我有一个通用的方法，将采取任何类型包括浮动，它已经过单元测试，希望它有帮助:

    /// <summary>
    /// Breaks the list into groups with each group containing no more than the specified group size
    /// </summary>
    /// <typeparam name="T"></typeparam>
    /// <param name="values">The values.</param>
    /// <param name="groupSize">Size of the group.</param>
    /// <returns></returns>
    public static List<List<T>> SplitList<T>(IEnumerable<T> values, int groupSize, int? maxCount = null)
    {
        List<List<T>> result = new List<List<T>>();
        // Quick and special scenario
        if (values.Count() <= groupSize)
        {
            result.Add(values.ToList());
        }
        else
        {
            List<T> valueList = values.ToList();
            int startIndex = 0;
            int count = valueList.Count;
            int elementCount = 0;

            while (startIndex < count && (!maxCount.HasValue || (maxCount.HasValue && startIndex < maxCount)))
            {
                elementCount = (startIndex + groupSize > count) ? count - startIndex : groupSize;
                result.Add(valueList.GetRange(startIndex, elementCount));
                startIndex += elementCount;
            }
        }


        return result;
    }

我也遇到过同样的需求，我使用了Linq的Skip()和Take()方法的组合。我用我取的数乘以到目前为止的迭代次数，这就得到了要跳过的项目数，然后我取下一组。

        var categories = Properties.Settings.Default.MovementStatsCategories;
        var items = summariesWithinYear
            .Select(s =>  s.sku).Distinct().ToList();

        //need to run by chunks of 10,000
        var count = items.Count;
        var counter = 0;
        var numToTake = 10000;

        while (count > 0)
        {
            var itemsChunk = items.Skip(numToTake * counter).Take(numToTake).ToList();
            counter += 1;

            MovementHistoryUtilities.RecordMovementHistoryStatsBulk(itemsChunk, categories, nLogger);

            count -= numToTake;
        }

public static IEnumerable<IEnumerable<T>> Batch<T>(this IEnumerable<T> items, int maxItems)
{
    return items.Select((item, index) => new { item, index })
                .GroupBy(x => x.index / maxItems)
                .Select(g => g.Select(x => x.item));
}

如果你想用条件而不是固定的数字来分割它:

///<summary>
/// splits a list based on a condition (similar to the split function for strings)
///</summary>
public static IEnumerable<List<T>> Split<T>(this IEnumerable<T> src, Func<T, bool> pred)
{
    var list = new List<T>();
    foreach(T item in src)
    {   
        if(pred(item))
        {
            if(list != null && list.Count > 0)
                yield return list;
                
            list = new List<T>();
        }
        else
        {
            list.Add(item);
        }
    }
}

在。net 6中，你可以使用source.Chunk(chunkSize)

一个基于Serj-Tm的公认答案的更通用的版本。

    public static IEnumerable<IEnumerable<T>> Split<T>(IEnumerable<T> source, int size = 30)
    {
        var count = source.Count();
        for (int i = 0; i < count; i += size)
        {
            yield return source
                .Skip(Math.Min(size, count - i))
                .Take(size);
        }
    }

库MoreLinq有一个方法叫做Batch

List<int> ids = new List<int>() { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 }; // 10 elements
int counter = 1;
foreach(var batch in ids.Batch(2))
{
    foreach(var eachId in batch)
    {
        Console.WriteLine("Batch: {0}, Id: {1}", counter, eachId);
    }
    counter++;
}

结果是

Batch: 1, Id: 1
Batch: 1, Id: 2
Batch: 2, Id: 3
Batch: 2, Id: 4
Batch: 3, Id: 5
Batch: 3, Id: 6
Batch: 4, Id: 7
Batch: 4, Id: 8
Batch: 5, Id: 9
Batch: 5, Id: 0

id被分成5个包含2个元素的块。