首頁猿問查詢字典以獲取前 3 個(gè)最常見的元素

查詢字典以獲取前 3 個(gè)最常見的元素

PHP

DIEA 2024-01-20 15:37:35

所以我有一個(gè) IEnumerable CallsForProperty 我需要找到特定時(shí)間段的前 3 個(gè)問題，所以我創(chuàng)建了 3 個(gè)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 個(gè)問題，我試圖找出最常見的 3 個(gè)問題，這個(gè)查詢正確嗎？var threeMonthFaqs = FAQfor3Months.GroupBy(x => x.Value).OrderByDescending(x => x.Count()).Take(3); 這也太低效了吧？有沒有更好的方法來查詢集合？

查看完整描述

2 回答

慕的地6264312

TA貢獻(xiàn)1817條經(jīng)驗(yàn) 獲得超6個(gè)贊

反轉(zhuǎn)Dictionarys 并計(jì)算遇到的每個(gè)問題，然后可以使用 LINQ 找出最重要的 3 個(gè)問題：

var FAQfor3Months = new Dictionary<string, int>();

var FAQfor4to12Months = new Dictionary<string, int>();

var FAQfor12plusMonths = new Dictionary<string, int>();

foreach (Activity call in CallsForProperty) {

if ( /*call is in 3 months*/ ) {

//do some unrelated stuff

FAQfor3Months.TryGetValue(call.callQuestion, out int count);

FAQfor3Months[call.callQuestion] = count+1;

}

var top3ThreeMonthFaqs = FAQfor3Months.OrderByDescending(kvp => kvp.Value).Take(3).Select(kvp => kvp.Key).ToList();

反對回復(fù) 2024-01-20

繁星點(diǎn)點(diǎn)滴滴

TA貢獻(xiàn)1803條經(jīng)驗(yàn) 獲得超3個(gè)贊

您的代碼將返回組，而不是實(shí)際的問題。嘗試這個(gè)：

var threeMonthFaqs = FAQfor3Months.Values
                                  .GroupBy(q => q)
                                  .OrderByDescending(g => g.Count())
                                  .SelectMany(g => g)
                                  .Distinct()
                                  .Take(3);

反對回復(fù) 2024-01-20