Question

我有一个customObject列表，我想将＆＃34; CustomObject＆＃34;通过CustomObject对象的List属性。

 public class CustomObject
{
    public string Name { get; set; }

    public List<string> List { get; set; }

    public CustomObject(string name, List<string> list)
    {
        this.Name = name;
        this.List = list;
    }
}

.....................

List<CustomObject> listCustomObject = new List<CustomObject>()
        {
            new  CustomObject("A", new List<string>(){ "1","2","3", "4"} ),
            new  CustomObject("B", new List<string>(){ "4","8","5"}),
            new  CustomObject("C", new List<string>(){ "5","1","2", "4"})
        };

期望的结果：

＆＃34; A＆＃34; /＆＃34; C＆＃34; =＆GT;列表中的相同项目（＆＃34; 1＆＃34;，＆＃34; 2＆＃34;）

＆＃34; A＆＃34; /＆＃34; B＆＃34; /＆＃34; C＆＃34; =＆GT;列表中的相同项目（＆＃34; 4＆＃34;）

＆＃34; B＆＃34; /＆＃34; C＆＃34; =＆GT;列表中的相同项目（＆＃34; 5＆＃34;）

Answer 1

使用某些扩展方法，您可以生成至少包含两个成员的所有输入组合：

public static IEnumerable<IEnumerable<T>> AtLeastCombinations<T>(this IEnumerable<T> elements, int minK) => Enumerable.Range(minK, elements.Count()+1-minK).SelectMany(k => elements.Combinations(k));
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k) {
    return k == 0 ? new[] { new T[0] } :
      elements.SelectMany((e, i) =>
        elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] { e }).Concat(c)));
}

现在，您只需测试每个组合，看看它们是否有任何共同元素：

var ans = listCustomObject.AtLeastCombinations(2)
                          .Select(c => new { CombinationNames = c.Select(co => co.Name).ToList(), CombinationIntersect = c.Select(co => co.List).Aggregate((sofar, coList) => sofar.Intersect(coList).ToList()) })
                          .Where(ci => ci.CombinationIntersect.Count > 0)
                          .ToList();

具有列表属性的分组自定义对象

1 个答案: