Question

我有以下简单的Ruby哈希：

{ 
  "builder_rule_0_filter"=>"artist_name", 
  "builder_rule_0_operator"=>"contains", 
  "builder_rule_0_value_0"=>"New Found Glory", 

  "builder_rule_1_filter"=>"bpm", 
  "builder_rule_1_operator"=>"less", 
  "builder_rule_1_value_0"=>"150", 

  "builder_rule_2_filter"=>"days_ago", 
  "builder_rule_2_operator"=>"less", 
  "builder_rule_2_value_0"=>"40", 

  "builder_rule_3_filter"=>"release_date_start", 
  "builder_rule_3_operator"=>"between", 
  "builder_rule_3_value_0"=>"2019-01-01", 
  "builder_rule_3_value_1"=>"2019-12-31", 
}

我想将其转换为更有条理/更有用的嵌套哈希：

{
  "0"=>
    {
      "filter"=>"artist_name",
      "operator"=>"contains",
      "values"=>
      {
        "0"=>"New Found Glory"
      }
    },
  "1"=>
    {
      "filter"=>"bpm",
      "operator"=>"less",
      "values"=>
      {
        "0"=>"150"
      }
    }
  "2"=>
    {
      "filter"=>"days_ago",
      "operator"=>"less",
      "values"=>
      {
        "0"=>"40"
      }
    }
  "3"=>
    {
      "filter"=>"release_date_start",
      "operator"=>"between",
      "values"=>
      {
        "0"=>"2019-01-01"
        "1"=>"2019-12-31"
      }
    }
}

那么，如何转换一个平面哈希（从表单中获取）并将其转换为基于这些键名的嵌套？

Answer 1

首先，您可以根据哈希的前缀编号将其拆分为多个哈希（可以随意运行以查看返回值）

groups = input.
  group_by { |k,v| k.match(/builder_rule_(\d+)/)[1] }.
  transform_values(&:to_h)

在这一点上，如果仅使用一些硬编码来构建键值，则创建内部对象会更容易：

result = groups.each_with_object({}) do |(prefix, hash), memo|
  memo[prefix] = {
    "filter" => hash["builder_rule_#{prefix}_filter"],
    "operator" => hash["builder_rule_#{prefix}_operator"],
    "values" => hash.select do |key, val|
      key =~ /builder_rule_#{prefix}_value/
    end.sort_by { |key, val| key }.map { |(key, val)| val }
  }
end

这可能会混淆.sort_by { |key, val| key }.map { |(key, val)| val }的含义。我可以清楚地说明：

hash.select { |key, val| key =~ /builder_rule_#{prefix}_value/ }获取将用于“值”数组的键值。它返回一个哈希。
.sort_by { |key, val| key }将哈希值转换为[key, val]元组的数组，并按键排序。这样可以使值以正确的顺序显示。
.map { |(key, val)| val }将嵌套数组变成单级数组，并丢弃键

您也可以使用sort_by(&:first).map(&:second)，尽管您需要积极的支持才能使用Array#second

Answer 2

假设深度固定为3，并且规则是一致的（如示例中所示），则可以如下进行一次单次转换：

h.each_with_object(Hash.new {|h,k| h[k]= {}}) do |(k,v),obj|
  new_k, k2, k3 = k[/(?<=builder_rule_).*/].split('_')
  k2 == 'value' ? (obj[new_k]["values"] ||= {}).merge!({k3 => v}) :  obj[new_k][k2] = v
end

这里我们正在构造一个新的Hash，其中第一级键默认为空的Hash（h.each_with_object(Hash.new {|h,k| h[k]= {}}) do |(k,v),obj|）。

然后，我们根据'builder_rule_'new_k, k2, k3 = k[/(?<=builder_rule_).*/].split('_')

后的下划线拆分密钥

然后，如果第二级密钥（k2）等于“值”，则需要将其设置为值，并将第三级密钥（k3）和该值合并到新的哈希中，否则我们将k2分配给该值。 k2 == 'value' ? (obj[new_k]["values"] ||= {}).merge!({k3 => v}) : obj[new_k][k2] = v

这实际上将返回请求的要求，其中values是具有关键索引的Hash

{
  "0"=>{
      "filter"=>"artist_name",
      "operator"=>"contains",
      "values"=>
      {
        "0"=>"New Found Glory"
      }
    },
  "1"=>{
      "filter"=>"bpm",
      "operator"=>"less",
      "values"=>
      {
        "0"=>"150"
      }
    },
  "2"=>{
      "filter"=>"days_ago",
      "operator"=>"less",
      "values"=>
      {
        "0"=>"40"
      }
    },
  "3"=>{
      "filter"=>"release_date_start",
      "operator"=>"between",
      "values"=>
      {
        "0"=>"2019-01-01",
        "1"=>"2019-12-31"
      }
    }
}

Answer 3

可接受的答案是好的，但是比严格必要的工作要多得多。我会用each_with_object完成所有工作：

input.each_with_object({}}) do |(key, val), hsh|
  _, num, name = *key.match(/^builder_rule_(\d+)_(.*)/)

  hsh[num] = {} unless hsh.key?(num)

  if name.start_with?(/value_\d/)
    hsh[num]["values"] = [] unless hsh[num].key?("values")
    hsh[num]["values"] << val
  else
    hsh[num][name] = val
  end
end
# => {
#      "0" => {
#        "filter" => "artist_name",
#        "operator" => "contains",
#        "values" => ["New Found Glory"]
#      },
#      "1" => {
#        "filter" => "bpm",
#        "operator" => "less",
#        "values" => ["150"]
#      },
#      "2" => {
#        "filter" => "days_ago",
#        "operator" => "less",
#        "values" => ["40"]
#      },
#      "3" => {
#        "filter" => "release_date_start",
#        "operator" => "between",
#        "values" => ["2019-01-01", "2019-12-31"]
#      }
#    }

查看有关repl.it的实际操作：https://repl.it/@jrunning/GiddyEnragedLinks

如何将平面Ruby哈希解析为嵌套哈希？

3 个答案: