所以我写了这个简单的控制台应用程序在我的问题,要求提供帮助。什么是使用的方法的第3行的lambda EX pression得到普遍成员的正确方法。尝试了加入(),但无法找出正确的语法。作为跟进...有一种非LINQ的方式来做到这一条线,我错过了?
类节目
{
静态无效的主要(字串[] args)
{
名单< INT> C =新的名单,其中,INT>(){1,2,3};
名单< INT> A =新的名单,其中,INT>(){5,3,2,4};
IEnumerable的< INT> J = c.Union&其中; INT>(一);
//你只要给我伯爵
Console.Write(j.ToList&其中; int的>()Count.ToString());
}
}
解决方案
您想要相交()
:
的IEnumerable< INT> J = c.Intersect(一);
下面是基于在评论中提到的思想的 OrderedIntersect()
的例子。如果你知道你的序列是有序的,应该跑得更快— O(N)
,而不是什么 .Intersect()
通常是(不记得了我的头顶部)。但是,如果你不知道他们是有序的,它可能不会返回正确的结果在所有的:
公共静态的IEnumerable< T> OrderedIntersect< T>(这IEnumerable的< T>源,IEnumerable的< T>其他),其中T:IComparable的
{
使用(VAR XE = source.GetEnumerator())
使用(VAR你们= other.GetEnumerator())
{
而(xe.MoveNext())
{
而(ye.MoveNext()&安培;&安培; ye.Current.CompareTo(xe.Current)℃,)
{
//什么也不做 - 我们所关心这里要说的是,我们提前在y枚举
}
如果(ye.Current.Equals(xe.Current))
收益回报xe.Current;
其他
{// y为现在和GT; X,因此得到对X陷入再次
而(xe.MoveNext()&安培;&安培; xe.Current.CompareTo(ye.Current)℃,)
{} //又:只是前进,不要做任何事情
如果(xe.Current.Equals(ye.Current))收益回报xe.Current;
}
}
}
}
So I wrote this simple console app to aid in my question asking. What is the proper way to use a lambda expression on line 3 of the method to get the common members. Tried a Join() but couldn't figure out the correct syntax. As follow up... is there a non-LINQ way to do this in one line that I missed?
class Program
{
static void Main(string[] args)
{
List<int> c = new List<int>() { 1, 2, 3 };
List<int> a = new List<int>() { 5, 3, 2, 4 };
IEnumerable<int> j = c.Union<int>(a);
// just show me the Count
Console.Write(j.ToList<int>().Count.ToString());
}
}
解决方案
You want Intersect()
:
IEnumerable<int> j = c.Intersect(a);
Here's an OrderedIntersect()
example based on the ideas mentioned in the comments. If you know your sequences are ordered it should run faster — O(n)
rather than whatever .Intersect()
normally is (don't remember off the top of my head). But if you don't know they are ordered, it likely won't return correct results at all:
public static IEnumerable<T> OrderedIntersect<T>(this IEnumerable<T> source, IEnumerable<T> other) where T : IComparable
{
using (var xe = source.GetEnumerator())
using (var ye = other.GetEnumerator())
{
while (xe.MoveNext())
{
while (ye.MoveNext() && ye.Current.CompareTo(xe.Current) < 0 )
{
// do nothing - all we care here is that we advanced the y enumerator
}
if (ye.Current.Equals(xe.Current))
yield return xe.Current;
else
{ // y is now > x, so get x caught up again
while (xe.MoveNext() && xe.Current.CompareTo(ye.Current) < 0 )
{ } // again: just advance, do do anything
if (xe.Current.Equals(ye.Current)) yield return xe.Current;
}
}
}
}