How can I split a list vertically in n parts with LINQ

Question

I would like to split a list in parts, without knowing how much items I will have in that list. The question is different from those of you who wants to split a list into chunk of fixed size.

int[] a = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };

I would like the values to be splitted vertically.

Splitted in 2 :

-------------------
| item 1 | item 6 |
| item 2 | item 7 |
| item 3 | item 8 |
| item 4 | item 9 |
| item 5 |        |

Splitted in 3:

| item 1 | item 4 | item 7 |
| item 2 | item 5 | item 8 |
| item 3 | item 6 | item 9 |

Splitted in 4:

| item 1 | item 4 | item 6 | item 8 |
| item 2 | item 5 | item 7 | item 9 |
| item 3 |        |        |        |

I've found a few c# extensions that can do that but it doesn't distribute the value the way I want. Here's what I've found:

// this technic is an horizontal distribution
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
{
    int i = 0;
    var splits = from item in list
                    group item by i++ % parts into part
                    select part.AsEnumerable();
    return splits;
}

the result is this but my problem is that the value are distributed horizontally:

| item 1 | item 2 |
| item 3 | item 4 |
| item 5 | item 6 |
| item 7 | item 8 |
| item 9 |        |

or

| item 1 | item 2 | item 3 |
| item 4 | item 5 | item 6 |
| item 7 | item 8 | item 9 |

any idea how I can distribute my values vertically and have the possibility to choose the number of parts that i want?

In real life

For those of you who want to know in which situation I would like to split a list vertically, here's a screenshot of a section of my website:

Answer 1

With .Take() and .Skip() you can:

int[] a = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };

int splitIndex = 4; // or (a.Length / 2) to split in the middle.

var list1 = a.Take(splitIndex).ToArray(); // Returns a specified number of contiguous elements from the start of a sequence.
var list2 = a.Skip(splitIndex).ToArray(); // Bypasses a specified number of elements in a sequence and then returns the remaining elements.

You can use .ToList() instead of .ToArray() if you want a List<int>.

---

EDIT:

After you changed (clarified maybe) your question a bit, I guess this is what you needed:

public static class Extensions
{
    public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> source, int parts)
    {
        var list = new List<T>(source);
        int defaultSize = (int)((double)list.Count / (double)parts);
        int offset = list.Count % parts;
        int position = 0;

        for (int i = 0; i < parts; i++)
        {
            int size = defaultSize;
            if (i < offset)
                size++; // Just add one to the size (it's enough).

            yield return list.GetRange(position, size);

            // Set the new position after creating a part list, so that it always start with position zero on the first yield return above.
            position += size;
        }
    }
}

Using it:

int[] a = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
var lists = a.Split(2);

This would generate:

split in 2 : a.Split(2);

| item 1 | item 6 |
| item 2 | item 7 |
| item 3 | item 8 |
| item 4 | item 9 |
| item 5 |        |

split in 3 : a.Split(3);

| item 1 | item 4 | item 7 |
| item 2 | item 5 | item 8 |
| item 3 | item 6 | item 9 |

split in 4 : a.Split(4);

| item 1 | item 4 | item 6 | item 8 |
| item 2 | item 5 | item 7 | item 9 |
| item 3 |        |        |        |

Also, if you would have:

int[] b = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; // 10 items

and split in 4 : b.Split(4);

| item 1 | item 4 | item 7 | item 9 |
| item 2 | item 5 | item 8 | item 10|
| item 3 | item 6 |        |        |

Answer 2

This seems to do the trick quite nicely. It can probably be done even more efficient, but this was puzzling enough... It's much easier to do:

1|4|7|10
2|5|8
3|6|9

Than:

1|4|7|9
2|5|8|10
3|6|

I ignored the LINQ request first, since I had trouble wrapping my head around it. The solution using normal array manipulation could result in something like this:

    public static IEnumerable<IEnumerable<TListItem>> Split<TListItem>(this IEnumerable<TListItem> items, int parts)
        where TListItem : struct
    {
        var itemsArray = items.ToArray();
        int itemCount = itemsArray.Length;
        int itemsOnlastRow = itemCount - ((itemCount / parts) * parts);
        int numberOfRows = (int)(itemCount / (decimal)parts) + 1;

        for (int row = 0; row < numberOfRows; row++)
        {
            yield return SplitToRow(itemsArray, parts, itemsOnlastRow, numberOfRows, row);
        }
    }

    private static IEnumerable<TListItem> SplitToRow<TListItem>(TListItem[] items, int itemsOnFirstRows, int itemsOnlastRow,
                                                                int numberOfRows, int row)
    {
        for (int column = 0; column < itemsOnFirstRows; column++)
        {
            // Are we on the last row?
            if (row == numberOfRows - 1)
            {
                // Are we within the number of items on that row?
                if (column < itemsOnlastRow)
                {
                    yield return items[(column + 1) * numberOfRows -1];
                }
            }
            else
            {
                int firstblock = itemsOnlastRow * numberOfRows;
                int index;

                // are we in the first block?
                if (column < itemsOnlastRow)
                {
                    index = column*numberOfRows + ((row + 1)%numberOfRows) - 1;
                }
                else
                {
                    index = firstblock + (column - itemsOnlastRow)*(numberOfRows - 1) + ((row + 1)%numberOfRows) - 1;
                }

                yield return
                    items[index];
            }
        }
    }

The LINQ pseudo code would be:

//WARNING: DOES NOT WORK
public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
{
    int itemOnIndex = 0;
    var splits = from item in list
                 group item by MethodToDefineRow(itemOnIndex++) into row
                 select row.AsEnumerable();
    return splits;
}

But without knowing the number of items, there is no way to calculate the place where to put it.

So by doing a little pre-calculation, you can use LINQ to achieve the same thing as above this requires going through the IEnumerable twice, there doesn't seem to be a way around that. The trick is to calculate the row each value will be assigned to.

    //WARNING: Iterates the IEnumerable twice
    public static IEnumerable<IEnumerable<T>> Split<T>(this IEnumerable<T> list, int parts)
    {
        int itemOnIndex = 0;
        int itemCount = list.Count();
        int itemsOnlastRow = itemCount - ((itemCount / parts) * parts);
        int numberOfRows = (int)(itemCount / (decimal)parts) + 1;
        int firstblock = (numberOfRows*itemsOnlastRow);

        var splits = from item in list
                     group item by (itemOnIndex++ < firstblock) ? ((itemOnIndex -1) % numberOfRows) : ((itemOnIndex - 1 - firstblock) % (numberOfRows - 1)) into row
                     orderby row.Key
                     select row.AsEnumerable();
        return splits;
    }

Answer 3

Use .Take(#OfElements) to specify number of elements you want to take.

You can also use .First and .Last