我正在寻找一种更好的模式,用于处理每个元素的列表,然后根据结果从列表中删除。

你不能在foreach (var element in X)中使用.Remove(element)(因为它会导致Collection被修改;枚举操作可能无法执行。例外)……你也不能使用for (int I = 0;i < elements.Count();i++)和. removeat (i),因为它破坏了你在集合中相对于i的当前位置。

有没有一种优雅的方式来做到这一点?


当前回答

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);

其他回答

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.

在c#中,一个简单的方法是标记你想要删除的,然后创建一个新的列表来迭代…

foreach(var item in list.ToList()){if(item.Delete) list.Remove(item);}  

或者更简单的使用linq....

list.RemoveAll(p=>p.Delete);

但是值得考虑的是,如果其他任务或线程在你忙着删除的同时访问同一个列表,那么可以使用ConcurrentList来代替。

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)
        }

     }

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));
myList.RemoveAt(i--);

simples;