我比较了3种不同的方法来比较不同的数据集。下面的测试创建了一个包含从0到length - 1的所有数字的字符串集合，然后是另一个具有相同范围但包含偶数的集合。然后我从第一个集合中挑出奇数。

使用Linq除外

public void TestExcept()
{
    WriteLine($"Except {DateTime.Now}");
    int length = 20000000;
    var dateTime = DateTime.Now;
    var array = new string[length];
    for (int i = 0; i < length; i++)
    {
        array[i] = i.ToString();
    }
    Write("Populate set processing time: ");
    WriteLine(DateTime.Now - dateTime);
    var newArray = new string[length/2];
    int j = 0;
    for (int i = 0; i < length; i+=2)
    {
        newArray[j++] = i.ToString();
    }
    dateTime = DateTime.Now;
    Write("Count of items: ");
    WriteLine(array.Except(newArray).Count());
    Write("Count processing time: ");
    WriteLine(DateTime.Now - dateTime);
}

输出

Except 2021-08-14 11:43:03 AM
Populate set processing time: 00:00:03.7230479
2021-08-14 11:43:09 AM
Count of items: 10000000
Count processing time: 00:00:02.9720879

使用HashSet。添加

public void TestHashSet()
{
    WriteLine($"HashSet {DateTime.Now}");
    int length = 20000000;
    var dateTime = DateTime.Now;
    var hashSet = new HashSet<string>();
    for (int i = 0; i < length; i++)
    {
        hashSet.Add(i.ToString());
    }
    Write("Populate set processing time: ");
    WriteLine(DateTime.Now - dateTime);
    var newHashSet = new HashSet<string>();
    for (int i = 0; i < length; i+=2)
    {
        newHashSet.Add(i.ToString());
    }
    dateTime = DateTime.Now;
    Write("Count of items: ");
    // HashSet Add returns true if item is added successfully (not previously existing)
    WriteLine(hashSet.Where(s => newHashSet.Add(s)).Count());
    Write("Count processing time: ");
    WriteLine(DateTime.Now - dateTime);
}

输出

HashSet 2021-08-14 11:42:43 AM
Populate set processing time: 00:00:05.6000625
Count of items: 10000000
Count processing time: 00:00:01.7703057

特殊HashSet测试:

public void TestLoadingHashSet()
{
    int length = 20000000;
    var array = new string[length];
    for (int i = 0; i < length; i++)
    {
       array[i] = i.ToString();
    }
    var dateTime = DateTime.Now;
    var hashSet = new HashSet<string>(array);
    Write("Time to load hashset: ");
    WriteLine(DateTime.Now - dateTime);
}
> TestLoadingHashSet()
Time to load hashset: 00:00:01.1918160

使用.Contains

public void TestContains()
{
    WriteLine($"Contains {DateTime.Now}");
    int length = 20000000;
    var dateTime = DateTime.Now;
    var array = new string[length];
    for (int i = 0; i < length; i++)
    {
        array[i] = i.ToString();
    }
    Write("Populate set processing time: ");
    WriteLine(DateTime.Now - dateTime);
    var newArray = new string[length/2];
    int j = 0;
    for (int i = 0; i < length; i+=2)
    {
        newArray[j++] = i.ToString();
    }
    dateTime = DateTime.Now;
    WriteLine(dateTime);
    Write("Count of items: ");
    WriteLine(array.Where(a => !newArray.Contains(a)).Count());
    Write("Count processing time: ");
    WriteLine(DateTime.Now - dateTime);
}

输出

Contains 2021-08-14 11:19:44 AM
Populate set processing time: 00:00:03.1046998
2021-08-14 11:19:49 AM
Count of items: Hosting process exited with exit code 1.
(Didnt complete. Killed it after 14 minutes)

结论:

Linq Except在我的设备上运行大约比使用HashSets慢1秒(n=20,000,000)。 使用Where和Contains运行了很长时间

哈希集的总结:

独特的数据 确保为类类型重写GetHashCode(正确地) 如果您复制数据集，可能需要高达2倍的内存，这取决于实现 HashSet是为使用IEnumerable构造函数克隆其他HashSet而优化的，但是将其他集合转换为HashSet比较慢(参见上面的特殊测试)

使用除外:

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

我怀疑有一些方法实际上会比这个稍微快一点，但即使是这个方法也会比O(N * M)方法快得多。

如果你想把它们结合起来，你可以用上面的方法创建一个方法，然后创建一个return语句:

return !firstNotSecond.Any() && !secondNotFirst.Any();

需要注意的一点是，问题中的原始代码和这里的解决方案之间的结果存在差异:在我的代码中，只在一个列表中出现的任何重复元素将只报告一次，而在原始代码中出现的次数与它们相同。

例如，对于[1,2,2,2,3]和[1]的列表，原始代码中的“元素在list1但不是list2”结果将是[2,2,2,3]。在我的代码中，它就是[2,3]。在许多情况下，这不是一个问题，但这是值得注意的。

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

Jon Skeet和miguelmpn的回答都很好。这取决于列表元素的顺序是否重要:

// take order into account
bool areEqual1 = Enumerable.SequenceEqual(list1, list2);

// ignore order
bool areEqual2 = !list1.Except(list2).Any() && !list2.Except(list1).Any();

可列举的。SequenceEqual方法 根据相等比较器确定两个序列是否相等。 MS.Docs

Enumerable.SequenceEqual(list1, list2);

这适用于所有基本数据类型。如果你需要在自定义对象上使用它，你需要实现IEqualityComparer

定义方法以支持相等的对象比较。

IEqualityComparer接口 定义方法以支持相等的对象比较。 MS.Docs for IEqualityComparer

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!