【C#】List求并集、交集、差集

寒冰的暖2024-12-20 20:10

值类型List

csharp 复制代码

            List<int> intList1 = new List<int>() { 1, 2, 3 };
            List<int> intList2 = new List<int>() { 3, 4, 5 };
            var result = intList1.Union(intList2);
            Console.WriteLine($"并 {string.Join(',',result)}");
             result = intList1.Intersect(intList2);
            Console.WriteLine($"交 {string.Join(',', result)}");
             result = intList1.Except(intList2);
            Console.WriteLine($"差 {string.Join(',', result)}");

结果：

对象类型List

csharp 复制代码

       List<Person> people = new List<Person>
        {
            new Person { Name = "Alice" },
            new Person { Name = "Bob" },
            new Person { Name = "Charlie" }
        };
        List<Person> people2 = new List<Person>
        {
            new Person { Name = "Alice" },
            new Person { Name = "Joan" }
        };
        var abc = people.Union(people2).ToList();
        Console.WriteLine($"并 { string.Join(',', abc.Select(s => s.Name))}");
        abc = people.Where(s => people2.Any(x => x.Name == s.Name)).ToList();
        Console.WriteLine($"Name交 {string.Join(',', abc.Select(s=>s.Name))}");
        abc = people.Where(s =>!  people2.Any(x => x.Name == s.Name)).ToList();
        Console.WriteLine($"Name差 {string.Join(',', abc.Select(s => s.Name))}");

结果

对象类型的还可以利用LINQ 左连接求交集、差集

csharp 复制代码

    var leftJoinQuery = from p in people
                        join pp in people2 on p.Name equals pp.Name into temp
                        from co in temp.DefaultIfEmpty()
                        where co is not null
                        select new { p.Name };

    Console.WriteLine($"Name交 {string.Join(',', leftJoinQuery.Select(s => s.Name))}");

    leftJoinQuery = from p in people
                        join pp in people2 on p.Name equals pp.Name into temp
                        from co in temp.DefaultIfEmpty()
                        where co is null 
                        select new { p.Name };

   Console.WriteLine($"Name差 {string.Join(',', leftJoinQuery.Select(s => s.Name))}");

结果：