LINQ Partition List into Lists of 8 members [duplicate]

Question

This question already has an answer here:

• Split List into Sublists with LINQ 27 answers

How would one take a List (using LINQ) and break it into a List of Lists partitioning the original list on every 8th entry?

I imagine something like this would involve Skip and/or Take, but I'm still pretty new to LINQ.

Edit: Using C# / .Net 3.5

Edit2: This question is phrased differently than the other "duplicate" question. Although the problems are similar, the answers in this question are superior: Both the "accepted" answer is very solid (with the yield statement) as well as Jon Skeet's suggestion to use MoreLinq (which is not recommended in the "other" question.) Sometimes duplicates are good in that they force a re-examination of a problem.

Answer 1

Use the following extension method to break the input into subsets

public static class IEnumerableExtensions
{
    public static IEnumerable<List<T>> InSetsOf<T>(this IEnumerable<T> source, int max)
    {
        List<T> 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;
        }
    }
}

Answer 2

We have just such a method in MoreLINQ as the Batch method:

// As IEnumerable<IEnumerable<T>>
var items = list.Batch(8);

or

// As IEnumerable<List<T>>
var items = list.Batch(8, seq => seq.ToList());

Answer 3

You're better off using a library like MoreLinq, but if you really had to do this using "plain LINQ", you can use GroupBy:

var sequence = new[] {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};

var result = sequence.Select((x, i) => new {Group = i/8, Value = x})
                     .GroupBy(item => item.Group, g => g.Value)
                     .Select(g => g.Where(x => true));

// result is: { {1,2,3,4,5,6,7,8}, {9,10,11,12,13,14,15,16} }

Basically, we use the version of Select() that provides an index for the value being consumed, we divide the index by 8 to identify which group each value belongs to. Then we group the sequence by this grouping key. The last Select just reduces the IGrouping<> down to an IEnumerable<IEnumerable<T>> (and isn't strictly necessary since IGrouping is an IEnumerable).

It's easy enough to turn this into a reusable method by factoring our the constant 8 in the example, and replacing it with a specified parameter. It's not necessarily the most elegant solution, and it is not longer a lazy, streaming solution ... but it does work.

You could also write your own extension method using iterator blocks (yield return) which could give you better performance and use less memory than GroupBy. This is what the Batch() method of MoreLinq does IIRC.

Answer 4

It's not at all what the original Linq designers had in mind, but check out this misuse of GroupBy:

public static IEnumerable<IEnumerable<T>> BatchBy<T>(this IEnumerable<T> items, int batchSize)
{
    var count = 0;
    return items.GroupBy(x => (count++ / batchSize)).ToList();
}

[TestMethod]
public void BatchBy_breaks_a_list_into_chunks()
{
    var values = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    var batches = values.BatchBy(3);
    batches.Count().ShouldEqual(4);
    batches.First().Count().ShouldEqual(3);
    batches.Last().Count().ShouldEqual(1);
}

I think it wins the "golf" prize for this question. The ToList is very important since you want to make sure the grouping has actually been performed before you try doing anything with the output. If you remove the ToList, you will get some weird side effects.

Answer 5

Take won't be very efficient, because it doesn't remove the entries taken.

why not use a simple loop:

public IEnumerable<IList<T>> Partition<T>(this/* <-- see extension methods*/ IEnumerable<T> src,int num)  
{  
    IEnumerator<T> enu=src.getEnumerator();  
    while(true)  
    {  
        List<T> result=new List<T>(num);  
        for(int i=0;i<num;i++)  
        {  
            if(!enu.MoveNext())  
            {  
                if(i>0)yield return result;  
                yield break;  
            }  
            result.Add(enu.Current);  
        }  
        yield return result;  
    }  
}

Answer 6

from b in Enumerable.Range(0,8) select items.Where((x,i) => (i % 8) == b);

Answer 7

The simplest solution is given by Mel:

public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> items, 
                                                       int partitionSize)
{
    int i = 0;
    return items.GroupBy(x => i++ / partitionSize).ToArray();
}

Concise but slower. The above method splits an IEnumerable into chunks of desired fixed size with total number of chunks being unimportant. To split an IEnumerable into N number of chunks of equal sizes or close to equal sizes, you could do:

public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> items, 
                                                   int numOfParts)
{
    int i = 0;
    return items.GroupBy(x => i++ % numOfParts);
}

To speed up things, a straightforward approach would do:

public static IEnumerable<IEnumerable<T>> Partition<T>(this IEnumerable<T> items, 
                                                       int partitionSize)
{
    if (partitionSize <= 0)
        throw new ArgumentOutOfRangeException("partitionSize");

    int innerListCounter = 0;
    int numberOfPackets = 0;
    foreach (var item in items)
    {
        innerListCounter++;
        if (innerListCounter == partitionSize)
        {
            yield return items.Skip(numberOfPackets * partitionSize).Take(partitionSize);
            innerListCounter = 0;
            numberOfPackets++;
        }
    }

    if (innerListCounter > 0)
        yield return items.Skip(numberOfPackets * partitionSize);
}

This is faster than anything currently on planet now :) The equivalent methods for a Split operation here