[javascript] Transposing a 2D-array in JavaScript

I've got an array of arrays, something like:

[
    [1,2,3],
    [1,2,3],
    [1,2,3],
]

I would like to transpose it to get the following array:

[
    [1,1,1],
    [2,2,2],
    [3,3,3],
]

It's not difficult to programmatically do so using loops:

function transposeArray(array, arrayLength){
    var newArray = [];
    for(var i = 0; i < array.length; i++){
        newArray.push([]);
    };

    for(var i = 0; i < array.length; i++){
        for(var j = 0; j < arrayLength; j++){
            newArray[j].push(array[i][j]);
        };
    };

    return newArray;
}

This, however, seems bulky, and I feel like there should be an easier way to do it. Is there?

Just another variation using Array.map. Using indexes allows to transpose matrices where M != N:

// Get just the first row to iterate columns first
var t = matrix[0].map(function (col, c) {
    // For each column, iterate all rows
    return matrix.map(function (row, r) { 
        return matrix[r][c]; 
    }); 
});

All there is to transposing is mapping the elements column-first, and then by row.

I think this is slightly more readable. It uses Array.from and logic is identical to using nested loops:

var arr = [_x000D_
  [1, 2, 3, 4],_x000D_
  [1, 2, 3, 4],_x000D_
  [1, 2, 3, 4]_x000D_
];_x000D_
_x000D_
/*_x000D_
 * arr[0].length = 4 = number of result rows_x000D_
 * arr.length = 3 = number of result cols_x000D_
 */_x000D_
_x000D_
var result = Array.from({ length: arr[0].length }, function(x, row) {_x000D_
  return Array.from({ length: arr.length }, function(x, col) {_x000D_
    return arr[col][row];_x000D_
  });_x000D_
});_x000D_
_x000D_
console.log(result);

If you are dealing with arrays of unequal length you need to replace arr[0].length with something else:

var arr = [_x000D_
  [1, 2],_x000D_
  [1, 2, 3],_x000D_
  [1, 2, 3, 4]_x000D_
];_x000D_
_x000D_
/*_x000D_
 * arr[0].length = 4 = number of result rows_x000D_
 * arr.length = 3 = number of result cols_x000D_
 */_x000D_
_x000D_
var result = Array.from({ length: arr.reduce(function(max, item) { return item.length > max ? item.length : max; }, 0) }, function(x, row) {_x000D_
  return Array.from({ length: arr.length }, function(x, col) {_x000D_
    return arr[col][row];_x000D_
  });_x000D_
});_x000D_
_x000D_
console.log(result);

You can do it in in-place by doing only one pass:

function transpose(arr,arrLen) {
  for (var i = 0; i < arrLen; i++) {
    for (var j = 0; j <i; j++) {
      //swap element[i,j] and element[j,i]
      var temp = arr[i][j];
      arr[i][j] = arr[j][i];
      arr[j][i] = temp;
    }
  }
}

A library-free implementation in TypeScript that works for any matrix shape that won't truncate your arrays:

const rotate2dArray = <T>(array2d: T[][]) => {
    const rotated2d: T[][] = []

    return array2d.reduce((acc, array1d, index2d) => {
        array1d.forEach((value, index1d) => {
            if (!acc[index1d]) acc[index1d] = []

            acc[index1d][index2d] = value
        })

        return acc
    }, rotated2d)
}

If you have an option of using Ramda JS and ES6 syntax, then here's another way to do it:

const transpose = a => R.map(c => R.map(r => r[c], a), R.keys(a[0]));_x000D_
_x000D_
console.log(transpose([_x000D_
  [1, 2, 3, 4],_x000D_
  [5, 6, 7, 8],_x000D_
  [9, 10, 11, 12]_x000D_
])); // =>  [[1,5,9],[2,6,10],[3,7,11],[4,8,12]]

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.22.1/ramda.min.js"></script>

Many good answers here! I consolidated them into one answer and updated some of the code for a more modern syntax:

One-liners inspired by Fawad Ghafoor and Óscar Gómez Alcañiz

function transpose(matrix) {
  return matrix[0].map((col, i) => matrix.map(row => row[i]));
}

function transpose(matrix) {
  return matrix[0].map((col, c) => matrix.map((row, r) => matrix[r][c]));
}

Functional approach style with reduce by Andrew Tatomyr

function transpose(matrix) {
  return matrix.reduce((prev, next) => next.map((item, i) =>
    (prev[i] || []).concat(next[i])
  ), []);
}

Lodash/Underscore by marcel

function tranpose(matrix) {
  return _.zip(...matrix);
}

// Without spread operator.
function transpose(matrix) {
  return _.zip.apply(_, [[1,2,3], [1,2,3], [1,2,3]])
}

Vanilla approach

function transpose(matrix) {
  const rows = matrix.length, cols = matrix[0].length;
  const grid = [];
  for (let j = 0; j < cols; j++) {
    grid[j] = Array(rows);
  }
  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      grid[j][i] = matrix[i][j];
    }
  }
  return grid;
}

Vanilla in-place ES6 approach inspired by Emanuel Saringan

function transpose(matrix) {
  for (var i = 0; i < matrix.length; i++) {
    for (var j = 0; j < i; j++) {
      const temp = matrix[i][j];
      matrix[i][j] = matrix[j][i];
      matrix[j][i] = temp;
    }
  }
}

// Using destructing
function transpose(matrix) {
  for (var i = 0; i < matrix.length; i++) {
    for (var j = 0; j < i; j++) {
      [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
    }
  }
}

You can achieve this without loops by using the following.

• Array
• Array.prototype.map
• Array.prototype.reduce
• Array.prototype.join
• String.prototype.split

It looks very elegant and it does not require any dependencies such as jQuery of Underscore.js.

function transpose(matrix) {  
    return zeroFill(getMatrixWidth(matrix)).map(function(r, i) {
        return zeroFill(matrix.length).map(function(c, j) {
            return matrix[j][i];
        });
    });
}

function getMatrixWidth(matrix) {
    return matrix.reduce(function (result, row) {
        return Math.max(result, row.length);
    }, 0);
}

function zeroFill(n) {
    return new Array(n+1).join('0').split('').map(Number);
}

Minified

function transpose(m){return zeroFill(m.reduce(function(m,r){return Math.max(m,r.length)},0)).map(function(r,i){return zeroFill(m.length).map(function(c,j){return m[j][i]})})}function zeroFill(n){return new Array(n+1).join("0").split("").map(Number)}

Here is a demo I threw together. Notice the lack of loops :-)

// Create a 5 row, by 9 column matrix._x000D_
var m = CoordinateMatrix(5, 9);_x000D_
_x000D_
// Make the matrix an irregular shape._x000D_
m[2] = m[2].slice(0, 5);_x000D_
m[4].pop();_x000D_
_x000D_
// Transpose and print the matrix._x000D_
println(formatMatrix(transpose(m)));_x000D_
_x000D_
function Matrix(rows, cols, defaultVal) {_x000D_
    return AbstractMatrix(rows, cols, function(r, i) {_x000D_
        return arrayFill(cols, defaultVal);_x000D_
    });_x000D_
}_x000D_
function ZeroMatrix(rows, cols) {_x000D_
    return AbstractMatrix(rows, cols, function(r, i) {_x000D_
        return zeroFill(cols);_x000D_
    });_x000D_
}_x000D_
function CoordinateMatrix(rows, cols) {_x000D_
    return AbstractMatrix(rows, cols, function(r, i) {_x000D_
        return zeroFill(cols).map(function(c, j) {_x000D_
            return [i, j];_x000D_
        });_x000D_
    });_x000D_
}_x000D_
function AbstractMatrix(rows, cols, rowFn) {_x000D_
    return zeroFill(rows).map(function(r, i) {_x000D_
        return rowFn(r, i);_x000D_
    });_x000D_
}_x000D_
/** Matrix functions. */_x000D_
function formatMatrix(matrix) {_x000D_
    return matrix.reduce(function (result, row) {_x000D_
        return result + row.join('\t') + '\n';_x000D_
    }, '');_x000D_
}_x000D_
function copy(matrix) {  _x000D_
    return zeroFill(matrix.length).map(function(r, i) {_x000D_
        return zeroFill(getMatrixWidth(matrix)).map(function(c, j) {_x000D_
            return matrix[i][j];_x000D_
        });_x000D_
    });_x000D_
}_x000D_
function transpose(matrix) {  _x000D_
    return zeroFill(getMatrixWidth(matrix)).map(function(r, i) {_x000D_
        return zeroFill(matrix.length).map(function(c, j) {_x000D_
            return matrix[j][i];_x000D_
        });_x000D_
    });_x000D_
}_x000D_
function getMatrixWidth(matrix) {_x000D_
    return matrix.reduce(function (result, row) {_x000D_
        return Math.max(result, row.length);_x000D_
    }, 0);_x000D_
}_x000D_
/** Array fill functions. */_x000D_
function zeroFill(n) {_x000D_
  return new Array(n+1).join('0').split('').map(Number);_x000D_
}_x000D_
function arrayFill(n, defaultValue) {_x000D_
    return zeroFill(n).map(function(value) {_x000D_
        return defaultValue || value;_x000D_
    });_x000D_
}_x000D_
/** Print functions. */_x000D_
function print(str) {_x000D_
    str = Array.isArray(str) ? str.join(' ') : str;_x000D_
    return document.getElementById('out').innerHTML += str || '';_x000D_
}_x000D_
function println(str) {_x000D_
    print.call(null, [].slice.call(arguments, 0).concat(['<br />']));_x000D_
}

#out {_x000D_
    white-space: pre;_x000D_
}

<div id="out"></div>

I found the above answers either hard to read or too verbose, so I write one myself. And I think this is most intuitive way to implement transpose in linear algebra, you don't do value exchange, but just insert each element into the right place in the new matrix:

function transpose(matrix) {
  const rows = matrix.length
  const cols = matrix[0].length

  let grid = []
  for (let col = 0; col < cols; col++) {
    grid[col] = []
  }
  for (let row = 0; row < rows; row++) {
    for (let col = 0; col < cols; col++) {
      grid[col][row] = matrix[row][col]
    }
  }
  return grid
}

You could use underscore.js

_.zip.apply(_, [[1,2,3], [1,2,3], [1,2,3]])

shortest way with lodash/underscore and es6:

_.zip(...matrix)

where matrix could be:

const matrix = [[1,2,3], [1,2,3], [1,2,3]];

function invertArray(array,arrayWidth,arrayHeight) {
  var newArray = [];
  for (x=0;x<arrayWidth;x++) {
    newArray[x] = [];
    for (y=0;y<arrayHeight;y++) {
        newArray[x][y] = array[y][x];
    }
  }
  return newArray;
}

reverseValues(values) {
        let maxLength = values.reduce((acc, val) => Math.max(val.length, acc), 0);
        return [...Array(maxLength)].map((val, index) => values.map((v) => v[index]));
}

One-liner that does not change given array.

a[0].map((col, i) => a.map(([...row]) => row[i]))

I didn't find an answer that satisfied me, so I wrote one myself, I think it is easy to understand and implement and suitable for all situations.

    transposeArray: function (mat) {
        let newMat = [];
        for (let j = 0; j < mat[0].length; j++) {  // j are columns
            let temp = [];
            for (let i = 0; i < mat.length; i++) {  // i are rows
                temp.push(mat[i][j]);  // so temp will be the j(th) column in mat
            }
            newMat.push(temp);  // then just push every column in newMat
        }
        return newMat;
    }

here is my implementation in modern browser (without dependency):

transpose = m => m[0].map((x,i) => m.map(x => x[i]))

Another approach by iterating the array from outside to inside and reduce the matrix by mapping inner values.

const_x000D_
    transpose = array => array.reduce((r, a) => a.map((v, i) => [...(r[i] || []), v]), []),_x000D_
    matrix = [[1, 2, 3], [1, 2, 3], [1, 2, 3]];_x000D_
_x000D_
console.log(transpose(matrix));

Edit: This answer would not transpose the matrix, but rotate it. I didn't read the question carefully in the first place :D

clockwise and counterclockwise rotation:

    function rotateCounterClockwise(a){
        var n=a.length;
        for (var i=0; i<n/2; i++) {
            for (var j=i; j<n-i-1; j++) {
                var tmp=a[i][j];
                a[i][j]=a[j][n-i-1];
                a[j][n-i-1]=a[n-i-1][n-j-1];
                a[n-i-1][n-j-1]=a[n-j-1][i];
                a[n-j-1][i]=tmp;
            }
        }
        return a;
    }

    function rotateClockwise(a) {
        var n=a.length;
        for (var i=0; i<n/2; i++) {
            for (var j=i; j<n-i-1; j++) {
                var tmp=a[i][j];
                a[i][j]=a[n-j-1][i];
                a[n-j-1][i]=a[n-i-1][n-j-1];
                a[n-i-1][n-j-1]=a[j][n-i-1];
                a[j][n-i-1]=tmp;
            }
        }
        return a;
    }

If using RamdaJS is an option, this can be achieved in one line: R.transpose(myArray)

ES6 1liners as :

let invert = a => a[0].map((col, c) => a.map((row, r) => a[r][c]))

so same as Óscar's, but as would you rather rotate it clockwise :

let rotate = a => a[0].map((col, c) => a.map((row, r) => a[r][c]).reverse())

Neat and pure:

[[0, 1], [2, 3], [4, 5]].reduce((prev, next) => next.map((item, i) =>
    (prev[i] || []).concat(next[i])
), []); // [[0, 2, 4], [1, 3, 5]]

Previous solutions may lead to failure in case an empty array is provided.

Here it is as a function:

function transpose(array) {
    return array.reduce((prev, next) => next.map((item, i) =>
        (prev[i] || []).concat(next[i])
    ), []);
}

console.log(transpose([[0, 1], [2, 3], [4, 5]]));

Update. It can be written even better with spread operator:

const transpose = matrix => matrix.reduce(
    ($, row) => row.map((_, i) => [...($[i] || []), row[i]]), 
    []
)

Since nobody so far mentioned a functional recursive approach here is my take. An adaptation of Haskell's Data.List.transpose.

var transpose = as => as.length ? as[0].length ? [as.reduce((rs, a) => a.length ? (rs.push(a[0]), rs) :
    rs, []
  ), ...transpose(as.map(a => a.slice(1)))] :
  transpose(as.slice(1)) :
  [],
  mtx = [
    [1],
    [1, 2],
    [1, 2, 3]
  ];

console.log(transpose(mtx))

.as-console-wrapper {
  max-height: 100% !important
}