ruby-on-rails - 根据排序值数组对哈希数组进行排序

Question

我有一个散列数组，如下所示:

user_quizzes = [{:id => 3897, :quiz_id => 1793, :user_id => 252}, {:id => 3897, :quiz_id => 1793, :user_id => 475}, {:id => 3897, :quiz_id => 1793, :user_id => 880}, {:id => 3897, :quiz_id => 1793, :user_id => 881}, {:id => 3897, :quiz_id => 1793, :user_id => 882}, {:id => 3897, :quiz_id => 1793, :user_id => 883}, {:id => 3897, :quiz_id => 1793, :user_id => 884}]

此外，根据特定条件，我从相同的散列中获取“user_id”键的值并对其进行排序，下面给出了相同的数组:

sorted_user_ids = [880, 881, 882, 883, 884, 475, 252]

现在，我需要根据 sorted_user_ids 数组中 user_id 的顺序重新排列 user_quizzes。

谁能帮我解决这个问题。 :)

Answer 1

使用 Enumerable#sort_by或 Array#sort_by! ，您可以指定将用于比较的键:

user_quizzes = [
  {:id => 3897, :quiz_id => 1793, :user_id => 252},
  {:id => 3897, :quiz_id => 1793, :user_id => 475},
  {:id => 3897, :quiz_id => 1793, :user_id => 880},
  {:id => 3897, :quiz_id => 1793, :user_id => 881},
  {:id => 3897, :quiz_id => 1793, :user_id => 882},
  {:id => 3897, :quiz_id => 1793, :user_id => 883},
  {:id => 3897, :quiz_id => 1793, :user_id => 884}
]
sorted_user_ids = [880, 881, 882, 883, 884, 475, 252]
user_quizzes.sort_by { |x| sorted_user_ids.index(x[:user_id]) }
# => [{:id=>3897, :quiz_id=>1793, :user_id=>880},
#     {:id=>3897, :quiz_id=>1793, :user_id=>881},
#     {:id=>3897, :quiz_id=>1793, :user_id=>882},
#     {:id=>3897, :quiz_id=>1793, :user_id=>883},
#     {:id=>3897, :quiz_id=>1793, :user_id=>884},
#     {:id=>3897, :quiz_id=>1793, :user_id=>475},
#     {:id=>3897, :quiz_id=>1793, :user_id=>252}]

旁注:如果数组很大，sorted_user_ids.index(x[:user_id]) 可能成为瓶颈(重复 O(n) 操作)。

在这种情况下构建一个将 user_id 映射到订单的散列:

sorted_user_ids = [880, 881, 882, 883, 884, 475, 252]
order = Hash[sorted_user_ids.each_with_index.to_a]
# => {880=>0, 881=>1, 882=>2, 883=>3, 884=>4, 475=>5, 252=>6}
user_quizzes.sort_by { |x| order[x[:user_id]] }
# => same as above.