使用LINQ，从一个列表<int>，我怎么能检索一个列表，包含重复不止一次的条目和他们的值?
当前回答
所有的GroupBy答案都是最简单的，但不是最有效的。它们对内存性能尤其不利，因为构建大型内部集合需要分配成本。
一个不错的替代方案是HuBeZa的HashSet。基于添加的方法。它表现得更好。
如果你不关心空值，就我所知，像这样的东西是最有效的(CPU和内存):
public static IEnumerable<TProperty> Duplicates<TSource, TProperty>(
this IEnumerable<TSource> source,
Func<TSource, TProperty> duplicateSelector,
IEqualityComparer<TProperty> comparer = null)
{
comparer ??= EqualityComparer<TProperty>.Default;
Dictionary<TProperty, int> counts = new Dictionary<TProperty, int>(comparer);
foreach (var item in source)
{
TProperty property = duplicateSelector(item);
counts.TryGetValue(property, out int count);
switch (count)
{
case 0:
counts[property] = ++count;
break;
case 1:
counts[property] = ++count;
yield return property;
break;
}
}
}
这里的技巧是在重复数达到1时避免额外的查找成本。当然，如果您还想知道每个项重复出现的次数，则可以使用count不断更新字典。对于null，你只需要一些额外的处理，仅此而已。
其他回答
你可以这样做:
var list = new[] {1,2,3,1,4,2};
var duplicateItems = list.Duplicates();
使用这些扩展方法:
public static class Extensions
{
public static IEnumerable<TSource> Duplicates<TSource, TKey>(this IEnumerable<TSource> source, Func<TSource, TKey> selector)
{
var grouped = source.GroupBy(selector);
var moreThan1 = grouped.Where(i => i.IsMultiple());
return moreThan1.SelectMany(i => i);
}
public static IEnumerable<TSource> Duplicates<TSource, TKey>(this IEnumerable<TSource> source)
{
return source.Duplicates(i => i);
}
public static bool IsMultiple<T>(this IEnumerable<T> source)
{
var enumerator = source.GetEnumerator();
return enumerator.MoveNext() && enumerator.MoveNext();
}
}
在duplicate方法中使用IsMultiple()比Count()更快，因为这不会迭代整个集合。
我创建了一个扩展来响应这个，你可以把它包括在你的项目中，我认为这返回的大多数情况下，当你在列表或Linq中搜索重复。
例子:
//Dummy class to compare in list
public class Person
{
public int Id { get; set; }
public string Name { get; set; }
public string Surname { get; set; }
public Person(int id, string name, string surname)
{
this.Id = id;
this.Name = name;
this.Surname = surname;
}
}
//The extention static class
public static class Extention
{
public static IEnumerable<T> getMoreThanOnceRepeated<T>(this IEnumerable<T> extList, Func<T, object> groupProps) where T : class
{ //Return only the second and next reptition
return extList
.GroupBy(groupProps)
.SelectMany(z => z.Skip(1)); //Skip the first occur and return all the others that repeats
}
public static IEnumerable<T> getAllRepeated<T>(this IEnumerable<T> extList, Func<T, object> groupProps) where T : class
{
//Get All the lines that has repeating
return extList
.GroupBy(groupProps)
.Where(z => z.Count() > 1) //Filter only the distinct one
.SelectMany(z => z);//All in where has to be retuned
}
}
//how to use it:
void DuplicateExample()
{
//Populate List
List<Person> PersonsLst = new List<Person>(){
new Person(1,"Ricardo","Figueiredo"), //fist Duplicate to the example
new Person(2,"Ana","Figueiredo"),
new Person(3,"Ricardo","Figueiredo"),//second Duplicate to the example
new Person(4,"Margarida","Figueiredo"),
new Person(5,"Ricardo","Figueiredo")//third Duplicate to the example
};
Console.WriteLine("All:");
PersonsLst.ForEach(z => Console.WriteLine("{0} -> {1} {2}", z.Id, z.Name, z.Surname));
/* OUTPUT:
All:
1 -> Ricardo Figueiredo
2 -> Ana Figueiredo
3 -> Ricardo Figueiredo
4 -> Margarida Figueiredo
5 -> Ricardo Figueiredo
*/
Console.WriteLine("All lines with repeated data");
PersonsLst.getAllRepeated(z => new { z.Name, z.Surname })
.ToList()
.ForEach(z => Console.WriteLine("{0} -> {1} {2}", z.Id, z.Name, z.Surname));
/* OUTPUT:
All lines with repeated data
1 -> Ricardo Figueiredo
3 -> Ricardo Figueiredo
5 -> Ricardo Figueiredo
*/
Console.WriteLine("Only Repeated more than once");
PersonsLst.getMoreThanOnceRepeated(z => new { z.Name, z.Surname })
.ToList()
.ForEach(z => Console.WriteLine("{0} -> {1} {2}", z.Id, z.Name, z.Surname));
/* OUTPUT:
Only Repeated more than once
3 -> Ricardo Figueiredo
5 -> Ricardo Figueiredo
*/
}
所有的GroupBy答案都是最简单的，但不是最有效的。它们对内存性能尤其不利，因为构建大型内部集合需要分配成本。
一个不错的替代方案是HuBeZa的HashSet。基于添加的方法。它表现得更好。
如果你不关心空值，就我所知，像这样的东西是最有效的(CPU和内存):
public static IEnumerable<TProperty> Duplicates<TSource, TProperty>(
this IEnumerable<TSource> source,
Func<TSource, TProperty> duplicateSelector,
IEqualityComparer<TProperty> comparer = null)
{
comparer ??= EqualityComparer<TProperty>.Default;
Dictionary<TProperty, int> counts = new Dictionary<TProperty, int>(comparer);
foreach (var item in source)
{
TProperty property = duplicateSelector(item);
counts.TryGetValue(property, out int count);
switch (count)
{
case 0:
counts[property] = ++count;
break;
case 1:
counts[property] = ++count;
yield return property;
break;
}
}
}
这里的技巧是在重复数达到1时避免额外的查找成本。当然，如果您还想知道每个项重复出现的次数，则可以使用count不断更新字典。对于null，你只需要一些额外的处理，仅此而已。
这种更简单的方法，不使用组，只是获得区域元素，然后迭代它们，并检查它们在列表中的计数，如果它们的计数是>1，这意味着它出现超过1项，所以将其添加到Repeteditemlist
var mylist = new List<int>() { 1, 1, 2, 3, 3, 3, 4, 4, 4 };
var distList= mylist.Distinct().ToList();
var Repeteditemlist = new List<int>();
foreach (var item in distList)
{
if(mylist.Count(e => e == item) > 1)
{
Repeteditemlist.Add(item);
}
}
foreach (var item in Repeteditemlist)
{
Console.WriteLine(item);
}
预期的输出:
1 3 4
找出一个枚举对象是否包含任何重复项:
var anyDuplicate = enumerable.GroupBy(x => x.Key).Any(g => g.Count() > 1);
找出一个枚举对象中的所有值是否都是唯一的:
var allUnique = enumerable.GroupBy(x => x.Key).All(g => g.Count() == 1);
