Copy 2D array from jagged 2D array in into another jagged 2D array

Question

I have jagged 2D array like so:

static void Main(string[] args)
    {
        int[][,] newArray = new int[2][,];

        int[][,] waypoints = new int[4][,]   
        {
            new int[,] {{6,3,4,5,6}},
            new int[,] {{1,3,4,5,6}},
            new int[,] {{1,4,3,2,1}},
            new int[,] {{6,3,4,5,6}}
        };

        int l = 0;
        int m = 0;

        for (int i = 0; i < waypoints.Length; i++)
        {
            for (int j = 0; j < waypoints[i].GetLength(0); j++)
            {
                for (int k = 0; k < waypoints[i].GetLength(1); k++)
                {
                    if (k == 1 || k == 3)
                    {
                        // waypoints[i][j,k].CopyTo(newArray[i][j,k]);
                    }
                    l++;
                    m++;
                }   
            }
        }
        Console.ReadKey();
    }

And I need to extract from each jagged array only [0,1] and [0,3] 2D array and store that in new jagged array - newArray. Please, could you help me, how to do that. Many thanks in advance.

Desired output should look like:

int[][,] newArray = new int[2][,];
{
     new int[,] {{3,5}},
     new int[,] {{3,5}},
     new int[,] {{4,2}},
     new int[,] {{3,5}}
 };

Try something like this Foreach (int i in newArray [0,1]{//do stuff} — Kerry White
– Kerry White, Commented Jul 25, 2016 at 9:33
It's not clear how you want to store the data in newArray. Can you describe what the resultant array should be? — Matthew Watson
– Matthew Watson, Commented Jul 25, 2016 at 9:38
Your arrays are somewhat unusual but it looks to me as if you are handling 3D arrays (arrays of 2D arrays), even though you don't need them because your 2D array elements all only have one element each (one row), i.e. they are 1-dimensional from a logical POV; as indicated by the double curly braces. — Peter - Reinstate Monica
– Peter - Reinstate Monica, Commented Jul 25, 2016 at 10:28
I know what you mean Peter but I need to keep this solution for another purposes. — SmithiM
– SmithiM, Commented Jul 25, 2016 at 10:30

Peter - Reinstate Monica · Accepted Answer · 2016-07-25 10:49:58Z

Try this. Note that I expanded the size of newArray to accomodate 4 2D-arrays.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication26
{
    class Program
    {
        static void Print3DArr(int[][,] arr)
        {
            foreach(var TwoDArr in arr)
            {
                for (int lineInd = 0; lineInd < TwoDArr.GetLength(0); lineInd++)
                {
                    for (int elemInd = 0; elemInd < TwoDArr.GetLength(1); elemInd++)
                    {
                        Console.Write(TwoDArr[lineInd, elemInd] + " ");
                    }
                    Console.WriteLine();
                }
                Console.WriteLine();
                Console.WriteLine();
            }
        }

        static void Main(string[] args)
        {
            int[][,] newArray = new int[4][,];

            int[][,] waypoints = new int[4][,]
            {
                new int[,] {{6,3,4,5,6}},
                new int[,] {{1,3,4,5,6}},
                new int[,] {{1,4,3,2,1}},
                new int[,] {{6,3,4,5,6}}
            };

            Print3DArr(waypoints);

            for (int TwoDArrIndex = 0; TwoDArrIndex < waypoints.Length; TwoDArrIndex++)
            {
                newArray[TwoDArrIndex] = new int[waypoints[TwoDArrIndex].GetLength(0), 2];

                for (int LineIn2DArr = 0; LineIn2DArr < waypoints[TwoDArrIndex].GetLength(0); LineIn2DArr++)
                {
                    newArray[TwoDArrIndex][LineIn2DArr, 0] = waypoints[TwoDArrIndex][LineIn2DArr, 1];
                    newArray[TwoDArrIndex][LineIn2DArr, 1] = waypoints[TwoDArrIndex][LineIn2DArr, 3];
                }
            }

            Print3DArr(newArray);
            Console.ReadKey();
        }
    }
}

This is exactly what I was looking for. Thank you very much for your time Peter!!

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