Question

我有两个字典，如Dic1<string,string>和Dic2<string,string> 我想为list和Dic1中存在的密钥生成新的Dic2值例如，如果

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

然后列表应为："hi" , "hello"

我尝试与Dic1.Keys.Intersect合作，但还无法弄明白。

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

Answer 1

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();

Answer 2

你在这里：

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();

结果列表包含{ 12, "hi" }和{ 12, "hello" }

HashSet对十字路口非常有用。

出于好奇，我比较了所有六种解决方案（希望没有错过任何解决方案），时间如下：

@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 3

您可以使用Where过滤Dic1中的键，然后将它们转换为如下值：

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

这应该与Dic1和Dic2上的查找操作一样高效，这通常是log（n）或更好，并且不需要构建任何临时哈希集或查找表。

这是一个版本，它避免了一个字典查找，但代价是不那么漂亮：

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

<强>更新

我的时间测试（使用t3chb0t方便的测试工具）显示第一个版本实际上运行得更快。它更简单，所以在这两者中，更喜欢这样。但到目前为止，我发现的最快速度根本没有使用Linq，在100毫秒交叉点为7毫秒，而在Linq版本为13：

    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;
    }

虽然它是旧式的，所以你可能不希望这样。

Answer 4

您只需要使用索引器获取值：

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

找到两个词典的交集

4 个答案: