我认为这是一个简单易行的方法来逐个元素比较两个列表

x=[1,2,3,5,4,8,7,11,12,45,96,25]
y=[2,4,5,6,8,7,88,9,6,55,44,23]

tmp = []


for i in range(len(x)) and range(len(y)):
    if x[i]>y[i]:
        tmp.append(1)
    else:
        tmp.append(0)
print(tmp)

我已经使用这段代码来比较两个有百万记录的列表。

这种方法不会花很多时间

    //Method to compare two list of string
    private List<string> Contains(List<string> list1, List<string> list2)
    {
        List<string> result = new List<string>();

        result.AddRange(list1.Except(list2, StringComparer.OrdinalIgnoreCase));
        result.AddRange(list2.Except(list1, StringComparer.OrdinalIgnoreCase));

        return result;
    }

试试这个方法:

var difList = list1.Where(a => !list2.Any(a1 => a1.id == a.id))
            .Union(list2.Where(a => !list1.Any(a1 => a1.id == a.id)));

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!

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();

第一种方法:

if (list1 != null && list2 != null && list1.Select(x => list2.SingleOrDefault(y => y.propertyToCompare == x.propertyToCompare && y.anotherPropertyToCompare == x.anotherPropertyToCompare) != null).All(x => true))
   return true;

第二种方法，如果你对重复值没有问题:

if (list1 != null && list2 != null && list1.Select(x => list2.Any(y => y.propertyToCompare == x.propertyToCompare && y.anotherPropertyToCompare == x.anotherPropertyToCompare)).All(x => true))
   return true;