我的第一个想法是压缩值并调用uniq，然后想办法返回到初始形式：

h['options'].zip(h['option_ids']).uniq(&:first).transpose
#=> [["Language", "Question", "Answer"], ["12345", "23456", "45678"]]

然后，通过并行分配：

h['options'], h['option_ids'] = h['options'].zip(h['option_ids']).uniq(&:first).transpose

h #=> {"id"=>"sjfdkjfd", "name"=>"Field Name", "type"=>"field", "options"=>["Language", "Question", "Answer"], "option_ids"=>["12345", "23456", "45678"]}

这些是步骤：

h['options'].zip(h['option_ids'])
#=> [["Language", "12345"], ["Question", "23456"], ["Question", "34567"], ["Answer", "45678"], ["Answer", "56789"]]

h['options'].zip(h['option_ids']).uniq(&:first)
#=> [["Language", "12345"], ["Question", "23456"], ["Answer", "45678"]]

hash = { 
  "id" => "sjfdkjfd",
  "name" => "Field Name",
  "type" => "field",
  "options" => ["L", "Q", "Q", "Q", "A", "A", "Q"],
  "option_ids" => ["12345", "23456", "34567", "dog", "45678", "56789", "cat"]
}

我假设“重复的元素”指的是连续的相等元素（仅在[1,2,2,1]中有2），而不是“重复的元素”（在前面的例子中有1和2）。我展示了如果第二种解释适用，代码将如何被修改（实际上是简化）。

idx = hash["options"].
  each_with_index.
  chunk_while { |(a,_),(b,_)| a==b }.
  map { |(_,i),*| i }
  #=> [0, 1, 4, 6]

hash.merge(
  ["options", "option_ids"].each_with_object({}) { |k,h| h[k] = hash[k].values_at(*idx) }
)
  #=> {"id"=>"sjfdkjfd",
  #    "name"=>"Field Name",
  #    "type"=>"field",
  #    "options"=>["L", "Q", "A", "Q"],
  #    "option_ids"=>["12345", "23456", "45678", "cat"]}

如果“重复元素”被解释为"options"和"option_ids"的值仅具有上面显示的前三个元素，则计算idx如下：

idx = hash["options"].
  each_with_index.
  uniq { |s,_| s }.
  map(&:last)
    #=> [0, 1, 4]

请参阅EQUIPMENT #chunk_while（EQUIPMENT #slice_when可以改为）和Array#values_at。其步骤如下。

a = hash["options"]
  #=> ["L", "Q", "Q", "Q", "A", "A", "Q"] 
e0 = a.each_with_index
  #=> #<Enumerator: ["L", "Q", "Q", "Q", "A", "A", "Q"]:each_with_index> 
e1 = e0.chunk_while { |(a,_),(b,_)| a==b }
  #=> #<Enumerator: #<Enumerator::Generator:0x000055e4bcf17740>:each>

我们可以看到枚举器e1将生成并通过将其转换为数组传递给map的值：

e1.to_a
  #=> [[["L", 0]],
  #    [["Q", 1], ["Q", 2], ["Q", 3]],
  #    [["A", 4], ["A", 5]], [["Q", 6]]]

继续

idx = e1.map { |(_,i),*| i }
  #=> [0, 1, 4, 6] 

c = ["options", "option_ids"].
      each_with_object({}) { |k,h| h[k] = hash[k].values_at(*idx) } 
  #=> {"options"=>["L", "Q", "A", "Q"],
  #    "option_ids"=>["12345", "23456", "45678", "cat"]} 
hash.merge(c)
  #=> {"id"=>"sjfdkjfd",
  #    "name"=>"Field Name",
  #    "type"=>"field",
  #    "options"=>["L", "Q", "A", "Q"],
  #    "option_ids"=>["12345", "23456", "45678", "cat"]}

使用Array#transpose

hash = {
  "options" => ["Language", "Question", "Question", "Answer", "Answer"],
  "option_ids" => ["12345", "23456", "34567", "45678", "56789"]
}

hash.values.transpose.uniq(&:first).transpose.map.with_index {|v,i| [hash.keys[i], v]}.to_h
#=> {"options"=>["Language", "Question", "Answer"], "option_ids"=>["12345", "23456", "45678"]}

OP编辑后：

hash = {
  "id" => "sjfdkjfd",
  "name" => "Field Name",
  "type" => "field",
  "options" => ["Language", "Question", "Question", "Answer", "Answer"],
  "option_ids" => ["12345", "23456", "34567", "45678", "56789"]
}

hash_array = hash.to_a.select {|v| v.last.is_a?(Array)}.transpose
hash.merge([hash_array.first].push(hash_array.last.transpose.uniq(&:first).transpose).transpose.to_h)
#=> {"id"=>"sjfdkjfd", "name"=>"Field Name", "type"=>"field", "options"=>["Language", "Question", "Answer"], "option_ids"=>["12345", "23456", "45678"]}