更有效的方法是使用Enumerable。除了:

var inListButNotInList2 = list.Except(list2);
var inList2ButNotInList = list2.Except(list);

该方法是通过使用延迟执行实现的。这意味着你可以这样写:

var first10 = inListButNotInList2.Take(10);

它也很有效，因为它在内部使用Set<T>来比较对象。它的工作原理是首先从第二个序列中收集所有不同的值，然后将第一个序列的结果流式传输，检查它们是否之前没有出现过。

不是针对这个问题，但是这里有一些代码来比较相等和不相等的列表!相同的对象:

public class EquatableList<T> : List<T>, IEquatable<EquatableList<T>> where    T : IEquatable<T>

/// <summary>
/// True, if this contains element with equal property-values
/// </summary>
/// <param name="element">element of Type T</param>
/// <returns>True, if this contains element</returns>
public new Boolean Contains(T element)
{
    return this.Any(t => t.Equals(element));
}

/// <summary>
/// True, if list is equal to this
/// </summary>
/// <param name="list">list</param>
/// <returns>True, if instance equals list</returns>
public Boolean Equals(EquatableList<T> list)
{
    if (list == null) return false;
    return this.All(list.Contains) && list.All(this.Contains);
}

这是你能找到的最好的解决办法

var list3 = list1.Where(l => list2.ToList().Contains(l));

using System.Collections.Generic;
using System.Linq;

namespace YourProject.Extensions
{
    public static class ListExtensions
    {
        public static bool SetwiseEquivalentTo<T>(this List<T> list, List<T> other)
            where T: IEquatable<T>
        {
            if (list.Except(other).Any())
                return false;
            if (other.Except(list).Any())
                return false;
            return true;
        }
    }
}

有时，您只需要知道两个列表是否不同，而不需要知道这些差异是什么。在这种情况下，考虑将此扩展方法添加到项目中。注意，你列出的对象应该实现IEquatable!

用法:

public sealed class Car : IEquatable<Car>
{
    public Price Price { get; }
    public List<Component> Components { get; }

    ...
    public override bool Equals(object obj)
        => obj is Car other && Equals(other);

    public bool Equals(Car other)
        => Price == other.Price
            && Components.SetwiseEquivalentTo(other.Components);

    public override int GetHashCode()
        => Components.Aggregate(
            Price.GetHashCode(),
            (code, next) => code ^ next.GetHashCode()); // Bitwise XOR
}

无论Component类是什么，这里为Car显示的方法应该几乎相同地实现。

注意我们如何编写GetHashCode是非常重要的。为了正确地实现IEquatable, Equals和GetHashCode必须以逻辑兼容的方式操作实例的属性。

Two lists with the same contents are still different objects, and will produce different hash codes. Since we want these two lists to be treated as equal, we must let GetHashCode produce the same value for each of them. We can accomplish this by delegating the hashcode to every element in the list, and using the standard bitwise XOR to combine them all. XOR is order-agnostic, so it doesn't matter if the lists are sorted differently. It only matters that they contain nothing but equivalent members.

注意:这个奇怪的名字是为了暗示这个方法不考虑列表中元素的顺序。如果您确实关心列表中元素的顺序，则此方法不适合您!

如果只需要组合结果，这也可以工作:

var set1 = new HashSet<T>(list1);
var set2 = new HashSet<T>(list2);
var areEqual = set1.SetEquals(set2);

其中T是列表元素的类型。

如果你想让结果不区分大小写，下面的方法可以工作:

List<string> list1 = new List<string> { "a.dll", "b1.dll" };
List<string> list2 = new List<string> { "A.dll", "b2.dll" };

var firstNotSecond = list1.Except(list2, StringComparer.OrdinalIgnoreCase).ToList();
var secondNotFirst = list2.Except(list1, StringComparer.OrdinalIgnoreCase).ToList();

firstNotSecond包含b1.dll

secondNotFirst将包含b2.dll