组合两个哈希数组

Question

寻找一种优雅的方式以特殊的方式合并两个散列数组:

new_data = [{"name" => "a"}, {"name" => "b"}, {"name" => "c"}]
old_data = [{"name" => "a", "data" => "extra1"}, {"name" => "d", "data2" => "extra"}]
result = [{"name" => "a", "data" => "extra1"}, {"name" => "b"}, {"name" => "c"}]

如果名称关键字匹配，则结果必须包含< code>new_data的所有名称散列，仅包含< code>old_data的额外数据。

我的第一次尝试是这样的，但是它创建了一个额外的散列:

def combine(new_data, old_data)
  int = []
  new_data.each do |s|
    old_data.each do |e|
      (int << (s.merge e)) if e["name"] == s["name"]
    end
    int << s
  end
  int
end

# => [{"name"=>"a", "data"=>"extra1"}, {"name"=>"a"}, {"name"=>"b"}, {"name"=>"c"}]

Answer 1

假设

我假设：

• 在new_data old_data中，“名称”的值都是唯一的;
• new_data 是一个包含一个键值对的哈希数组;和
• old_data 是一个包含至少两个键值对的哈希数组。

密码

def extract_elements(new_data, old_data)
  (new_data+old_data).group_by { |h| h["name"] }.
                      values.
                      select {  |a| a.size > 1 || a.first.size == 1 }.
                      map(&:last)
end

实例

new_data = [{"name" => "a"}, {"name" => "b"}, {"name" => "c"}]
old_data = [{"name" => "a", "data" => "extra1"},
            {"name" => "d", "data2" => "extra"}]
extract_elements(new_data, old_data)
  #=> [{"name"=>"a", "data"=>"extra1"}, {"name"=>"b"}, {"name"=>"c"}] 

解释

对于上面的例子:

a = (new_data+old_data)
  #=> [{"name"=>"a"}, {"name"=>"b"}, {"name"=>"c"},
  #    {"name"=>"a", "data"=>"extra1"}, {"name"=>"d", "data2"=>"extra"}] 
b = a.group_by { |h| h["name"] }
  #=> {"a"=>[{"name"=>"a"}, {"name"=>"a", "data"=>"extra1"}],
  #    "b"=>[{"name"=>"b"}],
  #    "c"=>[{"name"=>"c"}],
  #    "d"=>[{"name"=>"d", "data2"=>"extra"}]} 
c = b.values
  #=> [[{"name"=>"a"}, {"name"=>"a", "data"=>"extra1"}],
  #    [{"name"=>"b"}],
  #    [{"name"=>"c"}],
  #    [{"name"=>"d", "data2"=>"extra"}]] 
d = c.select {  |a| a.size > 1 || a.first.size == 1 }
  #=> [[{"name"=>"a"}, {"name"=>"a", "data"=>"extra1"}],
  #    [{"name"=>"b"}],
  #    [{"name"=>"c"}]] 
e = d.map(&:last)
  #=> [{"name"=>"a", "data"=>"extra1"},
  #    {"name"=>"b"},
  #    {"name"=>"c"}] 

但书

这需要Ruby v1.9，因为它需要为哈希＜code＞b＜/code＞维护密钥插入顺序。对于早期版本，请替换：

map(&:last)

跟：

map { |h| h.max_by(&:size) }

Answer 2

这有点棘手，但是您可以通过从new_data转换为查找哈希来解决它，然后遍历old_data以合并内容：

new_data = [{"name" => "a"}, {"name" => "b"}, {"name" => "c"}]
old_data = [{"name" => "a", "data" => "extra1"}, {"name" => "d", "data2" => "extra"}]

# Transform into a lookup table using "name" as a key
lookup = Hash[new_data.collect { |v| [ v['name'], v ] }]

old_data.each do |data|
  # Match based on "name"
  found = lookup[data['name']]

  next unless (found)

  # If found, swap out the element with a merged version
  # so the original is preserved as-is, not mangled.
  lookup[data['name']] = found.merge(data)
end

lookup.values
# => [{"name"=>"a", "data"=>"extra1"}, {"name"=>"b"}, {"name"=>"c"}]

Answer 3

一行程序，但在大型场景中可能不太有效。

new_data.map{ |e| e.merge(old_data.detect{ |e2| e2['name'] == e['name'] } || {}) }