如何从列表中选择不同的项目

时间:2016-08-12 15:42:38

标签: c# .net c#-4.0

我有这个班级

class Test
    {
        public string Property { get; set; }
        public string ID { get; set; }
        public List<string> MyProperty { get; set; } 

    }

我创建了它的实例

 List<Test> t = new List<Test>() {
                new Test() {
                     ID = "1",
                      Property = "2"

                },
                new Test() {
                     ID = "2",
                      Property = "3"
                },
                new Test() {
                     ID = "2",
                     Property = "5"
                }

            };

我希望有一个List,其中包含按ID过滤的不同元素,以及公共List MyProperty {get;组;应填充公共字符串Property {get;组;数据。

所以最后的结果应该是

List<Test> = {
   1.   ID = "1",List<MyProperty> = "2"
   2.   ID = "2",List<MyProperty> = "2"                        

};

3 个答案:

答案 0 :(得分:4)

您可以使用GroupByFirst删除重复项:

t.GroupBy(x => x.Id)
    .Select(g => g.First())
    .ToList();

答案 1 :(得分:2)

我会使用GroupBy() LINQ扩展名:

t.GroupBy(x => x.ID)
 .Select(x => new Test {
    ID = x.Key,
    MyProperty = x.Select(y => y.Property).ToList()
 })
 .ToList();

GroupBy的参数是您想要分组的密钥,在您的情况下是ID。

然后Select会将它们投射到新的Test

以下是一些有用的链接:

https://msdn.microsoft.com/en-us/library/bb545971.aspx

https://msdn.microsoft.com/en-us/library/bb534304(v=vs.110).aspx

结果将是:

[
    {
        "ID": "1",
        "MyProperty": [ "2" ],
        "Property": null
    },
    {
        "ID": "2",
        "MyProperty": [ "3", "5" ],
        "Property": null
    },
]

答案 2 :(得分:2)

t.Distinct(new TestComparer());

TestComparer是比较器的实现。 这是sample

// Custom comparer for the Test class
class ProductComparer : IEqualityComparer<Test>
{
    // Tests are equal if their IDs are equal.
    public bool Equals(Test x, Test y)
    {
        //Check whether the compared objects reference the same data.
        if (Object.ReferenceEquals(x, y)) return true;

        //Check whether any of the compared objects is null.
        if (Object.ReferenceEquals(x, null) || Object.ReferenceEquals(y, null))
            return false;

        //Check whether the products' properties are equal.
        return x.Id == y.Id;
    }

    // If Equals() returns true for a pair of objects 
    // then GetHashCode() must return the same value for these objects.

    public int GetHashCode(Test test)
    {
        //Check whether the object is null
        if (Object.ReferenceEquals(test, null)) return 0;

        //Get hash code for the Name field if it is not null.
        int hashId = test.Id == null ? 0 : test.Id.GetHashCode();

        //Calculate the hash code for the test.
        return hashId;

        //Should be enough, but you can merge hashcodes of other fields in some way, for example:
        //int hashProperty = test.Property == null ? 0 : test.Property.GetHashCode();
        //return hashId ^ hashProperty;
    }
}