Question

我正在尝试使用LINQ来返回最多次出现的元素及其出现的次数。

例如：我有一个字符串数组：

string[] words = { "cherry", "apple", "blueberry", "cherry", "cherry", "blueberry" };

//...
Some LINQ statement here
//...

在此数组中，查询将返回cherry作为最大发生元素，并3作为发生的次数。如果有必要，我也愿意将它们分成两个查询（ ie，第一个查询获取cherry，第二个返回3的数量。 / p>

Answer 1

到目前为止提出的解决方案是O(n log n)。这是一个O(n)解决方案：

var max = words.GroupBy(w => w)
               .Select(g => new { Word = g.Key, Count = g.Count() })
               .MaxBy(g => g.Count);
Console.WriteLine(
    "The most frequent word is {0}, and its frequency is {1}.",
    max.Word,
    max.Count
);

这需要MaxBy的定义。这是一个：

public static TSource MaxBy<TSource>(
    this IEnumerable<TSource> source,
    Func<TSource, IComparable> projectionToComparable
) {
    using (var e = source.GetEnumerator()) {
        if (!e.MoveNext()) {
            throw new InvalidOperationException("Sequence is empty.");
        }
        TSource max = e.Current;
        IComparable maxProjection = projectionToComparable(e.Current);
        while (e.MoveNext()) {
            IComparable currentProjection = projectionToComparable(e.Current);
            if (currentProjection.CompareTo(maxProjection) > 0) {
                max = e.Current;
                maxProjection = currentProjection;
            }
        }
        return max;                
    }
}

Answer 2

var topWordGroup = words.GroupBy(word => word).OrderByDescending(group => group.Count()).FirstOrDefault();
// topWordGroup might be a null!
string topWord = topWordGroup.Key;
int topWordCount = topWordGroup.Count;

如果我们不喜欢O(N log N)：

var topWordGroup = words.GroupBy(word => word).Aggregate((current, acc) => current.Count() < acc.Count() ? acc : current);

Answer 3

首先想到的是（意味着可能有更有效的方式）

var item = words.GroupBy(x => x).OrderByDescending(x => x.Count()).First()
//item.Key is "cherry", item.Count() is 3

编辑：忘了我想要名字和计数

Answer 4

string[] words = { "cherry", "apple", "blueberry", "cherry", "cherry", "blueberry" };

var topWordAndCount=words
    .GroupBy(w=>w)
    .OrderByDescending(g=>g.Count())
    .Select(g=>new {Word=g.Key,Count=g.Count()})
    .FirstOrDefault();

//if(topWordAndCount!=null)
//{
//    topWordAndCount.Word
//    topWordAndCount.Count

Answer 5

试试这个：

Converting SQL containing top, count, group and order to LINQ (2 Entities)

Answer 6

string[] words = { "cherry", "apple", "blueberry", "cherry", "cherry", "blueberry" };

var r = words.GroupBy (x => x)
             .OrderByDescending (g => g.Count ())
             .FirstOrDefault ();
Console.WriteLine (String.Format ("The element {0} occurs {1} times.", r.Key, r.Count ()));

Answer 7

更简单的O（n）解决方案：

var groups = words.GroupBy(x => x);
var max = groups.Max(x => x.Count());
var top = groups.First(y => y.Count() == max).Key;

Answer 8

这是一行（！）中非常快速的O（n）解决方案：

s.GroupBy(x => x).Aggregate((IGrouping<string,string>)null, (x, y) =>  (x != null && y != null && x.Count() >= y.Count()) || y == null ? x : y, x => x);

或者这个：

s.GroupBy(x => x).Select(x => new { Key = x.Key, Count = x.Count() }).Aggregate(new { Key = "", Count = 0 }, (x, y) => x.Count >= y.Count ? x : y, x => x);

C＃中简单的LINQ问题

8 个答案: