谁有一个快速的方法去重复在c#的泛型列表?


当前回答

简单地用相同类型的List初始化HashSet:

var noDupes = new HashSet<T>(withDupes);

或者,如果你想返回一个List:

var noDupsList = new HashSet<T>(withDupes).ToList();

其他回答

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 }
 */

扩展方法可能是一种不错的方式……就像这样:

public static List<T> Deduplicate<T>(this List<T> listToDeduplicate)
{
    return listToDeduplicate.Distinct().ToList();
}

然后像这样调用,例如:

List<int> myFilteredList = unfilteredList.Deduplicate();

简单地用相同类型的List初始化HashSet:

var noDupes = new HashSet<T>(withDupes);

或者,如果你想返回一个List:

var noDupsList = new HashSet<T>(withDupes).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;
    }
}

这里有一个扩展的方法来删除相邻的副本原位。首先调用Sort()并传入相同的ic比较器。这应该比Lasse V. Karlsen的版本更有效,后者重复调用RemoveAt(导致多次块内存移动)。

public static void RemoveAdjacentDuplicates<T>(this List<T> List, IComparer<T> Comparer)
{
    int NumUnique = 0;
    for (int i = 0; i < List.Count; i++)
        if ((i == 0) || (Comparer.Compare(List[NumUnique - 1], List[i]) != 0))
            List[NumUnique++] = List[i];
    List.RemoveRange(NumUnique, List.Count - NumUnique);
}