C#在列表中获取非重复项
有这样的列表:
int[] numbers = {1,2,2,3,3,4,4,5};
我可以使用Distinct()函数删除重复项,因此列表将显示为:1,2,3,4,5
但是,我想反过来。 我希望它删除所有重复的数字,留下我独特的数字。
所以列表如下:1,5。
这怎么办?
一种方法是
var singles = numbers.GroupBy(n => n) .Where(g => g.Count() == 1) .Select(g => g.Key); // add .ToArray() etc as required
对于它的价值,一个检查序列是否包含多于N
元素的扩展:
public static bool CountMoreThan(this IEnumerable source, int num) { if (source == null) throw new ArgumentNullException("source"); if (num < 0) throw new ArgumentException("num must be greater or equal 0", "num"); ICollection collection = source as ICollection ; if (collection != null) { return collection.Count > num; } ICollection collection2 = source as ICollection; if (collection2 != null) { return collection2.Count > num; } int count = 0; using (IEnumerator enumerator = source.GetEnumerator()) { while (++count <= num + 1) if (!enumerator.MoveNext()) return false; } return true; }
现在它简单而有效:
var allUniques = numbers.GroupBy(i => i) .Where(group => !group.CountMoreThan(1)) .Select(group => group.Key).ToList();
DEMO
或者,正如@KingKing对Jon的回答所评论:
var allUniques = numbers.GroupBy(i => i) .Where(group => !group.Skip(1).Any()) .Select(group => group.Key).ToList();
var cleanArray = numbers.GroupBy(x=>x) .Where(x=>x.Count() == 1) .SelectMany(x=>x) .ToArray();