Simple Matrix Rotation

Question

I needed to rotate a non-square matrix in my program to transpose some spreadsheet data and decided to make a function to rotate 90, 180, or 270 degrees.

It doesn't rotate the matrix in place, but instead generates a copy and returns that. I want to go for processing efficiency, let me know what I can do better:

//Rotates the provided matrix by provided degrees counterclockwise
function RotateMatrix(matrix, degrees){ 
 if(degrees == 90){
 //For 90 degrees swap the height/width and swap the location of each element
 var output = GenerateMatrix(matrix[0].length, matrix.length, 0); //Swapping the width and height for non square matrices
 for(var i = 0; i < matrix[0].length; i++){
 for(var j = 0; j < matrix.length; j++){
 output[i][j] = matrix[j][i];
 }
 } 
 } else if(degrees == 180) {
 //For 180 degrees, rebuild array backwards
 var output = GenerateMatrix(matrix.length, matrix[0].length, 0);
 for(var i = matrix.length - 1; i >= 0; i--){
 for(var j = matrix[0].length - 1; j >=0; j--){
 output[matrix.length - 1 - i][matrix[0].length - 1 - j] = matrix[i][j];
 }
 } 
 } else if(degrees == 270) {
 //For 270 degrees, not sure how to make short description
 var output = GenerateMatrix(matrix[0].length, matrix.length, 0); //Swapping the width and height for non square matrices
 for(var i = 0; i < matrix[0].length; i++){
 for(var j = matrix.length - 1; j >=0; j--){
 output[i][matrix.length - 1 - j] = matrix[j][i];
 }
 } 
 } 
 return output;
}
//Generates a matrix with the requested length and width and value
function GenerateMatrix(length, width, value){
 var output = [];
 for(var i = 0; i < length; i++){
 width > 0 ? output.push([]) : output.push(value); //If matrix has 0 width
 for(var j = 0; j < width; j++){
 output[i].push(value)
 }
 }
 return output;
}

Answer 1

Unnecessary matrix generation

In JavaScript, Arrays don't have a fixed size. For example, you could do this:

var arr = [];
arr[324] = 91;

with no problem.

Your GenerateMatrix function is really over-doing it. All you need to do is create the arrays for the rows/columns then change their values.

And, you can do that with a neat trick:

Array.apply(null, Array(rowCount)).map(function(){return []});

• Douglas Gaskell: Thanks for the reply. I've typically programmed in C#, so JS is pretty new to me. That's good to know that I don't need to generate the "blank" array first. I initially avoided map due to the performance hit, looping ended up being more performance friendly when this is being executed in a cloud environment with very limited processing time.

Answer 2

Here's a version to where it does not flip, but instead reads or writes into the correct location. Fastest way I know to "rotate" it, but does add some overhead when reading and writing.

function addC(row, i) {
 Object.defineProperty(row.self, i, {
 get: function() { 
 if(!(row.matrix.degrees % 360)) {
 return row.data[i];
 }
 else if(!(row.matrix.degrees % 270)) {
 }
 else if(!(row.matrix.degrees % 180)) {
 }
 else if(!(row.matrix.degrees % 90)) {
 return row.matrix.rows[i].data[row.matrix.column];
 }
 },
 set: function(v) { 
 if(!(row.matrix.degrees % 360)) {
 row.data[i] = v;
 }
 else if(!(row.matrix.degrees % 270)) {
 }
 else if(!(row.matrix.degrees % 180)) {
 }
 else if(!(row.matrix.degrees % 90)) {
 row.matrix.rows[i].data[row.matrix.column] = v;
 }
 },
 enumerable: true
 });
}
function Row(matrix) {
 var members = {
 self: this,
 matrix: matrix,
 data: {}
 }
 for(var i = matrix.width; i--;) {
 addC(members, i);
 }
 Object.defineProperty(members.self, "data", { value: members.data });
}
function addR(matrix, i) {
 matrix.rows[i] = new Row(matrix);
 Object.defineProperty(matrix.self, i, {
 get: function() { matrix.column = i; return matrix.rows[i]; },
 enumerable: true
 });
}
function Matrix(width, height) {
 var members = {
 self: this,
 width: width,
 height: height,
 degrees: 0,
 column: -1,
 rows: {}
 };
 for(var i = height; i--;) {
 addR(members, i);
 }
 Object.defineProperty(members.self, "degrees", { value: function(v) {
 if(v === undefined) return members.degrees;
 // Can add more checks if need be.
 if(!(v % 90)) members.degrees = v;
 return members.self;
 }});
}

---

You probably know more about the rotations than I do so I did the ones that I was 100% sure were correct. It is pretty straight forward, when the matrix object is accessed it returns a row object that then figures out what data you actually meant.

var matrix = new Matrix(2, 2);
matrix[0][1] = 12;
matrix.degrees(90);
console.log(matrix[1][0]); // => 12

---

Hope this helps!