如何将项目从列表移动到C＃中的另一个列表？

Question

What is the preferable way for transferring some items (not all) from one list to another.

What I am doing is the following:

var selected = from item in items
               where item.something > 10
               select item;

otherList.AddRange(selected);

items.RemoveAll(item => selected.Contains(item));

In the interest of having the fastest/best code there is, is there a better way?

Answer 1

I'd try @Mehrdad's answer, and maybe test it against this one too...

var selected = items.Where(item => item.Something > 10).ToList();
selected.ForEach(item => items.Remove(item));
otherList.AddRange(selected);

Answer 2

I suggest:

var selected = items.Where(item => item.Something > 10).ToList();
items = items.Except(selected).ToList();
otherList.AddRange(selected);

Answer 3

RemoveAll goes trough each item and enumerates all values of your selected list every time. This will take longer than it should...

What I'd do is put the condition directly in the RemoveAll parameter:

items.RemoveAll(item => item.something > 10);

If you do this and don't change the rest of your code there would be code duplication, which is not good. I'd do the following to avoid it:

Func<ItemType, bool> selectedCondition = (item => item.something > 10);

otherList.AddRange(items.Where(selectedCondition));

items.RemoveAll(new Predicate<ItemType>(selectedCondition));

Answer 4

That is quite bad performance wise - it actually enumerates a query n times (for n items in items ). It would be better if you built (for example) a HashSet<T> of the items to manipulate.

To give a simple example just with int values:

    var items = new List<int> { 1, 2, 3, 4, 5, 6 };
    var otherList = new List<int>();
    var selected = new HashSet<int>(items.Where(
        item => item > 3));
    otherList.AddRange(selected);
    items.RemoveAll(selected.Contains);

Answer 5

How about a partition:

int[] items = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };
var partition = items.ToLookup(x => x > 5);
var part1 = partition[true];
var part2 = partition[false];