Question

我有一系列哈希：

data = [{"user_id"=>1, "answer"=>"cupcakes"},
 {"user_id"=>1, "answer"=>"Colorado"},
 {"user_id"=>1, "answer"=>"newspaper"},
 {"user_id"=>2, "answer"=>"fruitcake"},
 {"user_id"=>2, "answer"=>"Louisiana"},
 {"user_id"=>2, "answer"=>"tv"}]

如何重新组织它以便按"user_id"分组并在一个哈希中列出所有"answer"？类似的东西：

output_data = [{"user_id" => 1, "answer1"=>"cupcakes", "answer2"=>"Colorado", "answer3"=>"newspaper"},
{"user_id" => 2, "answer1"=>"fruitcake", "answer2"=>"Louisiana", "answer3"=>"tv"}]

或者可能在数组中包含所有答案：

output_data = [{"user_id" => 1, "answers"=>["cupcakes", "Colorado", "newspaper"]},
{"user_id" => 2, "answers"=>["fruitcake", "Louisiana", "tv"]}]

我与此特定输出无关。我需要将"user_id"作为关键，并将所有答案组织在一起。有什么建议吗？

Answer 1

你可以这样做：

<强>代码

def convert(arr)
  arr.each_with_object({}) do |g,h|
    h.update(g["user_id"]=>[g["answer"]]) { |_,o,n| o+n }
  end.map { |k,v| { "user_id"=>k, "answer"=>v } }
end

示例

convert(data) #=> [{"user_id"=>1, "answer"=>["cupcakes", "Colorado", "newspaper"]}, # {"user_id"=>2, "answer"=>["fruitcake", "Louisiana", "tv"]}]

<强>解释

我们有：

enum = data.each_with_object(Hash.new { |h,k| h[k] = [] }) #=> #<Enumerator: [{"user_id"=>1, "answer"=>"cupcakes"}, # {"user_id"=>1, "answer"=>"Colorado"}, # {"user_id"=>1, "answer"=>"newspaper"}, # {"user_id"=>2, "answer"=>"fruitcake"}, # {"user_id"=>2, "answer"=>"Louisiana"}, # {"user_id"=>2, "answer"=>"tv"}]: # each_with_object({})>

我们可以将枚举器转换为数组，以查看将传递给块的值：

a = enum.to_a #=> [[{"user_id"=>1, "answer"=>"cupcakes"}, {}], # [{"user_id"=>1, "answer"=>"Colorado"}, {}], # [{"user_id"=>1, "answer"=>"newspaper"}, {}], # [{"user_id"=>2, "answer"=>"fruitcake"}, {}], # [{"user_id"=>2, "answer"=>"Louisiana"}, {}], # [{"user_id"=>2, "answer"=>"tv"}, {}]]

如您所见，枚举器包含六个元素，每个元素都包含一个由data元素组成的双元素数组和一个最初为空的哈希。

关键是我正在使用Hash#update（又名merge!）的形式，当两个哈希值合并时，使用一个块来确定键的值。< / p>
enum的第一个元素被传递给块并分配给块变量，如下所示：

g, h = enum.next #=> [{"user_id"=>1, "answer"=>"cupcakes"}, {}] g #=> {"user_id"=>1, "answer"=>"cupcakes"} h #=> {}

因此，块计算是：

h.update(g["user_id"]=>[g["answer"]]) # {}.update(1=>["cupcakes"]) #=> {1=>["cupcakes"]} h #=> {1=>["cupcakes"]}

update的块未用于此第一次合并操作，因为（合并之前）h没有键1。在稍后的操作中再次g["user_id"] #=> 1。此时，该块将用于确定键1的值。

这导致：

h = data.each_with_object({}) do |g,h| h.update(g["user_id"]=>[g["answer"]]) { |_,o,n| o+n } end #=> { 1=>["cupcakes", "Colorado", "newspaper"], # 2=>["fruitcake", "Louisiana", "tv"] }

将h的键元素对映射到所需的哈希数组是一件简单的事情。

<强>替代

通过合并哈希来实现此目的的另一种方法如下：

data.each_with_object(Hash.new { |h,k| h[k]=[] }) do |g,h| h[g["user_id"]] << g["answer"] end.map { |k,v| { "user_id"=>k, "answer"=>v } } #=> [{"user_id"=>1, "answer"=>["cupcakes", "Colorado", "newspaper"]}, # {"user_id"=>2, "answer"=>["fruitcake", "Louisiana", "tv"]}]

当h[k]没有键h时要修改k时，这会为哈希提供一个空数组的默认值。例如：

h = Hash.new { |h,k| h[k]=[] } #=> {} h[:cat] << 'boots' #=> ["boots"] h #=> {:cat=>["boots"]}

Answer 2

您的预期结果没有意义。要保留"answer"信息，您需要将它们保存为数组。

data.group_by{|h| h["user_id"]}.each{|_, v| v.map!{|h| h["answer"]}}
# =>
# {
#   1=>["cupcakes", "Colorado", "newspaper"],
#   2=>["fruitcake", "Louisiana", "tv"]
# }

"user_id"和"answer"这样的字符串是多余的，您应该避免它们存在于数据中，除非它有助于以任何方式使它们清晰。

如何重新组织/过滤哈希值

2 个答案: