完全懒惰的，没有计数或复制的:

public static class EnumerableExtensions
{

  public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> source, int len)
  {
     if (len == 0)
        throw new ArgumentNullException();

     var enumer = source.GetEnumerator();
     while (enumer.MoveNext())
     {
        yield return Take(enumer.Current, enumer, len);
     }
  }

  private static IEnumerable<T> Take<T>(T head, IEnumerator<T> tail, int len)
  {
     while (true)
     {
        yield return head;
        if (--len == 0)
           break;
        if (tail.MoveNext())
           head = tail.Current;
        else
           break;
     }
  }
}

我认为下面的建议是最快的。为了能够使用数组，我牺牲了源Enumerable的惰性。复制和提前知道每个子列表的长度。

public static IEnumerable<T[]> Chunk<T>(this IEnumerable<T> items, int size)
{
    T[] array = items as T[] ?? items.ToArray();
    for (int i = 0; i < array.Length; i+=size)
    {
        T[] chunk = new T[Math.Min(size, array.Length - i)];
        Array.Copy(array, i, chunk, 0, chunk.Length);
        yield return chunk;
    }
}

好吧，以下是我的看法:

完全懒惰:工作在无限枚举上 没有中间复制/缓冲 O(n)执行时间 当内部序列仅被部分消耗时也适用

public static IEnumerable<IEnumerable<T>> Chunks<T>(this IEnumerable<T> enumerable, int chunkSize) { if (chunkSize < 1) throw new ArgumentException("chunkSize must be positive"); using (var e = enumerable.GetEnumerator()) while (e.MoveNext()) { var remaining = chunkSize; // elements remaining in the current chunk var innerMoveNext = new Func<bool>(() => --remaining > 0 && e.MoveNext()); yield return e.GetChunk(innerMoveNext); while (innerMoveNext()) {/* discard elements skipped by inner iterator */} } } private static IEnumerable<T> GetChunk<T>(this IEnumerator<T> e, Func<bool> innerMoveNext) { do yield return e.Current; while (innerMoveNext()); } Example Usage var src = new [] {1, 2, 3, 4, 5, 6}; var c3 = src.Chunks(3); // {{1, 2, 3}, {4, 5, 6}}; var c4 = src.Chunks(4); // {{1, 2, 3, 4}, {5, 6}}; var sum = c3.Select(c => c.Sum()); // {6, 15} var count = c3.Count(); // 2 var take2 = c3.Select(c => c.Take(2)); // {{1, 2}, {4, 5}} Explanations The code works by nesting two yield based iterators. The outer iterator must keep track of how many elements have been effectively consumed by the inner (chunk) iterator. This is done by closing over remaining with innerMoveNext(). Unconsumed elements of a chunk are discarded before the next chunk is yielded by the outer iterator. This is necessary because otherwise you get inconsistent results, when the inner enumerables are not (completely) consumed (e.g. c3.Count() would return 6). Note: The answer has been updated to address the shortcomings pointed out by @aolszowka.

没有办法在一个解决方案中结合所有理想的特性，如完全懒惰、无复制、完全通用性和安全性。最根本的原因是不能保证在访问块之前输入不发生变化。 假设我们有一个如下签名的函数:

public static IEnumerable<IEnumerable<T>> Chunk<T>(this IEnumerable<T> source, int chunkSize)
{
    // Some implementation
}

那么下面的使用方式就有问题了:

var myList = new List<int>()
{
    1,2,3,4
};
var myChunks = myList.Chunk(2);
myList.RemoveAt(0);
var firstChunk = myChunks.First();    
Console.WriteLine("First chunk:" + String.Join(',', firstChunk));
myList.RemoveAt(0);
var secondChunk = myChunks.Skip(1).First();
Console.WriteLine("Second chunk:" + String.Join(',', secondChunk));
// What outputs do we see for first and second chunk? Probably not what you would expect...

根据具体的实现，代码将失败并产生运行时错误或产生不直观的结果。

所以，至少有一个属性需要减弱。如果你想要一个无懈无击的惰性解决方案，你需要将输入类型限制为不可变类型，即使这样也不能直接覆盖所有用例。但是，如果您可以控制使用，您仍然可以选择最通用的解决方案，只要您确保以一种有效的方式使用它。否则，你可能会放弃懒惰，接受一定数量的复制。

最后，这完全取决于您的用例和需求，哪种解决方案最适合您。

我们发现David B的解决方案效果最好。但我们把它改成了一个更普遍的解决方案:

list.GroupBy(item => item.SomeProperty) 
   .Select(group => new List<T>(group)) 
   .ToArray();

旧代码，但这是我一直在使用的:

    public static IEnumerable<List<T>> InSetsOf<T>(this IEnumerable<T> source, int max)
    {
        var toReturn = new List<T>(max);
        foreach (var item in source)
        {
            toReturn.Add(item);
            if (toReturn.Count == max)
            {
                yield return toReturn;
                toReturn = new List<T>(max);
            }
        }
        if (toReturn.Any())
        {
            yield return toReturn;
        }
    }