我有一个基于行的多维数组：
/** [row][column]. */
public int[][] tiles;
我想将此数组转换为基于列的数组，如下所示：
/** [column][row]. */
public int[][] tiles;
......但我真的不知道从哪里开始
答案 0 :(得分：12)
我看到所有答案都会创建一个新的结果矩阵。这很简单：matrix[i][j] = matrix[j][i];但是，如果是方形矩阵，您也可以就地执行此操作。
// 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;
}
}
这对于较大的矩阵更好，在这种矩阵中，创建新的结果矩阵在内存方面是浪费的。如果它不是正方形，您可以创建一个具有NxM维度的新方法并执行不合适的方法。注意：对于就地，请注意j = i + 1;它不是0。
答案 1 :(得分：10)
试试这个：
@Test
public void transpose() {
final int[][] original = new int[][] { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 } };
for (int i = 0; i < original.length; i++) {
for (int j = 0; j < original[i].length; j++) {
System.out.print(original[i][j] + " ");
}
System.out.print("\n");
}
System.out.print("\n\n matrix transpose:\n");
// transpose
if (original.length > 0) {
for (int i = 0; i < original[0].length; i++) {
for (int j = 0; j < original.length; j++) {
System.out.print(original[j][i] + " ");
}
System.out.print("\n");
}
}
}
输出：
1 2 3 4
5 6 7 8
9 10 11 12
matrix transpose:
1 5 9
2 6 10
3 7 11
4 8 12
答案 2 :(得分：4)
我正在挖掘这个帖子，因为我没有在答案中找到一个有效的解决方案，因此我会发布一个来帮助搜索一个人的人：
public int[][] transpose (int[][] array) {
if (array == null || array.length == 0)//empty or unset array, nothing do to here
return array;
int width = array.length;
int height = array[0].length;
int[][] array_new = new int[height][width];
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
array_new[y][x] = array[x][y];
}
}
return array_new;
}
你应该通过以下方式调用它：
int[][] a = new int[][] {{1,2,3,4},{5,6,7,8}};
for (int i = 0; i < a.length; i++) {
System.out.print("[");
for (int y = 0; y < a[0].length; y++) {
System.out.print(a[i][y] + ",");
}
System.out.print("]\n");
}
a = transpose(a); // call
System.out.println("");
for (int i = 0; i < a.length; i++) {
System.out.print("[");
for (int y = 0; y < a[0].length; y++) {
System.out.print(a[i][y] + ",");
}
System.out.print("]\n");
}
将按预期输出：
[1,2,3,4,]
[5,6,7,8,]
[1,5,]
[2,6,]
[3,7,]
[4,8,]
答案 3 :(得分：3)
更通用的方式：
/**
* Transposes the given array, swapping rows with columns. The given array might contain arrays as elements that are
* not all of the same length. The returned array will have {@code null} values at those places.
*
* @param <T>
* the type of the array
*
* @param array
* the array
*
* @return the transposed array
*
* @throws NullPointerException
* if the given array is {@code null}
*/
public static <T> T[][] transpose(final T[][] array) {
Objects.requireNonNull(array);
// get y count
final int yCount = Arrays.stream(array).mapToInt(a -> a.length).max().orElse(0);
final int xCount = array.length;
final Class<?> componentType = array.getClass().getComponentType().getComponentType();
@SuppressWarnings("unchecked")
final T[][] newArray = (T[][]) Array.newInstance(componentType, yCount, xCount);
for (int x = 0; x < xCount; x++) {
for (int y = 0; y < yCount; y++) {
if (array[x] == null || y >= array[x].length) break;
newArray[y][x] = array[x][y];
}
}
return newArray;
}
答案 4 :(得分：1)
如果你想进行矩阵的转置（在这种情况下为row count = col count），你可以这样跟随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];
}
}
}
答案 5 :(得分：0)
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;
}
}
答案 6 :(得分：0)
public int[][] getTranspose() {
int[][] transpose = new int[row][column];
for (int i = 0; i < row; i++) {
for (int j = 0; j < column; j++) {
transpose[i][j] = original[j][i];
}
}
return transpose;
}
答案 7 :(得分：0)
使用此函数（如有必要，用int替换String）。它以矩阵作为字符串数组，并返回一个新的矩阵，即转置矩阵。它也检查空数组的边缘情况。没有照片。
/// Complete
private void issueBWworker_RunWorkerCompleted_1(object sender, RunWorkerCompletedEventArgs e)
{
if (e.Cancelled)
{
//toolStripStatusLabel1.Text = "Cancelled by User Intentionally...";
pbar.Value = 0;
pbar.Visible = false;
}
// Check to see if an error occurred in the background process.
else if (e.Error != null)
{
//toolStripStatusLabel1.Text = e.Error.Message;
pbar.Visible = false;
}
else
{
// BackGround Task Completed with out Error
pbar.Visible = false;
//toolStripStatusLabel1.Text = " All Records Loaded...";
}
}
答案 8 :(得分：0)
这是我的50美分：用于转换多维数组的实用方法和测试（在我的情况下为双精度）：
/**
* 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)));
}
答案 9 :(得分：-1)
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
那应该为你做。
答案 10 :(得分：-1)
import java.util.*;
public class TestClass {
public static void main(String args[] ) throws Exception {
Scanner in=new Scanner(System.in);
int isize=in.nextInt();
int jsize=in.nextInt();
int arr[][]=new int[isize][jsize];
int array[][]=new int[jsize][isize];
for(int i=0;i<isize;i++) {
for(int j=0;j<jsize;j++) {
arr[i][j]=in.nextInt();
}
System.out.println("\n");
}
for(int n=0;n<arr[0].length;n++) {
for(int m=0;m<arr.length;m++) {
array[n][m]=arr[m][n];
System.out.print(array[n][m]+" ");
}
System.out.print("\n");
}
}
}