foreach(var item in list.ToList())

{

if(item.Delete) list.Remove(item);

}

只需从第一个列表创建一个全新的列表。我说“简单”而不是“正确”，因为创建一个全新的列表可能比之前的方法具有更高的性能(我没有费心进行任何基准测试)。我通常更喜欢这种模式，它在克服Linq-To-Entities限制方面也很有用。

for(i = list.Count()-1;i>=0;i--)

{

item=list[i];

if (item.Delete) list.Remove(item);

}

这种方法使用普通的For循环向后遍历列表。如果集合的大小发生了变化，那么向前执行这个操作可能会有问题，但是向后执行应该总是安全的。

我将从一个过滤掉您不想保留的元素的LINQ查询中重新分配列表。

list = list.Where(item => ...).ToList();

除非列表非常大，否则这样做应该不会有明显的性能问题。

使用for循环反向迭代列表:

for (int i = safePendingList.Count - 1; i >= 0; i--)
{
    // some code
    // safePendingList.RemoveAt(i);
}

例子:

var list = new List<int>(Enumerable.Range(1, 10));
for (int i = list.Count - 1; i >= 0; i--)
{
    if (list[i] > 5)
        list.RemoveAt(i);
}
list.ForEach(i => Console.WriteLine(i));

或者，你可以使用RemoveAll方法和一个谓词来测试:

safePendingList.RemoveAll(item => item.Value == someValue);

下面是一个简单的例子:

var list = new List<int>(Enumerable.Range(1, 10));
Console.WriteLine("Before:");
list.ForEach(i => Console.WriteLine(i));
list.RemoveAll(i => i > 5);
Console.WriteLine("After:");
list.ForEach(i => Console.WriteLine(i));

我发现自己遇到了类似的情况，我必须删除给定List<T>中的每n个元素。

for (int i = 0, j = 0, n = 3; i < list.Count; i++)
{
    if ((j + 1) % n == 0) //Check current iteration is at the nth interval
    {
        list.RemoveAt(i);
        j++; //This extra addition is necessary. Without it j will wrap
             //down to zero, which will throw off our index.
    }
    j++; //This will always advance the j counter
}

使用属性跟踪要删除的元素，并在处理后将它们全部删除。

using System.Linq;

List<MyProperty> _Group = new List<MyProperty>();
// ... add elements

bool cond = false;
foreach (MyProperty currObj in _Group)
{
    // here it is supposed that you decide the "remove conditions"...
    cond = true; // set true or false...
    if (cond) 
    {
        // SET - element can be deleted
        currObj.REMOVE_ME = true;
    }
}
// RESET
_Group.RemoveAll(r => r.REMOVE_ME);

Using Remove or RemoveAt on a list while iterating over that list has intentionally been made difficult, because it is almost always the wrong thing to do. You might be able to get it working with some clever trick, but it would be extremely slow. Every time you call Remove it has to scan through the entire list to find the element you want to remove. Every time you call RemoveAt it has to move subsequent elements 1 position to the left. As such, any solution using Remove or RemoveAt, would require quadratic time, O(n²).

如果可以，使用RemoveAll。否则，下面的模式将在线性时间O(n)内就地过滤列表。

// Create a list to be filtered
IList<int> elements = new List<int>(new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10});
// Filter the list
int kept = 0;
for (int i = 0; i < elements.Count; i++) {
    // Test whether this is an element that we want to keep.
    if (elements[i] % 3 > 0) {
        // Add it to the list of kept elements.
        elements[kept] = elements[i];
        kept++;
    }
}
// Unfortunately IList has no Resize method. So instead we
// remove the last element of the list until: elements.Count == kept.
while (kept < elements.Count) elements.RemoveAt(elements.Count-1);