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);

使用.ToList()将创建一个列表的副本，正如这个问题中所解释的: ToList()—它是否创建一个新列表?

通过使用ToList()，可以从原始列表中删除，因为实际上是在迭代一个副本。

foreach (var item in listTracked.ToList()) {    

        if (DetermineIfRequiresRemoval(item)) {
            listTracked.Remove(item)
        }

     }

List<T> TheList = new List<T>();

TheList.FindAll(element => element.Satisfies(Condition)).ForEach(element => TheList.Remove(element));

当您希望在迭代Collection时从其中删除元素时，首先想到的应该是反向迭代。

幸运的是，有一个比编写for循环更优雅的解决方案，因为for循环涉及不必要的输入，而且容易出错。

ICollection<int> test = new List<int>(new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10});

foreach (int myInt in test.Reverse<int>())
{
    if (myInt % 2 == 0)
    {
        test.Remove(myInt);
    }
}

从列表中删除一个项的成本与后面要删除的项的数量成正比。在前半部分的条目符合删除条件的情况下，任何基于单独删除条目的方法最终都将不得不执行大约N*N/4个条目复制操作，如果列表很大，这可能会非常昂贵。

A faster approach is to scan through the list to find the first item to be removed (if any), and then from that point forward copy each item which should be retained to the spot where it belongs. Once this is done, if R items should be retained, the first R items in the list will be those R items, and all of the items requiring deletion will be at the end. If those items are deleted in reverse order, the system won't end up having to copy any of them, so if the list had N items of which R items, including all of the first F, were retained, it will be necessary to copy R-F items, and shrink the list by one item N-R times. All linear time.

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);