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

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。

其他回答

一个简单而直接的解决方案:

在你的集合上使用一个标准的for循环,并使用RemoveAt(i)来删除元素。

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

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

我会这样做

using System.IO;
using System;
using System.Collections.Generic;

class Author
    {
        public string Firstname;
        public string Lastname;
        public int no;
    }

class Program
{
    private static bool isEven(int i) 
    { 
        return ((i % 2) == 0); 
    } 

    static void Main()
    {    
        var authorsList = new List<Author>()
        {
            new Author{ Firstname = "Bob", Lastname = "Smith", no = 2 },
            new Author{ Firstname = "Fred", Lastname = "Jones", no = 3 },
            new Author{ Firstname = "Brian", Lastname = "Brains", no = 4 },
            new Author{ Firstname = "Billy", Lastname = "TheKid", no = 1 }
        };

        authorsList.RemoveAll(item => isEven(item.no));

        foreach(var auth in authorsList)
        {
            Console.WriteLine(auth.Firstname + " " + auth.Lastname);
        }
    }
}

输出

Fred Jones
Billy TheKid

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。

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);