完整的集Linq到SQL扩展的重复功能检查在MS SQL Server。不使用. tolist()或IEnumerable。这些查询在SQL Server中执行，而不是在内存中。结果只在内存中返回。

public static class Linq2SqlExtensions {

    public class CountOfT<T> {
        public T Key { get; set; }
        public int Count { get; set; }
    }

    public static IQueryable<TKey> Duplicates<TSource, TKey>(this IQueryable<TSource> source, Expression<Func<TSource, TKey>> groupBy)
        => source.GroupBy(groupBy).Where(w => w.Count() > 1).Select(s => s.Key);

    public static IQueryable<TSource> GetDuplicates<TSource, TKey>(this IQueryable<TSource> source, Expression<Func<TSource, TKey>> groupBy)
        => source.GroupBy(groupBy).Where(w => w.Count() > 1).SelectMany(s => s);

    public static IQueryable<CountOfT<TKey>> DuplicatesCounts<TSource, TKey>(this IQueryable<TSource> source, Expression<Func<TSource, TKey>> groupBy)
        => source.GroupBy(groupBy).Where(w => w.Count() > 1).Select(y => new CountOfT<TKey> { Key = y.Key, Count = y.Count() });

    public static IQueryable<Tuple<TKey, int>> DuplicatesCountsAsTuble<TSource, TKey>(this IQueryable<TSource> source, Expression<Func<TSource, TKey>> groupBy)
        => source.GroupBy(groupBy).Where(w => w.Count() > 1).Select(s => Tuple.Create(s.Key, s.Count()));
}

这是另一种方法:

对于HasDuplicate:

bool hasAnyDuplicate = list.Count > list.Distinct().Count;

对于重复值

List<string> duplicates = new List<string>();
duplicates.AddRange(list);
list.Distinct().ToList().ForEach(x => duplicates.Remove(x));

// for unique duplicate values:
duplicates.Distinct():

只查找重复的值:

var duplicates = list.GroupBy(x => x.Key).Where(g => g.Count() > 1);

E.g.

var list = new[] {1,2,3,1,4,2};

GroupBy将根据它们的键对数字进行分组，并使用它维护计数(重复的次数)。在那之后，我们只是检查重复了不止一次的值。

要查找唯一的值:

var unique = list.GroupBy(x => x.Key).Where(g => g.Count() == 1);

E.g.

var list = new[] {1,2,3,1,4,2};

GroupBy将根据它们的键对数字进行分组，并使用它维护计数(重复的次数)。在此之后，我们只是检查那些只重复一次的值是否惟一。

另一种方法是使用HashSet:

var hash = new HashSet<int>();
var duplicates = list.Where(i => !hash.Add(i));

如果你想在你的重复列表中的唯一值:

var myhash = new HashSet<int>();
var mylist = new List<int>(){1,1,2,2,3,3,3,4,4,4};
var duplicates = mylist.Where(item => !myhash.Add(item)).Distinct().ToList();

下面是与通用扩展方法相同的解决方案:

public static class Extensions
{
  public static IEnumerable<TSource> GetDuplicates<TSource, TKey>(this IEnumerable<TSource> source, Func<TSource, TKey> selector, IEqualityComparer<TKey> comparer)
  {
    var hash = new HashSet<TKey>(comparer);
    return source.Where(item => !hash.Add(selector(item))).ToList();
  }

  public static IEnumerable<TSource> GetDuplicates<TSource>(this IEnumerable<TSource> source, IEqualityComparer<TSource> comparer)
  {
    return source.GetDuplicates(x => x, comparer);      
  }

  public static IEnumerable<TSource> GetDuplicates<TSource, TKey>(this IEnumerable<TSource> source, Func<TSource, TKey> selector)
  {
    return source.GetDuplicates(selector, null);
  }

  public static IEnumerable<TSource> GetDuplicates<TSource>(this IEnumerable<TSource> source)
  {
    return source.GetDuplicates(x => x, null);
  }
}

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。

有一个答案，但我不明白为什么不工作;

var anyDuplicate = enumerable.GroupBy(x => x.Key).Any(g => g.Count() > 1);

我的解是这样的;

var duplicates = model.list
                    .GroupBy(s => s.SAME_ID)
                    .Where(g => g.Count() > 1).Count() > 0;
if(duplicates) {
    doSomething();
}