排序基於排序的值

Question

我有散列數組的數組散列數組如下給出：

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}] 

另外，基於特定的條件我把「user_id」鍵的值從相同的散列和排序，並在同一陣列是如下：

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

現在，我需要的user_quizzes基礎的user_id在sorted_user_ids排列的順序重新排列。

任何人都可以請幫助我。 :)

Answer 1

使用Enumerable#sort_by或Array#sort_by!，您可以指定將用於比較的關鍵：

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}] 

旁註：sorted_user_ids.index(x[:user_id])可能成爲瓶頸（重複O（n）的操作），如果數組巨大。

構建一個user_id小號映射到在這種情況下的訂單的哈希：

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.