Question

假设我有以下整数数组：

int[] numbers = { 1, 6, 4, 10, 9, 12, 15, 17, 8, 3, 20, 21, 2, 23, 25, 27, 5, 67,33, 13, 8, 12, 41, 5 };

我怎样才能编写Linq查询，找到3个连续元素，例如大于10？此外，如果我可以指定我想要说出这些元素的第一，第二，第三等组，那将是很好的。

例如，Linq查询应该能够识别： 12,15,17为第一组连续元素 23,25,27为第二组 67,33,13为第三组

如果我指定我想要第二组3个连续元素，那么查询应该返回给我第二组。

感谢。

Answer 1

更新：虽然帕特里克在评论中指出，技术上不是“linq查询”，但此解决方案可重用，灵活且通用。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication32
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] numbers = { 1, 6, 4, 10, 9, 12, 15, 17, 8, 3, 20, 21, 2, 23, 25, 27, 5, 67,33, 13, 8, 12, 41, 5 };

            var consecutiveGroups = numbers.FindConsecutiveGroups((x) => x > 10, 3);

            foreach (var group in consecutiveGroups)
            {
                Console.WriteLine(String.Join(",", group));
            }
        }        
    }

    public static class Extensions
    {
        public static IEnumerable<IEnumerable<T>> FindConsecutiveGroups<T>(this IEnumerable<T> sequence, Predicate<T> predicate, int count)
        {
            IEnumerable<T> current = sequence;

            while (current.Count() > count)
            {
                IEnumerable<T> window = current.Take(count);

                if (window.Where(x => predicate(x)).Count() >= count)
                    yield return window;

                current = current.Skip(1);
            }
        }
    }
}

输出：

12,15,17
23,25,27
67,33,13

要获得第二组，请更改：

var consecutiveGroups = numbers.FindConsecutiveGroups((x) => x > 10, 3);

要：

var consecutiveGroups = numbers.FindConsecutiveGroups((x) => x > 10, 3).Skip(1).Take(1);

更新2 在我们的生产用途中对此进行调整后，随着数字数组中项目数量的增加，以下实现速度会快得多。

public static IEnumerable<IEnumerable<T>> FindConsecutiveGroups<T>(this IEnumerable<T> sequence, Predicate<T> predicate, int sequenceSize)
{
    IEnumerable<T> window = Enumerable.Empty<T>();

    int count = 0;

    foreach (var item in sequence)
    {
        if (predicate(item))
        {
            window = window.Concat(Enumerable.Repeat(item, 1));
            count++;

            if (count == sequenceSize)
            {
                yield return window;
                window = window.Skip(1);
                count--;
            }
        }
        else
        {
            count = 0;
            window = Enumerable.Empty<T>();
        }
    }
}

Answer 2

int[] numbers = { 1, 6, 4, 10, 9, 12, 15, 17, 8, 3, 20, 21, 2, 23, 25, 27, 5, 67, 33, 13, 8, 12, 41, 5 };

var numbersQuery = numbers.Select((x, index) => new { Index = index, Value = x});

var query = from n in numbersQuery
            from n2 in numbersQuery.Where(x => n.Index == x.Index - 1).DefaultIfEmpty()
            from n3 in numbersQuery.Where(x => n.Index == x.Index - 2).DefaultIfEmpty()
            where n.Value > 10
            where n2 != null && n2.Value > 10
            where n3 != null && n3.Value > 10
            select new
            {
              Value1 = n.Value,
              Value2 = n2.Value,
              Value3 = n3.Value
            };

为了指定哪个组，您可以调用Skip方法

query.Skip(1)

Answer 3

为什么不尝试这种扩展方法？

public static IEnumerable<IEnumerable<T>> Consecutives<T>(this IEnumerable<T> numbers, int ranges, Func<T, bool> predicate)
{
    IEnumerable<T> ordered = numbers.OrderBy(a => a).Where(predicate);
    decimal n = Decimal.Divide(ordered.Count(), ranges);
    decimal max = Math.Ceiling(n); // or Math.Floor(n) if you want
    return from i in Enumerable.Range(0, (int)max)
           select ordered.Skip(i * ranges).Take(ranges);
}

唯一要改进的可能是调用Count方法，因为导致numbers的枚举（因此查询失去了它的懒惰）。

无论如何，我确信这符合您的linqness要求。

编辑：或者这是 less words 版本（它没有使用Count方法）：

public static IEnumerable<IEnumerable<T>> Consecutives<T>(this IEnumerable<T> numbers, int ranges, Func<T, bool> predicate)
{
    var ordered = numbers.OrderBy(a => a);
    return ordered.Where(predicate)
                  .Select((element, i) => ordered.Skip(i * ranges).Take(ranges))
                  .TakeWhile(Enumerable.Any);
}

Answer 4

我必须为双打列表执行此操作。有上限和下限。这也不是真正的Linq解决方案，它只是我用脚本语言编写的一种实用方法，它只实现了C＃的一个子集。

var sequence =
 [0.25,0.5,0.5,0.5,0.7,0.8,0.7,0.9,0.5,0.5,0.8,0.8,0.5,0.5,0.65,0.65,0.65,0.65,0.65,0.65,0.65];
double lowerLimit = 0.1;
double upperLimit = 0.6;
int minWindowLength = 3;

// return type is a list of lists
var windows = [[0.0]];
windows.Clear();

int consec = 0;
int index = 0;

while (index < sequence.Count){

        // store segments here
        var window = new System.Collections.Generic.List<double>();

        while ((index < sequence.Count) && (sequence[index] > upperLimit || sequence[index] < lowerLimit)) {        
            window.Add(sequence[index]);
            consec = consec + 1;
            index = index +1;
        }

        if (consec > minWindowLength) {
            windows.Add(window);
        }

        window = new System.Collections.Generic.List<double>();
        consec = 0;

        index = index+1;
}

return windows;

使用Linq在列表中查找连续项

4 个答案: