How to change rows to columns in ArrayList<Integer[]>

Question

How to change rows to columns in ArrayList<Integer[]>? For instance:

ArrayList<Integer[]> arr = ArrayList<Integer[]>();
arr.add(new Integer[]{1,2,3});
arr.add(new Integer[]{4,5,6});

It should be:

[1]: 1 4
[2]: 2 5
[3]: 3 6

If it´s impossible with ArrayList, what are the other options to store 2D data and change rows to columns?

Solution 1

Anything wrong with int[][]? That would be the standard approach:

public static void main(String[] args) {
    int[][] table = new int[][] { { 1, 2, 3 }, { 4, 5, 6 } };

    // This code assumes all rows have same number of columns
    int[][] pivot = new int[table[0].length][];
    for (int row = 0; row < table[0].length; row++)
        pivot[row] = new int[table.length];

    for (int row = 0; row < table.length; row++)
        for (int col = 0; col < table[row].length; col++)
            pivot[col][row] = table[row][col];

    for (int row = 0; row < pivot.length; row++)
        System.out.println(Arrays.toString(pivot[row]));
}

Output:

[1, 4]
[2, 5]
[3, 6]

If you must use Collections, use as a start:

List<List<Integer>> table = new ArrayList<List<Integer>>();
table.add(Arrays.asList(1, 4));
table.add(Arrays.asList(2, 5));
table.add(Arrays.asList(3, 6));

Solution 2

This problem is called 'Matrix Transpose'. If you know the number of rows beforehand, you could just use the 2D matrix and just transpose it, as below:

Integer[][] matrix = new Integer[rows][cols];
//Let i = 2 (rows); j = 3 (cols)
matrix[0] = new Integer[]{1,2,3};
matrix[1] = new Integer[]{4,5,6};

Integer[][] transposedMatrix = new Integer[cols][rows];

for(int i=0;i<cols;i++) {
   for(int j=0;j<rows;j++) {
      transposedMatrix[i][j] = matrix[j][i];
   }
}

Even if you don't know the number of rows or columns beforehand, you can use other datastructures like ArrayList and then use the same logic as above.

Solution 3

You can use two loops - something like this should work:

ArrayList<Integer[]> res = ArrayList<Integer[]>();
int C = arr.get(0).length;
int R = arr.size();
for (int c = 0 ; c != C ; c++) {
    int[] row = new Integer[R];
    for (int r = 0 ; r != R ; r++) {
        row[r] = arr.get(R)[c];
    }
    res.add(row);
}

Solution 4

ArrayList<Integer[]> arr = new ArrayList<Integer[]>();
    arr.add(new Integer[]{1,2,3});
    arr.add(new Integer[]{4,5,6});

    ArrayList<Integer[]> res = new ArrayList<Integer[]>();

    int C = arr.get(0).length;
    int R = arr.size();
    for (int c = 0 ; c != C ; c++) {
        Integer[] row = new Integer[R];
        for (int r = 0 ; r < R ; r++) {
            row[r] = arr.get(r)[c];
        }
        res.add(row);
    }

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Comments

[4ndrew]

@Bohemian to preserve order you can use LinkedHashMap impl.