我有LIST1<>和LIST2<>并且喜欢比较这两个列表。以下是我的条件..
1-If LIST1 and LIST2 have the same items than add same items to LIST3
2-If LIST1 doesnt contain LIST2 items than add different items to LIST4
3-if LIST2 doesnt contain LIST1 items than add different items to LIST5
让我说我的结果如下,取决于条件;
LIST1<string> = A,B,C,D
LIST2<string> = A,K,F,C
LIST3<string> = A,C
LIST4<string> = B,D
LIST5<string> = K,F
这是我的代码;
foreach (string src in LIST1)
{
foreach (string trg in LIST2)
{
if (LIST1.ToString() == LIST2.ToString())
{
LIST3.Add(LIST1.ToString());
}
else
{
LIST4.Clear();
foreach (string l3 in LIST1)
{
if (!LIST2.Contains(l3))
LIST4.Add(l3);
}
LIST5.Clear();
foreach (string l4 in LIST2)
{
if (!LIST1.Contains(l4))
{
LIST5.Add(l4);
}
}
}
}
}
答案 0 :(得分:5)
快速执行此操作的方法是:
var list3 = list1.Intersect(list2).ToList();
var list4 = list1.Except(list2).ToList();
var list5 = list2.Except(list1).ToList();
更新:如果你需要处理更大的列表(和/或必须在多个地方写这个),你可以编写如下的扩展方法:
public static Tuple<IEnumerable<T>, IEnumerable<T>, IEnumerable<T>> Diff<T>(
this IEnumerable<T> first, IEnumerable<T> second)
{
var intersection = new List<T>();
var onlyInFirst = new HashSet<T>();
var onlyInSecond = new HashSet<T>(second);
foreach (var item in first)
{
if (onlyInSecond.Remove(item)) intersection.Add(item);
else onlyInFirst.Add(item);
}
return Tuple.Create<IEnumerable<T>, IEnumerable<T>, IEnumerable<T>>
(intersection, onlyInFirst, onlyInSecond);
}
此方法返回三个IEnumerable<T>
的元组,表示交集,仅在第一个集合中的项集,以及仅在第二个集合中的项集;分别。
用法:
var list1 = new[] { "A", "B", "C", "D" };
var list2 = new[] { "A", "K", "F", "C" };
var diff = list1.Diff(list2);
// diff.Item1 = A,C (intersection)
// diff.Item2 = B,D (only in first)
// diff.Item3 = K,F (only in second)
答案 1 :(得分:0)
不确定这与排序有什么关系,但这里是每个条件的Linq语句:
List3 = List1.Intersect(List2).ToList();
List4 = List1.Where(l1 => !List2.Any(l2 => l2 == l1)).ToList();
List5 = List2.Where(l2 => !List1.Any(l1 => l2 == l1)).ToList();
在评论Except
中指出也将有效:
List4 = List1.Except(List2).ToList();
List5 = List2.Except(List1).ToList();