谁有一个快速的方法去重复在c#的泛型列表?


当前回答

所有的答案要么复制列表,要么创建一个新列表,要么使用慢函数,要么就是慢得令人痛苦。

据我所知,这是我所知道的最快和最便宜的方法(同时,还得到了一个非常有经验的实时物理优化程序员的支持)。

// Duplicates will be noticed after a sort O(nLogn)
list.Sort();

// Store the current and last items. Current item declaration is not really needed, and probably optimized by the compiler, but in case it's not...
int lastItem = -1;
int currItem = -1;

int size = list.Count;

// Store the index pointing to the last item we want to keep in the list
int last = size - 1;

// Travel the items from last to first O(n)
for (int i = last; i >= 0; --i)
{
    currItem = list[i];

    // If this item was the same as the previous one, we don't want it
    if (currItem == lastItem)
    {
        // Overwrite last in current place. It is a swap but we don't need the last
       list[i] = list[last];

        // Reduce the last index, we don't want that one anymore
        last--;
    }

    // A new item, we store it and continue
    else
        lastItem = currItem;
}

// We now have an unsorted list with the duplicates at the end.

// Remove the last items just once
list.RemoveRange(last + 1, size - last - 1);

// Sort again O(n logn)
list.Sort();

最终成本为:

nlogn + n + nlogn = n + 2nlogn = O(nlogn)非常漂亮。

关于RemoveRange注意事项: 由于我们不能设置列表的计数并避免使用Remove函数,我不知道这个操作的确切速度,但我猜这是最快的方法。

其他回答

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);
}

如何:

var noDupes = list.Distinct().ToList();

在。net 3.5?

我认为最简单的方法是:

创建一个新列表并添加唯一的项目。

例子:

        class MyList{
    int id;
    string date;
    string email;
    }
    
    List<MyList> ml = new Mylist();

ml.Add(new MyList(){
id = 1;
date = "2020/09/06";
email = "zarezadeh@gmailcom"
});

ml.Add(new MyList(){
id = 2;
date = "2020/09/01";
email = "zarezadeh@gmailcom"
});

 List<MyList> New_ml = new Mylist();

foreach (var item in ml)
                {
                    if (New_ml.Where(w => w.email == item.email).SingleOrDefault() == null)
                    {
                        New_ml.Add(new MyList()
                        {
                          id = item.id,
     date = item.date,
               email = item.email
                        });
                    }
                }

使用HashSet可以很容易地做到这一点。

List<int> listWithDuplicates = new List<int> { 1, 2, 1, 2, 3, 4, 5 };
HashSet<int> hashWithoutDuplicates = new HashSet<int> ( listWithDuplicates );
List<int> listWithoutDuplicates = hashWithoutDuplicates.ToList();

所有的答案要么复制列表,要么创建一个新列表,要么使用慢函数,要么就是慢得令人痛苦。

据我所知,这是我所知道的最快和最便宜的方法(同时,还得到了一个非常有经验的实时物理优化程序员的支持)。

// Duplicates will be noticed after a sort O(nLogn)
list.Sort();

// Store the current and last items. Current item declaration is not really needed, and probably optimized by the compiler, but in case it's not...
int lastItem = -1;
int currItem = -1;

int size = list.Count;

// Store the index pointing to the last item we want to keep in the list
int last = size - 1;

// Travel the items from last to first O(n)
for (int i = last; i >= 0; --i)
{
    currItem = list[i];

    // If this item was the same as the previous one, we don't want it
    if (currItem == lastItem)
    {
        // Overwrite last in current place. It is a swap but we don't need the last
       list[i] = list[last];

        // Reduce the last index, we don't want that one anymore
        last--;
    }

    // A new item, we store it and continue
    else
        lastItem = currItem;
}

// We now have an unsorted list with the duplicates at the end.

// Remove the last items just once
list.RemoveRange(last + 1, size - last - 1);

// Sort again O(n logn)
list.Sort();

最终成本为:

nlogn + n + nlogn = n + 2nlogn = O(nlogn)非常漂亮。

关于RemoveRange注意事项: 由于我们不能设置列表的计数并避免使用Remove函数,我不知道这个操作的确切速度,但我猜这是最快的方法。