Question

我正在尝试将一个集合拆分成多个集合，同时保持我对集合的排序。我尝试使用以下扩展方法，但它错误地打破了它们。基本上，如果我要查看集合中的项目，与加入的分解集合相比，顺序应该相同。这是我正在使用的代码不起作用：

public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
        {
            int i = 0;
            var splits = from name in list
                         group name by i++ % parts into part
                         select part.AsEnumerable();
            return splits;
        }

int parts =子枚举数

Answer 1

我不得不利用这个来比较4个组中的对象列表... 它会使对象保持原始拥有的顺序。可以扩展为“List”之外的其他内容

/// <summary>
/// Partition a list of elements into a smaller group of elements
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="list"></param>
/// <param name="totalPartitions"></param>
/// <returns></returns>
public static List<T>[] Partition<T>(List<T> list, int totalPartitions)
{
    if (list == null)
        throw new ArgumentNullException("list");

    if (totalPartitions < 1)
        throw new ArgumentOutOfRangeException("totalPartitions");

    List<T>[] partitions = new List<T>[totalPartitions];

    int maxSize = (int)Math.Ceiling(list.Count / (double)totalPartitions);
    int k = 0;

    for (int i = 0; i < partitions.Length; i++)
    {
        partitions[i] = new List<T>();
        for (int j = k; j < k + maxSize; j++)
        {
            if (j >= list.Count)
                break;
            partitions[i].Add(list[j]);
        }
        k += maxSize;
    }

    return partitions;
}

Answer 2

对于这个相当古老的问题，

一种稍微干净的LINQ方法：

public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> source, int n)
{
    var count = source.Count();

    return source.Select((x, i) => new { value = x, index = i })
        .GroupBy(x => x.index / (int)Math.Ceiling(count / (double)n))
        .Select(x => x.Select(z => z.value));
}

Answer 3

Jon Skeet的MoreLINQ库可能会为您提供帮助：

https://code.google.com/p/morelinq/source/browse/MoreLinq/Batch.cs

var items = list.Batch(parts);  // gives you IEnumerable<IEnumerable<T>>
var items = list.Batch(parts, seq => seq.ToList()); // gives you IEnumerable<List<T>>
// etc...

另一个例子：

public class Program
{
    static void Main(string[] args)
    {
        var list = new List<int>();
        for (int i = 1; i < 10000; i++)
        {
            list.Add(i);
        }

        var batched = list.Batch(681);

        // will print 15. The 15th element has 465 items...
        Console.WriteLine(batched.Count().ToString());  
        Console.WriteLine(batched.ElementAt(14).Count().ToString());
        Console.WriteLine();
        Console.WriteLine("Press enter to exit.");
        Console.ReadLine();
    }
}

当我扫描批次的内容时，订单被保留。

Answer 4

    double partLength = list.Count() / (double)parts;

    int i = 0;
    var splits = from name in list
                 group name by Math.Floor((double)(i++ / partLength)) into part
                 select part;

Answer 5

据我所知，你希望在相同大小的几个部分上打破可枚举，而不会破坏元素的顺序。看起来唯一的选择是首先可以枚举输入的长度，因此您需要至少两次迭代才能通过可枚举。

Answer 6

    public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
    {
        int nGroups = (int)Math.Ceiling(list.Count() / (double)parts);

        var groups = Enumerable.Range(0, nGroups);

        return groups.Select(g => list.Skip(g * parts).Take(parts));
    }

Answer 7

这将完全按照要求执行。它也将迎合不均匀的分组，即10个组中的27个元素将产生3组3组和3组2个

        public static IEnumerable<IEnumerable<T>> SplitMaintainingOrder<T>(this IEnumerable<T> list, int parts)
    {
        if (list.Count() == 0) return Enumerable.Empty<IEnumerable<T>>();

        var toreturn = new List<IEnumerable<T>>();

        var splitFactor = Decimal.Divide((decimal)list.Count(), parts);
        int currentIndex = 0;

        for (var i = 0; i < parts; i++)
        {
            var toTake = Convert.ToInt32(
                i == 0 ? Math.Ceiling(splitFactor) : (
                    (Decimal.Compare(Decimal.Divide(Convert.ToDecimal(currentIndex), Convert.ToDecimal(i)), splitFactor) > 0) ? 
                        Math.Floor(splitFactor) : Math.Ceiling(splitFactor)));

            toreturn.Add(list.Skip(currentIndex).Take(toTake));
            currentIndex += toTake;
        }

        return toreturn;
    }

出于演示目的

        [TestMethod]
    public void splitlist()
    {
        var list = new decimal[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27 };

        var splitlists = list.SplitMaintainingOrder(10);

        int i = 1;

        foreach (var group in splitlists)
        {
            Console.WriteLine("Group {0} elements {1}", i++, String.Join(",", group));                 
        }
    }

以上演示产生

Test Name:  splitlist
Test Outcome:   Passed
Result StandardOutput:  
Group 1 elements 1,2,3
Group 2 elements 4,5
Group 3 elements 6,7,8
Group 4 elements 9,10,11
Group 5 elements 12,13
Group 6 elements 14,15,16
Group 7 elements 17,18,19
Group 8 elements 20,21
Group 9 elements 22,23,24
Group 10 elements 25,26,27

Answer 8

要将通用列表拆分为相等的块，请使用以下通用方法

＆＃13;

 private IEnumerable<IEnumerable<T>> SplitMaintainingOrder<T>(IEnumerable<T> list, int columnCount)
                {
                    var elementsCount = list.Count();
                    int rowCount = elementsCount / columnCount;
                    int noOfCells = elementsCount % columnCount;

                    int finalRowCount = rowCount;
                    if (noOfCells > 0)
                    {
                        finalRowCount++;
                    }

                    var toreturn = new List<IEnumerable<T>>();
                    var pushto = 0;
                    for (int j = 0; j < columnCount; j++)
                    {
                        var start = j;
                        int i = 0;
                        var end = i;
                        for (i = 0; i < finalRowCount; i++)
                        {
                            if ((i < rowCount) || ((i == rowCount) && (j < noOfCells)))
                            {
                                start = j;
                                end = i;
                            }
                        }
                        toreturn.Add(list.Skip(pushto).Take(end + 1));
                        pushto += end + 1;
                    }

                    return toreturn;

                }

＆＃13;

List<int> recordNumbers = new List<int>() { 1, 2, 3, 4, 5, 6,7,8,9,10,11};

var splitedItems = SplitMaintainingOrder<int>(recordNumbers , 4);

＆＃13;

Output will be:

List 1 : 1,2,3
List 2 : 4,5,6
List 3 : 7,8,9
List 4 : 10,11

＆＃13;

〜快乐的编码..

将C＃集合拆分为相等的部分，保持排序

8 个答案: