是否有任何方法可以将List<SomeObject>分离为SomeObject的几个单独的列表,使用项目索引作为每个分割的分隔符?

让我举个例子:

我有一个List<SomeObject>,我需要一个List<List<SomeObject>>或List<SomeObject>[],这样每个结果列表将包含一组原始列表的3个项目(依次)。

eg.:

原始列表:[a, g, e, w, p, s, q, f, x, y, i, m, c] 结果列表:[a、g e], [w、p, s], [q, f, x]、[y,我,m], [c]

我还需要结果列表的大小是这个函数的参数。


当前回答

几年前我写了一个集群扩展方法。工作得很好,是这里最快的实现。: P

/// <summary>
/// Clumps items into same size lots.
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="source">The source list of items.</param>
/// <param name="size">The maximum size of the clumps to make.</param>
/// <returns>A list of list of items, where each list of items is no bigger than the size given.</returns>
public static IEnumerable<IEnumerable<T>> Clump<T>(this IEnumerable<T> source, int size)
{
    if (source == null)
        throw new ArgumentNullException("source");
    if (size < 1)
        throw new ArgumentOutOfRangeException("size", "size must be greater than 0");

    return ClumpIterator<T>(source, size);
}

private static IEnumerable<IEnumerable<T>> ClumpIterator<T>(IEnumerable<T> source, int size)
{
    Debug.Assert(source != null, "source is null.");

    T[] items = new T[size];
    int count = 0;
    foreach (var item in source)
    {
        items[count] = item;
        count++;

        if (count == size)
        {
            yield return items;
            items = new T[size];
            count = 0;
        }
    }
    if (count > 0)
    {
        if (count == size)
            yield return items;
        else
        {
            T[] tempItems = new T[count];
            Array.Copy(items, tempItems, count);
            yield return tempItems;
        }
    }
}

其他回答

下面是我几个月前写的一个列表拆分例程:

public static List<List<T>> Chunk<T>(
    List<T> theList,
    int chunkSize
)
{
    List<List<T>> result = theList
        .Select((x, i) => new {
            data = x,
            indexgroup = i / chunkSize
        })
        .GroupBy(x => x.indexgroup, x => x.data)
        .Select(g => new List<T>(g))
        .ToList();

    return result;
}

下面这个解是我能想到的最紧凑的解是O(n)

public static IEnumerable<T[]> Chunk<T>(IEnumerable<T> source, int chunksize)
{
    var list = source as IList<T> ?? source.ToList();
    for (int start = 0; start < list.Count; start += chunksize)
    {
        T[] chunk = new T[Math.Min(chunksize, list.Count - start)];
        for (int i = 0; i < chunk.Length; i++)
            chunk[i] = list[start + i];

        yield return chunk;
    }
}

山姆·萨弗伦的做法是如此的做作。

public static IEnumerable<IEnumerable<T>> Batch<T>(this IEnumerable<T> source, int size)
{
    if (source == null) throw new ArgumentNullException(nameof(source));
    if (size <= 0) throw new ArgumentOutOfRangeException(nameof(size), "Size must be greater than zero.");

    return BatchImpl(source, size).TakeWhile(x => x.Any());
}

static IEnumerable<IEnumerable<T>> BatchImpl<T>(this IEnumerable<T> source, int size)
{
    var values = new List<T>();
    var group = 1;
    var disposed = false;
    var e = source.GetEnumerator();

    try
    {
        while (!disposed)
        {
            yield return GetBatch(e, values, group, size, () => { e.Dispose(); disposed = true; });
            group++;
        }
    }
    finally
    {
        if (!disposed)
            e.Dispose();
    }
}

static IEnumerable<T> GetBatch<T>(IEnumerator<T> e, List<T> values, int group, int size, Action dispose)
{
    var min = (group - 1) * size + 1;
    var max = group * size;
    var hasValue = false;

    while (values.Count < min && e.MoveNext())
    {
        values.Add(e.Current);
    }

    for (var i = min; i <= max; i++)
    {
        if (i <= values.Count)
        {
            hasValue = true;
        }
        else if (hasValue = e.MoveNext())
        {
            values.Add(e.Current);
        }
        else
        {
            dispose();
        }

        if (hasValue)
            yield return values[i - 1];
        else
            yield break;
    }
}

}

我刚刚写了这个,我认为它比其他提出的解决方案更优雅一点:

/// <summary>
/// Break a list of items into chunks of a specific size
/// </summary>
public static IEnumerable<IEnumerable<T>> Chunk<T>(this IEnumerable<T> source, int chunksize)
{
    while (source.Any())
    {
        yield return source.Take(chunksize);
        source = source.Skip(chunksize);
    }
}

我发现这个小片段做得很好。

public static IEnumerable<List<T>> Chunked<T>(this List<T> source, int chunkSize)
{
    var offset = 0;

    while (offset < source.Count)
    {
        yield return source.GetRange(offset, Math.Min(source.Count - offset, chunkSize));
        offset += chunkSize;
    }
}