这对我很管用。简单地使用

List<Type> liIDs = liIDs.Distinct().ToList<Type>();

将“类型”替换为所需的类型，例如int。

也许您应该考虑使用HashSet。

从MSDN链接:

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        HashSet<int> evenNumbers = new HashSet<int>();
        HashSet<int> oddNumbers = new HashSet<int>();

        for (int i = 0; i < 5; i++)
        {
            // Populate numbers with just even numbers.
            evenNumbers.Add(i * 2);

            // Populate oddNumbers with just odd numbers.
            oddNumbers.Add((i * 2) + 1);
        }

        Console.Write("evenNumbers contains {0} elements: ", evenNumbers.Count);
        DisplaySet(evenNumbers);

        Console.Write("oddNumbers contains {0} elements: ", oddNumbers.Count);
        DisplaySet(oddNumbers);

        // Create a new HashSet populated with even numbers.
        HashSet<int> numbers = new HashSet<int>(evenNumbers);
        Console.WriteLine("numbers UnionWith oddNumbers...");
        numbers.UnionWith(oddNumbers);

        Console.Write("numbers contains {0} elements: ", numbers.Count);
        DisplaySet(numbers);
    }

    private static void DisplaySet(HashSet<int> set)
    {
        Console.Write("{");
        foreach (int i in set)
        {
            Console.Write(" {0}", i);
        }
        Console.WriteLine(" }");
    }
}

/* This example produces output similar to the following:
 * evenNumbers contains 5 elements: { 0 2 4 6 8 }
 * oddNumbers contains 5 elements: { 1 3 5 7 9 }
 * numbers UnionWith oddNumbers...
 * numbers contains 10 elements: { 0 2 4 6 8 1 3 5 7 9 }
 */

你可以使用Union

obj2 = obj1.Union(obj1).ToList();

如果需要比较复杂的对象，则需要在Distinct()方法中传递一个compararer对象。

private void GetDistinctItemList(List<MyListItem> _listWithDuplicates)
{
    //It might be a good idea to create MyListItemComparer 
    //elsewhere and cache it for performance.
    List<MyListItem> _listWithoutDuplicates = _listWithDuplicates.Distinct(new MyListItemComparer()).ToList();
        
    //Choose the line below instead, if you have a situation where there is a chance to change the list while Distinct() is running.
    //ToArray() is used to solve "Collection was modified; enumeration operation may not execute" error.
    //List<MyListItem> _listWithoutDuplicates = _listWithDuplicates.ToArray().Distinct(new MyListItemComparer()).ToList();

    return _listWithoutDuplicates;
}

假设你有另外两个类，比如:

public class MyListItemComparer : IEqualityComparer<MyListItem>
{
    public bool Equals(MyListItem x, MyListItem y)
    {
        return x != null 
               && y != null 
               && x.A == y.A 
               && x.B.Equals(y.B); 
               && x.C.ToString().Equals(y.C.ToString());
    }

    public int GetHashCode(MyListItem codeh)
    {
        return codeh.GetHashCode();
    }
}

And:

public class MyListItem
{
    public int A { get; }
    public string B { get; }
    public MyEnum C { get; }

    public MyListItem(int a, string b, MyEnum c)
    {
        A = a;
        B = b;
        C = c;
    }
}

我喜欢使用这个命令:

List<Store> myStoreList = Service.GetStoreListbyProvince(provinceId)
                                                 .GroupBy(s => s.City)
                                                 .Select(grp => grp.FirstOrDefault())
                                                 .OrderBy(s => s.City)
                                                 .ToList();

我的列表中有这些字段:Id、StoreName、City、PostalCode 我想在一个有重复值的下拉列表中显示城市。 解决方案:按城市分组，然后选择列表中的第一个城市。

我认为最简单的方法是:

创建一个新列表并添加唯一的项目。

例子:

        class MyList{
    int id;
    string date;
    string email;
    }
    
    List<MyList> ml = new Mylist();

ml.Add(new MyList(){
id = 1;
date = "2020/09/06";
email = "zarezadeh@gmailcom"
});

ml.Add(new MyList(){
id = 2;
date = "2020/09/01";
email = "zarezadeh@gmailcom"
});

 List<MyList> New_ml = new Mylist();

foreach (var item in ml)
                {
                    if (New_ml.Where(w => w.email == item.email).SingleOrDefault() == null)
                    {
                        New_ml.Add(new MyList()
                        {
                          id = item.id,
     date = item.date,
               email = item.email
                        });
                    }
                }