6

现在，我有一个服务器调用来回退以下 Ruby 哈希：

{
  "id"=>"-ct",
  "factualId"=>"",
  "outOfBusiness"=>false,
  "publishedAt"=>"2012-03-09 11:02:01",
  "general"=>{
    "name"=>"A Cote",
    "timeZone"=>"EST",
    "desc"=>"À Côté is a small-plates restaurant in Oakland's charming
            Rockridge district. Cozy tables surround large communal tables in both
            the main dining room and on the sunny patio to create a festive atmosphere.
              Small plates reflecting the best of seasonal Mediterranean cuisine are served
            family-style by a friendly and knowledgeable staff.\nMenu items are paired with
            a carefully chosen selection of over 40 wines by the glass as well as a highly
            diverse bottled wine menu. Specialty drinks featuring fresh fruits, rare
            botaniques and fine liqueurs are featured at the bar.",
    "website"=>"http://acoterestaurant.com/"
  },
  "location"=>{
    "address1"=>"5478 College Ave",
    "address2"=>"",
    "city"=>"Oakland",
    "region"=>"CA",
    "country"=>"US",
    "postcode"=>"94618",
    "longitude"=>37.84235,
    "latitude"=>-122.25222
  },
  "phones"=>{
    "main"=>"510-655-6469",
    "fax"=>nil
  },
  "hours"=>{
    "mon"=>{"start"=>"", "end"=>""},
    "tue"=>{"start"=>"", "end"=>""},
    "wed"=>{"start"=>"", "end"=>""},
    "thu"=>{"start"=>"", "end"=>""},
    "fri"=>{"start"=>"", "end"=>""},
    "sat"=>{"start"=>"", "end"=>""},
    "sun"=>{"start"=>"", "end"=>""},
    "holidaySchedule"=>""
  },
  "businessType"=>"Restaurant"
}

它有几个嵌套的属性，例如：

"wed"=>{"start"=>"", "end"=>""}

我需要将此对象转换为 Ruby 中的未嵌套哈希。理想情况下，我想检测一个属性是否嵌套，并做出相应的响应，当 IE 确定属性“ wed”是嵌套的时，它会提取其数据并存储在“ wed-start”和“ wed-end”字段或类似的字段中。

有人对如何开始有任何提示吗？

4

4 回答 4

10

编辑：sparseify gem是作为这个问题的一般解决方案发布的。

这是我几个月前完成的一个实现。您需要将 JSON 解析为散列，然后使用 Sparsify 来稀疏散列。

# Extend into a hash to provide sparse and unsparse methods. 
# 
# {'foo'=>{'bar'=>'bingo'}}.sparse #=> {'foo.bar'=>'bingo'}
# {'foo.bar'=>'bingo'}.unsparse => {'foo'=>{'bar'=>'bingo'}}
# 
module Sparsify
  def sparse(options={})
    self.map do |k,v|
      prefix = (options.fetch(:prefix,[])+[k])
      next Sparsify::sparse( v, options.merge(:prefix => prefix ) ) if v.is_a? Hash
      { prefix.join(options.fetch( :separator, '.') ) => v}
    end.reduce(:merge) || Hash.new
  end
  def sparse!
    self.replace(sparse)
  end

  def unsparse(options={})
    ret = Hash.new
    sparse.each do |k,v|
      current = ret
      key = k.to_s.split( options.fetch( :separator, '.') )
      current = (current[key.shift] ||= Hash.new) until (key.size<=1)
      current[key.first] = v
    end
    return ret
  end
  def unsparse!(options={})
    self.replace(unsparse)
  end

  def self.sparse(hsh,options={})
    hsh.dup.extend(self).sparse(options)
  end

  def self.unsparse(hsh,options={})
    hsh.dup.extend(self).unsparse(options)
  end

  def self.extended(base)
    raise ArgumentError, "<#{base.inspect}> must be a Hash" unless base.is_a? Hash
  end
end

用法：

external_data = JSON.decode( external_json )
flattened = Sparsify.sparse( external_data, :separator => '-' )

这最初是因为我们在 Mongo 中存储一组东西而创建的，这允许我们在更新时使用稀疏键（点分隔）来更新嵌套哈希的某些内容，而不会覆盖不相关的键。

于 2012-08-22T00:24:11.433 回答
6

这是完整解决方案的第一步。我相信你可以写得更优雅，但这似乎很清楚。如果您将其保存在 Ruby 文件中并运行它，您将获得我在下面显示的输出。

class Hash
  def unnest
    new_hash = {}
    each do |key,val|
      if val.is_a?(Hash)
        new_hash.merge!(val.prefix_keys("#{key}-"))
      else
        new_hash[key] = val
      end
    end
    new_hash
  end

  def prefix_keys(prefix)
    Hash[map{|key,val| [prefix + key, val]}].unnest
  end
end

p ({"a" => 2, "f" => 5}).unnest
p ({"a" => {"b" => 3}, "f" => 5}).unnest
p ({"a" => {"b" => {"c" => 4}, "f" => 5}}).unnest

输出：

{"a"=>2, "f"=>5}
{"a-b"=>3, "f"=>5}
{"a-b-c"=>4, "a-f"=>5}
于 2012-08-22T00:14:41.430 回答
4

另一种选择：

class Hash
  def smash(prefix = nil)
    inject({}) do |acc, (k, v)|
      key = prefix.to_s + k
      if Hash === v
        acc.merge(v.smash(key + '-'))
      else
        acc.merge(key => v)
      end
    end
  end
end

hash = {
  'f' => 100,
  'z' => {'j' => 25},
  'a' => {'b' => {'c' => 1}}
}

puts hash.smash # => {"f"=>100, "z-j"=>25, "a-b-c"=>1}
于 2012-08-22T05:42:41.467 回答
2

解决这个问题的另一种方法不是将散列变平，而是像平展一样访问它。例如，给定这个哈希：

h = {
  'a' => 1,
  'b' => {
    'c' => 2,
    'd' => 3,
  },
}

那么这个函数：

NESTED_KEY_SEPARATOR = '-'
NESTED_KEY_REGEX = /^(.*?)(?:#{NESTED_KEY_SEPARATOR}(.*))?$/

def nested_fetch(key, hash)
  return hash if key.empty?
  first_part_of_key, rest_of_key = NESTED_KEY_REGEX.match(key).captures
  value = hash[first_part_of_key]
  if value.is_a?(Hash)
    nested_hash_fetch(value, rest_of_key || '')
  elsif rest_of_key
    nil
  else
    value
  end
end

将允许您通过将各个哈希键与 KEY_SEPARATOR 连接在一起来检索嵌套的哈希元素（在此处设置为破折号，但可以是任何在您需要搜索的哈希中从未作为键出现的字符）：

p nested_fetch('a', h)      # => 1
p nested_fetch('b-c', h)    # => 2

如果您提供部分限定的密钥，您将获得在该点匹配的哈希：

p nested_fetch('b', h)      # => {"c"=>2, "d"=>3}

如果你给出一个不存在的密钥，你会得到 nil：

p nested_fetch('b-x', h)    # => nil

如果需要，可以通过简单地将上述代码包含在类 Hash 中，并将 self 作为参数散列的默认值，将其猴子修补到散列上：

class Hash
  NESTED_KEY_SEPARATOR = '-'
  NESTED_KEY_REGEX = /^(.*?)(?:#{KEY_SEPARATOR}(.*))?$/

  def nested_fetch(key, hash = self)
  ...
end
于 2012-08-22T00:51:28.527 回答