c#得出兩個列表的交集

c#提供了Intersect來得到兩個列表的交集，它是通過使用默認的相等比較器對值進行比較生成兩個序列的交集，定義為：

public static IEnumerable<TSource> Intersect<TSource>(this IEnumerable<TSource> first, IEnumerable<TSource> second);

我們使用它來比較兩個列表試試：

            List<double> oneList = new List<double> { 9, 8, 5, 3 };
            List<double> twoList = new List<double> { 9, 3 };
            List<double> result = oneList.Intersect(twoList).ToList();

結果是{9,3}

我們再來試試對象列表

我們先定義一個模型

        public class TarGetValueList
        {           
            public double TargetValue { get; set; }       
          
        }

然后求交集

            List<TarGetValueList> ListOne = new List<TarGetValueList>();
            List<TarGetValueList> ListTwo= new List<TarGetValueList>();           
            //第一個列表增加的對象
            TarGetValueList Night = new TarGetValueList() { TargetValue = 9.0 };
            TarGetValueList Sex = new TarGetValueList() { TargetValue = 6.0 };
            TarGetValueList Seven = new TarGetValueList() { TargetValue = 7.0 };
            TarGetValueList Five = new TarGetValueList() { TargetValue = 5.0 };
            //第二個列表增加的對象
            TarGetValueList Three = new TarGetValueList() { TargetValue = 3.0 };//第一個列表增加對象
            ListOne.Add(Night); ListOne.Add(Sex); ListOne.Add(Five); ListOne.Add(Seven);//9 6 5 7
            //第二個列表增加對象
            ListTwo.Add(Five); ListTwo.Add(Three);           //5 3
            List<TarGetValueList> resultList = ListTwo.Intersect(ListOne).ToList();

第一個列表是{9，6，5，7}，第二個列表是{5,3}，結果交集是{5}，結果在預想之內，至今為止，一切都很順利。我們似乎已經掌握了用c#來求兩個列表的交集，但讓我們在以上的例子上進行以下的嘗試：

            List<TarGetValueList> ListOne = new List<TarGetValueList>();
            List<TarGetValueList> ListTwo= new List<TarGetValueList>();           
            //第一個列表增加的對象
            TarGetValueList Night = new TarGetValueList() { TargetValue = 9.0 };
            TarGetValueList Sex = new TarGetValueList() { TargetValue = 6.0 };
            TarGetValueList Seven = new TarGetValueList() { TargetValue = 7.0 };
            TarGetValueList Five = new TarGetValueList() { TargetValue = 5.0 };
            //第二個列表增加的對象
            TarGetValueList Three = new TarGetValueList() { TargetValue = 3.0 };
            TarGetValueList FiveTwo = new TarGetValueList() { TargetValue = 5.0 };
            //第一個列表增加對象
            ListOne.Add(Night); ListOne.Add(Sex); ListOne.Add(Five); ListOne.Add(Seven);//9 6 5 7
            //第二個列表增加對象
            ListTwo.Add(FiveTwo); ListTwo.Add(Three);           //5 3
            List<TarGetValueList> resultList = ListTwo.Intersect(ListOne).ToList();

第一個列表仍然是{9，6，5，7}，第二個列表仍然是{5,3}，得到的結果交集列表卻為空了。第二個列表的5跟第一個列表的5來自不同的對象，不過對象里的值都是一樣的。但Intersect函數判斷它們不是交集對象，這是因為判斷對象是否相等是根據它們的內存地址來判斷，那我們判斷對象之間相等就沒辦法了嗎？我們重寫IEqualityComparer即可：

 public class TarGetComparer : IEqualityComparer<TarGetValueList>
    {
        public bool Equals(TarGetValueList x, TarGetValueList y)
        {
            try
            {
                return x.TargetValue == y.TargetValue;
            }
           catch(Exception ex)
            {
                return base.Equals(x);
            }
        }

        public int GetHashCode(TarGetValueList obj)
        {
            return obj.TargetValue.GetHashCode();
        }
    }

List<TarGetValueList> ListOne = new List<TarGetValueList>();
            List<TarGetValueList> ListTwo= new List<TarGetValueList>();           
            //第一個列表增加的對象
            TarGetValueList Night = new TarGetValueList() { TargetValue = 9.0 };
            TarGetValueList Sex = new TarGetValueList() { TargetValue = 6.0 };
            TarGetValueList Seven = new TarGetValueList() { TargetValue = 7.0 };
            TarGetValueList Five = new TarGetValueList() { TargetValue = 5.0 };
            //第二個列表增加的對象
            TarGetValueList Three = new TarGetValueList() { TargetValue = 3.0 };
            TarGetValueList FiveTwo = new TarGetValueList() { TargetValue = 5.0 };
            //第一個列表增加對象
            ListOne.Add(Night); ListOne.Add(Sex); ListOne.Add(Five); ListOne.Add(Seven);//9 6 5 7
            //第二個列表增加對象
            ListTwo.Add(FiveTwo); ListTwo.Add(Three);           //5 3
            List<TarGetValueList> resultList = ListTwo.Intersect(ListOne,new TarGetComparer()).ToList();

這樣就得出了我們預期的結果{5}