GroupBy multidimensional of Int Array

Question

I have this multidimensional array:

int[][] x;

So, for each position I have:

x[0] = [10,10,10,10,10]
x[1] = [10,10,10,10,10]
x[2] = [10,10,10,10,10]
x[3] = [5,5,5]
x[4] = [5,5,5]
x[5] = [10,5,5,5]
x[6] = [10,5,5,5]
x[7] = [5,5,5]

I want group like this:

z[0] = 3 x [10,10,10,10,10]
z[1] = 3 x [5,5,5]
z[2] = 2 x [10,5,5,5]

Or if is possible:

y[0] = 3 | 5x10
y[1] = 3 | 3x5 
y[2] = 2 | 1x10, 3x5

So you basically want to make a unique filter? – Willem Van Onsem Mar 30 '15 at 12:27 — Willem Van Onsem, Mar 30 '15 at 12:27
Does the order within the arrays matter? – D Stanley Mar 30 '15 at 13:07 — D Stanley, Mar 30 '15 at 13:07

score 5 · Accepted Answer · edited May 23 '17 at 11:50

Well C#'s LINQ has a .Distinct() extension method that returns the list of distinct elements. The problem is that two arrays are not seen as equal if they have the same elements (in the same order). So one first needs to design an IEqualityComparer<T>:

public class ArrayEqual<T> : IEqualityComparer<T[]> {

    public bool Equals(T[] x, T[] y) {
        return Enumerable.SequenceEqual(x,y);
    }

    public int GetHashCode (T[] x) {
        int hash = 0;
        foreach(T t in x) {
            hash = 3*hash ^ t.GetHashCode();
        }
        return hash;
    }

}

And then call it with:

int[][] z = x.Distinct(new ArrayEqual<int>()).ToArray();

If you want to generate tuples of the count with an array, you can do this with:

Tuple<int,int[]>[] z = x.GroupBy(x => x,new ArrayEqual<int>()).Select(x => new Tuple<int,int[]>(x.Count(),x.Key)).ToArray();

The Enumerable.SequenceEqual is borrowed from here. The GetHashCode method I've invented myself, and can probably be improved. After all this is only a short demo.

Why did you shift to `Distinct` when the OP wants _groups_? – D Stanley Mar 30 '15 at 13:07 — D Stanley, Mar 30 '15 at 13:07
@DStanley: modified, better? – Willem Van Onsem Mar 30 '15 at 13:33 — Willem Van Onsem, Mar 30 '15 at 13:33

GroupBy multidimensional of Int Array

1 Answers1