c#得出两个列表的交集

Posted 2021-03-20 农村程序员

c#提供了Intersect来得到两个列表的交集，它是通过使用默认的相等比较器对值进行比较生成两个序列的交集，定义为：

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

我们使用它来比较两个列表试试：

            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();

结果是{9,3}

我们再来试试对象列表

我们先定义一个模型

        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>();           
            //第一个列表增加的对象
            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 };
            //第二个列表增加的对象
            TarGetValueList Three = new TarGetValueList() { TargetValue = 3.0 };//第一个列表增加对象
            ListOne.Add(Night); ListOne.Add(Sex); ListOne.Add(Five); ListOne.Add(Seven);//9 6 5 7
            //第二个列表增加对象
            ListTwo.Add(Five); ListTwo.Add(Three);           //5 3
            List<TarGetValueList> resultList = ListTwo.Intersect(ListOne).ToList();

第一个列表是{9，6，5，7}，第二个列表是{5,3}，结果交集是{5}，结果在预想之内，至今为止，一切都很顺利。我们似乎已经掌握了用c#来求两个列表的交集，但让我们在以上的例子上进行以下的尝试：

            List<TarGetValueList> ListOne = new List<TarGetValueList>();
            List<TarGetValueList> ListTwo= new List<TarGetValueList>();           
            //第一个列表增加的对象
            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 };
            //第二个列表增加的对象
            TarGetValueList Three = new TarGetValueList() { TargetValue = 3.0 };
            TarGetValueList FiveTwo = new TarGetValueList() { TargetValue = 5.0 };
            //第一个列表增加对象
            ListOne.Add(Night); ListOne.Add(Sex); ListOne.Add(Five); ListOne.Add(Seven);//9 6 5 7
            //第二个列表增加对象
            ListTwo.Add(FiveTwo); ListTwo.Add(Three);           //5 3
            List<TarGetValueList> resultList = ListTwo.Intersect(ListOne).ToList();

第一个列表仍然是{9，6，5，7}，第二个列表仍然是{5,3}，得到的结果交集列表却为空了。第二个列表的5跟第一个列表的5来自不同的对象，不过对象里的值都是一样的。但Intersect函数判断它们不是交集对象，这是因为判断对象是否相等是根据它们的内存地址来判断，那我们判断对象之间相等就没办法了吗？我们重写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>();           
            //第一个列表增加的对象
            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 };
            //第二个列表增加的对象
            TarGetValueList Three = new TarGetValueList() { TargetValue = 3.0 };
            TarGetValueList FiveTwo = new TarGetValueList() { TargetValue = 5.0 };
            //第一个列表增加对象
            ListOne.Add(Night); ListOne.Add(Sex); ListOne.Add(Five); ListOne.Add(Seven);//9 6 5 7
            //第二个列表增加对象
            ListTwo.Add(FiveTwo); ListTwo.Add(Three);           //5 3
            List<TarGetValueList> resultList = ListTwo.Intersect(ListOne,new TarGetComparer()).ToList();

这样就得出了我们预期的结果{5}