Question

如果名称相同，我试图弄清楚如何只选择ID最低的对象：

public class BasicInfo
{
    public int Id {get;set;}
    public string Name {get; set;}
}

List<BasicInfo> BasicInfos = new List<BasicInfo>();
BasicInfos.Add(new BasicInfo() { Id = 1, Name = "John" });
BasicInfos.Add(new BasicInfo() { Id = 2, Name = "John" });

BasicInfos = BasicInfos.GroupBy(y => y.Name)...(What goes here?)

Answer 1

BasicInfos = BasicInfos.GroupBy(y => y.Name)
    .Select(y => y.OrderBy(z => z.Id).First())

Answer 2

类似的东西：

BasicInfos = BasicInfos.GroupBy(y => y.Name)
                       .Select(y => new BasicInfo{ Name = y.Key, Id = y.Min(x => x.Id));

这将为您提供最低密钥的所有名称。

Answer 3

您可以使用MoreLINQ中的MinBy：

public void enqueue(T queueElement) throws QueueOverflowException
    {

Answer 4

List<BasicInfo> BasicInfos = new List<BasicInfo>();
BasicInfos.Add(new BasicInfo() { Id = 1, Name = "John" });
BasicInfos.Add(new BasicInfo() { Id = 2, Name = "John" });
BasicInfos.Add(new BasicInfo() { Id = 3, Name = "Sam" });
BasicInfos.Add(new BasicInfo() { Id = 4, Name = "Sam" });
BasicInfos.Add(new BasicInfo() { Id = 5, Name = "Igor" });
BasicInfos.Add(new BasicInfo() { Id = 6, Name = "joei" });

var usersWithSameName = BasicInfos.GroupBy(x => x.Name)
    .Where(x => x.Count() > 1)
    .Select(x => x.Single(y => y.Id == x.Min(z => z.Id)));

如果名称相同，则仅选择ID最低的对象

因此，这只返回重复的内容，如果有重复，则返回id最小的记录。上面的示例将返回2个项目，并忽略那些没有找到重复名称的项目。如果这不正确，那么使用没有where子句的这个版本。

var usersWithSameName = BasicInfos.GroupBy(x => x.Name)
    .Select(x => x.Single(y => y.Id == x.Min(z => z.Id)));

LINQ GroupBy - 按名称选择最低ID组

4 个答案: