java multi-dimensional array transposing

Question

I have a row-based multidimensional array:

/** [row][column]. */
public int[][] tiles;

I would like to transform this array to column-based

Answer 1

Here is my 50 cents: a utility method and test to transponse multidimensional array (for doubles in my case):

/**
  * Transponse bidimensional array.
  *
  * @param original Original table.
  * @return Transponsed.
  */
public static double[][] transponse(double[][] original) {

double[][] transponsed = new double
    [original[0].length]
    [original.length];

  for (int i = 0; i < original[0].length; i++) {

    for (int j = 0; j < original.length; j++) {

      transponsed[i][j] = original[j][i];
    }
  }

  return transponsed;
}


@Test
void aMatrix_OfTwoDimensions_ToBeTransponsed() {

  final double[][] original =
    new double[][] { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 } };

  double[][] transponsed = Analysis.transponse(original);

  assertThat(transponsed[1][2], is(equalTo(10)));
}

Answer 2

I saw that all of the answers create a new resultant matrix. This is simple: matrix[i][j] = matrix[j][i]; However, you can also do this in-place, in case of square matrix.

// Transpose, where m == n
for(int i = 0; i < m; i++) {
  for(int j = i+1; j < n; j++) {
    int temp = matrix[i][j];
    matrix[i][j] = matrix[j][i];
    matrix[j][i] = temp;
  }
}

This is better for larger matrices, where creating a new resultant matrix is wasteful in terms of memory. If its not square, you can create a new one with NxM dimensions and do the out of place method. Note: for in-place, take care of j = i+1; Its not 0.

Answer 3

import java.util.Arrays;
import java.util.Scanner;

public class Demo {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int rows = askArray("Enter number of rows :", input);//asking number of rows from user

        int columns = askArray("Enter number of columns :", input);//asking number of columns from user
        int[][] array = Array(rows, columns, input);
        DisplayArray(array, rows, columns);//displaying initial matrix
        System.out.println("Transpose array ");
        int[][] array2=TransposeArray(array, rows,columns );//calling Transpose array method
        for (int i = 0; i < array[0].length; i++) {
            System.out.println(Arrays.toString(array2[i]));
        }

    }
    //method to take number of rows and number of columns from the user
    public static int askArray(String s, Scanner in) {
        System.out.print(s);
        int value = in.nextInt();

        return value;
    }
    //feeding elements to the matrix
    public static int[][] Array(int x, int y, Scanner input) {
        int[][] array = new int[x][y];
        for (int j = 0; j < x; j++) {
            System.out.print("Enter row number " + (j + 1) + ":");
            for (int i = 0; i < y; i++) {
                array[j][i] = input.nextInt();
            }

        }

        return array;
    }
    //Method to display initial matrix
    public static void DisplayArray(int[][] arra, int x, int y) {

        for (int i = 0; i < x; i++) {
            System.out.println(Arrays.toString(arra[i]));
        }
    }
    //Method to transpose matrix
    public static int[][] TransposeArray(int[][] arr,int x,int y){
        int[][] Transpose_Array= new int [y][x];
        for (int i = 0; i < x; i++) {
            for (int j = 0; j <y ; j++) {
                Transpose_Array[j][i]=arr[i][j];
            }
        }

        return Transpose_Array;
    }
}

Answer 4

Use this function (replace String by int if necessary). It takes the matrix as string array and returns a new matrix which is the transpose. It checks for the edge case of an empty array, too. No prints.

private String[][] transposeTable(String[][] table) {
    // Transpose of empty table is empty table
    if (table.length < 1) {
        return table;
    }

    // Table isn't empty
    int nRows = table.length;
    int nCols = table[0].length;
    String[][] transpose = new String[nCols][nRows];

    // Do the transpose
    for (int i = 0; i < nRows; i++) {
        for (int j = 0; j < nCols; j++) {
            transpose[j][i] = table[i][j];
        }
    }

    return transpose;
}

Answer 5

If you want to go for in place transpose of a matrix(in which case row count = col count) you can so following in Java

public static void inPlaceTranspose(int [][] matrix){

    int rows = matrix.length;
    int cols = matrix[0].length;

    for(int i=0;i<rows;i++){
        for(int j=i+1;j<cols;j++){
            matrix[i][j] = matrix[i][j] + matrix[j][i];
            matrix[j][i] = matrix[i][j] - matrix[j][i];
            matrix[i][j] = matrix[i][j] - matrix[j][i];
        }
    }
}

Answer 6

public int[][] tiles, temp;

// Add values to tiles, wherever you end up doing that, then:
System.arraycopy(tiles, 0, temp, 0, tiles.length);

for(int row = 0; row < tiles.length; row++) // Loop over rows
  for(int col = 0; col < tiles[row].length; col++) // Loop over columns
    tiles[col][row] = temp[row][col]; // Rotate

That should do it for you.