比较两个庞大(>50.000项)的最快(和最少资源密集型)的方法是什么，从而得到如下所示的两个列表:
在第一个列表中出现但在第二个列表中没有出现的项目 出现在第二个列表中但不在第一个列表中的项目
目前，我正在使用列表或IReadOnlyCollection，并在linq查询中解决这个问题:
var list1 = list.Where(i => !list2.Contains(i)).ToList();
var list2 = list2.Where(i => !list.Contains(i)).ToList();
但这并不像我想的那样好。 有什么想法使这更快和更少的资源密集，因为我需要处理很多列表?
当前回答
我做了比较两个列表的泛型函数。
public static class ListTools
{
public enum RecordUpdateStatus
{
Added = 1,
Updated = 2,
Deleted = 3
}
public class UpdateStatu<T>
{
public T CurrentValue { get; set; }
public RecordUpdateStatus UpdateStatus { get; set; }
}
public static List<UpdateStatu<T>> CompareList<T>(List<T> currentList, List<T> inList, string uniqPropertyName)
{
var res = new List<UpdateStatu<T>>();
res.AddRange(inList.Where(a => !currentList.Any(x => x.GetType().GetProperty(uniqPropertyName).GetValue(x)?.ToString().ToLower() == a.GetType().GetProperty(uniqPropertyName).GetValue(a)?.ToString().ToLower()))
.Select(a => new UpdateStatu<T>
{
CurrentValue = a,
UpdateStatus = RecordUpdateStatus.Added,
}));
res.AddRange(currentList.Where(a => !inList.Any(x => x.GetType().GetProperty(uniqPropertyName).GetValue(x)?.ToString().ToLower() == a.GetType().GetProperty(uniqPropertyName).GetValue(a)?.ToString().ToLower()))
.Select(a => new UpdateStatu<T>
{
CurrentValue = a,
UpdateStatus = RecordUpdateStatus.Deleted,
}));
res.AddRange(currentList.Where(a => inList.Any(x => x.GetType().GetProperty(uniqPropertyName).GetValue(x)?.ToString().ToLower() == a.GetType().GetProperty(uniqPropertyName).GetValue(a)?.ToString().ToLower()))
.Select(a => new UpdateStatu<T>
{
CurrentValue = a,
UpdateStatus = RecordUpdateStatus.Updated,
}));
return res;
}
}
其他回答
也许这很有趣，但这对我来说很管用:
string.Join("",List1) != string.Join("", List2)
这是你能找到的最好的解决办法
var list3 = list1.Where(l => list2.ToList().Contains(l));
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。
我做了比较两个列表的泛型函数。
public static class ListTools
{
public enum RecordUpdateStatus
{
Added = 1,
Updated = 2,
Deleted = 3
}
public class UpdateStatu<T>
{
public T CurrentValue { get; set; }
public RecordUpdateStatus UpdateStatus { get; set; }
}
public static List<UpdateStatu<T>> CompareList<T>(List<T> currentList, List<T> inList, string uniqPropertyName)
{
var res = new List<UpdateStatu<T>>();
res.AddRange(inList.Where(a => !currentList.Any(x => x.GetType().GetProperty(uniqPropertyName).GetValue(x)?.ToString().ToLower() == a.GetType().GetProperty(uniqPropertyName).GetValue(a)?.ToString().ToLower()))
.Select(a => new UpdateStatu<T>
{
CurrentValue = a,
UpdateStatus = RecordUpdateStatus.Added,
}));
res.AddRange(currentList.Where(a => !inList.Any(x => x.GetType().GetProperty(uniqPropertyName).GetValue(x)?.ToString().ToLower() == a.GetType().GetProperty(uniqPropertyName).GetValue(a)?.ToString().ToLower()))
.Select(a => new UpdateStatu<T>
{
CurrentValue = a,
UpdateStatus = RecordUpdateStatus.Deleted,
}));
res.AddRange(currentList.Where(a => inList.Any(x => x.GetType().GetProperty(uniqPropertyName).GetValue(x)?.ToString().ToLower() == a.GetType().GetProperty(uniqPropertyName).GetValue(a)?.ToString().ToLower()))
.Select(a => new UpdateStatu<T>
{
CurrentValue = a,
UpdateStatus = RecordUpdateStatus.Updated,
}));
return res;
}
}
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);
}
