查询字典以获取前 3 个最常见的元素

Question

所以我有一个 IEnumerable CallsForProperty 我需要找到特定时间段的前 3 个问题，所以我创建了 3 个字典

 Dictionary<int, string> FAQfor3Months = null;

 Dictionary<int, string> FAQfor4to12Months = null;

 Dictionary<int, string> FAQfor12plusMonths = null;

 foreach(Activity call in CallsForProperty){

   if( /*call is in 3 months*/ ){

      //do some unrelated stuff

      FAQfor3Months.Add(counter, call.callQuestion)

      counter++;

   }

 }

我有 2 个问题，

我试图找出最常见的 3 个问题，这个查询正确吗？

var threeMonthFaqs = FAQfor3Months.GroupBy(x => x.Value).OrderByDescending(x => x.Count()).Take(3);   

这也太低效了吧？有没有更好的方法来查询集合？

Answer 1

反转Dictionarys 并计算遇到的每个问题，然后可以使用 LINQ 找出最重要的 3 个问题：var FAQfor3Months = new Dictionary<string, int>();var FAQfor4to12Months = new Dictionary<string, int>();var FAQfor12plusMonths = new Dictionary<string, int>();foreach (Activity call in CallsForProperty) {&nbsp; &nbsp; if ( /*call is in 3 months*/ ) {&nbsp; &nbsp; &nbsp; &nbsp; //do some unrelated stuff&nbsp; &nbsp; &nbsp; &nbsp; FAQfor3Months.TryGetValue(call.callQuestion, out int count);&nbsp; &nbsp; &nbsp; &nbsp; FAQfor3Months[call.callQuestion] = count+1;&nbsp; &nbsp; }}var top3ThreeMonthFaqs = FAQfor3Months.OrderByDescending(kvp => kvp.Value).Take(3).Select(kvp => kvp.Key).ToList();

Answer 2

您的代码将返回组，而不是实际的问题。尝试这个：var&nbsp;threeMonthFaqs&nbsp;=&nbsp;FAQfor3Months.Values &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.GroupBy(q&nbsp;=>&nbsp;q) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.OrderByDescending(g&nbsp;=>&nbsp;g.Count()) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.SelectMany(g&nbsp;=>&nbsp;g) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.Distinct() &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.Take(3);