Split List into Sublists with LINQ

Question

Is there any way I can separate a List<SomeObject> into several separate lists of SomeObject, using the item index as the delimiter of each split?

Let me exemplify:

I have a List<SomeObject> and I need a List<List<SomeObject>> or List<SomeObject>[], so that each of these resulting lists will contain a group of 3 items of the original list (sequentially).

eg.:

• Original List: [a, g, e, w, p, s, q, f, x, y, i, m, c]

• Resulting lists: [a, g, e], [w, p, s], [q, f, x], [y, i, m], [c]

I'd also need the resulting lists size to be a parameter of this function.

Answer 1

This question is a bit old, but I just wrote this, and I think it's a little more elegant than the other proposed solutions:

/// <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);
    }
}

Answer 2

This following solution is the most compact I could come up with that is 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;
    }
}

Answer 3

We found David B's solution worked the best. But we adapted it to a more general solution:

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

Answer 4

I think the following suggestion would be the fastest. I am sacrificing the lazyness of the source Enumerable for the ability to use Array.Copy and knowing ahead of the time the length of each of my sublists.

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;
    }
}

Answer 5

To insert my two cents...

By using the list type for the source to be chunked, I found another very compact solution:

public static IEnumerable<IEnumerable<TSource>> Chunk<TSource>(this IEnumerable<TSource> source, int chunkSize)
{
    // copy the source into a list
    var chunkList = source.ToList();

    // return chunks of 'chunkSize' items
    while (chunkList.Count > chunkSize)
    {
        yield return chunkList.GetRange(0, chunkSize);
        chunkList.RemoveRange(0, chunkSize);
    }

    // return the rest
    yield return chunkList;
}

Answer 6

You could use a number of queries that use Take and Skip, but that would add too many iterations on the original list, I believe.

Rather, I think you should create an iterator of your own, like so:

public static IEnumerable<IEnumerable<T>> GetEnumerableOfEnumerables<T>(
  IEnumerable<T> enumerable, int groupSize)
{
   // The list to return.
   List<T> list = new List<T>(groupSize);

   // Cycle through all of the items.
   foreach (T item in enumerable)
   {
     // Add the item.
     list.Add(item);

     // If the list has the number of elements, return that.
     if (list.Count == groupSize)
     {
       // Return the list.
       yield return list;

       // Set the list to a new list.
       list = new List<T>(groupSize);
     }
   }

   // Return the remainder if there is any,
   if (list.Count != 0)
   {
     // Return the list.
     yield return list;
   }
}

You can then call this and it is LINQ enabled so you can perform other operations on the resulting sequences.

---

In light of Sam's answer, I felt there was an easier way to do this without:

• Iterating through the list again (which I didn't do originally)
• Materializing the items in groups before releasing the chunk (for large chunks of items, there would be memory issues)
• All of the code that Sam posted

That said, here's another pass, which I've codified in an extension method to IEnumerable<T> called Chunk:

public static IEnumerable<IEnumerable<T>> Chunk<T>(this IEnumerable<T> source, 
    int chunkSize)
{
    // Validate parameters.
    if (source == null) throw new ArgumentNullException("source");
    if (chunkSize <= 0) throw new ArgumentOutOfRangeException("chunkSize",
        "The chunkSize parameter must be a positive value.");

    // Call the internal implementation.
    return source.ChunkInternal(chunkSize);
}

Nothing surprising up there, just basic error checking.

Moving on to ChunkInternal:

private static IEnumerable<IEnumerable<T>> ChunkInternal<T>(
    this IEnumerable<T> source, int chunkSize)
{
    // Validate parameters.
    Debug.Assert(source != null);
    Debug.Assert(chunkSize > 0);

    // Get the enumerator.  Dispose of when done.
    using (IEnumerator<T> enumerator = source.GetEnumerator())
    do
    {
        // Move to the next element.  If there's nothing left
        // then get out.
        if (!enumerator.MoveNext()) yield break;

        // Return the chunked sequence.
        yield return ChunkSequence(enumerator, chunkSize);
    } while (true);
}

Basically, it gets the IEnumerator<T> and manually iterates through each item. It checks to see if there any items currently to be enumerated. After each chunk is enumerated through, if there aren't any items left, it breaks out.

Once it detects there are items in the sequence, it delegates the responsibility for the inner IEnumerable<T> implementation to ChunkSequence:

private static IEnumerable<T> ChunkSequence<T>(IEnumerator<T> enumerator, 
    int chunkSize)
{
    // Validate parameters.
    Debug.Assert(enumerator != null);
    Debug.Assert(chunkSize > 0);

    // The count.
    int count = 0;

    // There is at least one item.  Yield and then continue.
    do
    {
        // Yield the item.
        yield return enumerator.Current;
    } while (++count < chunkSize && enumerator.MoveNext());
}

Since MoveNext was already called on the IEnumerator<T> passed to ChunkSequence, it yields the item returned by Current and then increments the count, making sure never to return more than chunkSize items and moving to the next item in the sequence after every iteration (but short-circuited if the number of items yielded exceeds the chunk size).

If there are no items left, then the InternalChunk method will make another pass in the outer loop, but when MoveNext is called a second time, it will still return false, as per the documentation (emphasis mine):

If MoveNext passes the end of the collection, the enumerator is positioned after the last element in the collection and MoveNext returns false. When the enumerator is at this position, subsequent calls to MoveNext also return false until Reset is called.

At this point, the loop will break, and the sequence of sequences will terminate.

This is a simple test:

static void Main()
{
    string s = "agewpsqfxyimc";

    int count = 0;

    // Group by three.
    foreach (IEnumerable<char> g in s.Chunk(3))
    {
        // Print out the group.
        Console.Write("Group: {0} - ", ++count);

        // Print the items.
        foreach (char c in g)
        {
            // Print the item.
            Console.Write(c + ", ");
        }

        // Finish the line.
        Console.WriteLine();
    }
}

Output:

Group: 1 - a, g, e,
Group: 2 - w, p, s,
Group: 3 - q, f, x,
Group: 4 - y, i, m,
Group: 5 - c,

An important note, this will not work if you don't drain the entire child sequence or break at any point in the parent sequence. This is an important caveat, but if your use case is that you will consume every element of the sequence of sequences, then this will work for you.

Additionally, it will do strange things if you play with the order, just as Sam's did at one point.