My question is closely related to this question but I'm looking for a solution in Javascript
How to Transpose 2D Matrix Stored as C 1D Array
Basically I have a 2D square matrix
1 2 3
4 5 6
7 8 9
Stored as follows
let anArray = [1 ,2, 3, 4, 5, 6, 7, 8, 9]
How can I transpose this matrix so that the elements of my source array are switched as follows?
let newArray = [1, 4, 7, 2, 5, 8, 3, 6, 9]
You could take the length for the dimension of the array and map items on a specific index for a new array.
var array = [1 ,2, 3, 4, 5, 6, 7, 8, 9],
n = Math.sqrt(array.length),
transposed = array.map((_, i, a) => a[(i % n) * n + Math.floor(i / n)]);
console.log(transposed.join(' '));The approach in the answer you linked to works well in JavaScript too.
For a 3 x 3:
const anArray = [1, 2, 3, 4, 5, 6, 7, 8, 9];
let newArray = [];
for (let i = 0; i < 3; i++) {
for (let j = 0; j < 3; j++) {
newArray[3 * i + j] = anArray[3 * j + i];
}
}
console.log(newArray);For an N x N, just replace the 3's with N.
This answer avoids division and flooring (integer division) and a decent optimizer should make the code relatively fast. You might also consider initializing the new array with
let newArray = new Array(9);
or
let newArray = new Array(N * N);
but profile the code before attempting "optimizations" such as this.
var arr1 = [];
var arr2 = [];
for(int i=0; i<mat.length; i++) {
for(int j=0; j<mat[i].length; j++) {
arr1.push(mat[i][j]);
}
}
for(int j=0; j<mat[i].length; j++) {
for(int i=0; i<mat.length; i++) {
arr2.push(mat[i][j]);
}
}
Set a max "width" for your matrix and insert into a new array in loops, offset by 1 for each run.
function transpose(list, width) {
if (width === void 0) {
width = 1;
}
var t = 0;
var transposed = [];
while (t < width) {
for (var index = t; index < list.length; index += width) {
transposed.push(list[index]);
}
t++;
}
return transposed;
}
//TEST
var list = [1, 2, 3, 4, 5, 6, 7, 8, 9];
var transposed = transpose(list, 3);
console.log(list.join());
console.log(transposed.join());