << { [val] => hsh.keys.first }··· end end Result: [{["a"]=>["k1"]}, {["b"]=>["k1"]}, {["c"]=>["k1"]}, {["d"]=>["k2"]}, {["e"]=>["k2"]}, {["f"]=>["k2"]}]