找到兩個詞典

Question

我有兩個字典一樣Dic1<string,string>和Dic2<string,string> 我想產生的值的同時存在於兩個Dic1和Dic2 因此，例如密鑰的新list的交集如果

Dic1: <12,"hi">, <14,"bye"> 
Dic2: <12,"hello">, <18,"bye"> 

然後列表應該是："hi" , "hello"

我試圖與Dic1.Keys.Intersect一起工作，但還不能完全弄清楚。

What I tried: Dic1.Keys.Intersect(Dic2.Keys).ToList(t => t.Values); 

Answer 1

給你：

var dic1 = new Dictionary<int, string> { { 12, "hi" }, { 14, "bye" } }; 
var dic2 = new Dictionary<int, string> { { 12, "hello" }, { 18, "bye" } }; 
HashSet<int> commonKeys = new HashSet<int>(dic1.Keys); 
commonKeys.IntersectWith(dic2.Keys); 
var result = 
    dic1 
    .Where(x => commonKeys.Contains(x.Key)) 
    .Concat(dic2.Where(x => commonKeys.Contains(x.Key))) 
    // .Select(x => x.Value) // With this additional select you'll get only the values. 
    .ToList(); 

結果列表包含{ 12, "hi" }和{ 12, "hello" }

的HashSet是交叉非常有用的。

---

剛走出curiostiy的我相比，所有六個解決方案（希望沒有錯過任何）和時間如下：

@EZI  Intersect2 GroupBy   ~149ms 
@Selman22 Intersect3 Keys.Intersect ~41ms 
@dbc  Intersect4 Where1   ~22ms 
@dbc  Intersect5 Where2   ~18ms 
@dbc  Intersect5 Classic   ~11ms 
@t3chb0t Intersect1 HashSet   ~66ms 

---

class Program 
{ 
    static void Main(string[] args) 
    { 
     var dic1 = new Dictionary<int, string>(); 
     var dic2 = new Dictionary<int, string>(); 

     Random rnd = new Random(DateTime.Now.Millisecond); 
     for (int i = 0; i < 100000; i++) 
     { 
      int id = 0; 

      do { id = rnd.Next(0, 1000000); } while (dic1.ContainsKey(id)); 
      dic1.Add(id, "hi"); 

      do { id = rnd.Next(0, 1000000); } while (dic2.ContainsKey(id)); 
      dic2.Add(id, "hello"); 
     } 

     List<List<string>> results = new List<List<string>>(); 

     using (new AutoStopwatch(true)) { results.Add(Intersect1(dic1, dic2)); } 
     Console.WriteLine("Intersect1 elapsed in {0}ms (HashSet)", AutoStopwatch.Stopwatch.ElapsedMilliseconds); 

     using (new AutoStopwatch(true)) { results.Add(Intersect2(dic1, dic2)); } 
     Console.WriteLine("Intersect2 elapsed in {0}ms (GroupBy)", AutoStopwatch.Stopwatch.ElapsedMilliseconds); 

     using (new AutoStopwatch(true)) { results.Add(Intersect3(dic1, dic2)); } 
     Console.WriteLine("Intersect3 elapsed in {0}ms (Keys.Intersect)", AutoStopwatch.Stopwatch.ElapsedMilliseconds); 

     using (new AutoStopwatch(true)) { results.Add(Intersect4(dic1, dic2)); } 
     Console.WriteLine("Intersect4 elapsed in {0}ms (Where1)", AutoStopwatch.Stopwatch.ElapsedMilliseconds); 

     using (new AutoStopwatch(true)) { results.Add(Intersect5(dic1, dic2)); } 
     Console.WriteLine("Intersect5 elapsed in {0}ms (Where2)", AutoStopwatch.Stopwatch.ElapsedMilliseconds); 

     using (new AutoStopwatch(true)) { results.Add(Intersect7(dic1, dic2)); } 
     Console.WriteLine("Intersect7 elapsed in {0}ms (Old style :-)", AutoStopwatch.Stopwatch.ElapsedMilliseconds); 

     Console.ReadKey(); 
    } 

    static List<string> Intersect1(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     HashSet<int> commonKeys = new HashSet<int>(dic1.Keys); 
     commonKeys.IntersectWith(dic2.Keys); 
     var result = 
      dic1 
      .Where(x => commonKeys.Contains(x.Key)) 
      .Concat(dic2.Where(x => commonKeys.Contains(x.Key))) 
      .Select(x => x.Value) 
      .ToList(); 
     return result; 
    } 

    static List<string> Intersect2(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     var result = dic1.Concat(dic2) 
        .GroupBy(x => x.Key) 
        .Where(g => g.Count() > 1) 
        .SelectMany(g => g.Select(x => x.Value)) 
        .ToList(); 
     return result; 
    } 

    static List<string> Intersect3(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     var result = 
      dic1 
      .Keys 
      .Intersect(dic2.Keys) 
      .SelectMany(key => new[] { dic1[key], dic2[key] }) 
      .ToList(); 
     return result; 
    } 

    static List<string> Intersect4(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     var result = 
      dic1. 
      Where(pair => dic2.ContainsKey(pair.Key)) 
      .SelectMany(pair => new[] { dic2[pair.Key], pair.Value }).ToList(); 
     return result; 
    } 

    static List<string> Intersect5(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     var result = 
      dic1 
      .Keys 
      .Where(dic2.ContainsKey).SelectMany(k => new[] { dic1[k], dic2[k] }) 
      .ToList(); 
     return result; 
    } 

    static List<string> Intersect7(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     var list = new List<string>(); 
     foreach (var key in dic1.Keys) 
     { 
      if (dic2.ContainsKey(key)) 
      { 
       list.Add(dic1[key]); 
       list.Add(dic2[key]); 
      } 
     } 
     return list; 
    } 
} 

class AutoStopwatch : IDisposable 
{ 
    public static readonly Stopwatch Stopwatch = new Stopwatch(); 

    public AutoStopwatch(bool start) 
    { 
     Stopwatch.Reset(); 
     if (start) Stopwatch.Start(); 
    } 
    public void Dispose() 
    { 
     Stopwatch.Stop(); 
    } 
} 

Answer 2

你只需要使用索引來獲取值：

var values = Dic1.Keys.Intersect(Dic2.Keys) 
      .SelectMany(key => new[] { Dic1[key], Dic2[key] }) 
      .ToList(); 

Answer 3

可以使用Where濾波器DIC1鑰匙，然後將其轉換爲值，像這樣：

var values = Dic1.Keys.Where(Dic2.ContainsKey).SelectMany(k => new[] { Dic1[k], Dic2[k] }) 
    .ToList(); 

這應該與Dic1和Dic2上的查找操作一樣高效，該操作通常是log（n）或更好，並且不需要構建任何臨時哈希集或查找表。

下面是避免了被少一點相當的成本字典查找的一個版本：

var values = Dic1.Where(pair => Dic2.ContainsKey(pair.Key)).SelectMany(pair => new[] { pair.Value, Dic2[pair.Key] }) 
    .ToList(); 

更新

我的時間測試（使用t3chb0t的方便測試工具）顯示第一個版本實際運行得更快這很簡單，所以在這兩個，更喜歡這一點。但到目前爲止，我已經找到了最快不使用Linq可言，在7毫秒達到1000000路交叉口對13對LINQ的版本：

static List<string> Intersect7(Dictionary<int, string> dic1, Dictionary<int, string> dic2) 
    { 
     var list = new List<string>(); 
     foreach (var key in dic1.Keys) 
     { 
      if (dic2.ContainsKey(key)) 
      { 
       list.Add(dic1[key]); 
       list.Add(dic2[key]); 
      } 
     } 
     return list; 
    } 

這是一個古老的風格，但這樣你可能不希望這樣。

Answer 4

var d1 = new Dictionary<int, string>() { { 12, "hi" }, { 14, "bye" } }; 
var d2 = new Dictionary<int, string>() { { 12, "hello" }, { 18, "bye" } }; 

var res = d1.Concat(d2) 
      .GroupBy(x => x.Key) 
      .Where(g => g.Count() > 1) 
      .SelectMany(g => g.Select(x=>x.Value)) 
      .ToList();