Swap rows with columns (transposition) of a matrix

问题:

This question already has answers here:

Transposing a 2D-array in JavaScript (21 answers)

Closed 2 years ago.

For instance I have a matrix like this:

|1 2 3|    
|4 5 6|
|7 8 9|

and I need it to convert into a matrix like this:

|1 4 7|    
|2 5 8|
|3 6 9|

What is the best and optimal way to achieve this goal?

回答1:

See article: Transpose An Array In JavaScript and jQuery

function transpose(a) {

  // Calculate the width and height of the Array
  var w = a.length || 0;
  var h = a[0] instanceof Array ? a[0].length : 0;

  // In case it is a zero matrix, no transpose routine needed.
  if(h === 0 || w === 0) { return []; }

  /**
   * @var {Number} i Counter
   * @var {Number} j Counter
   * @var {Array} t Transposed data is stored in this array.
   */
  var i, j, t = [];

  // Loop through every item in the outer array (height)
  for(i=0; i<h; i++) {

    // Insert a new row (array)
    t[i] = [];

    // Loop through every item per item in outer array (width)
    for(j=0; j<w; j++) {

      // Save transposed data.
      t[i][j] = a[j][i];
    }
  }

  return t;
}

console.log(transpose([[1,2,3],[4,5,6],[7,8,9]]));

回答2:

DuckDucking turned up this by Ken. Surprisingly, it's even more concise and complete than Nikita's answer. It retrieves column and row lengths implicitly within the guts of map().

function transpose(a) {
    return Object.keys(a[0]).map(function(c) {
        return a.map(function(r) { return r[c]; });
    });
}

console.log(transpose([
    [1,2,3],
    [4,5,6],
    [7,8,9]
]));

回答3:

Just like in any other language:

int[][] copy = new int[columns][rows];
for (int i = 0; i < rows; ++i) {
    for (int j = 0; j < columns; ++j) {
        copy[j][i] = original[i][j];
    }
}

You just have to construct the 2D array differently in JS. Like this:

function transpose(original) {
    var copy = [];
    for (var i = 0; i < original.length; ++i) {
        for (var j = 0; j < original[i].length; ++j) {
            // skip undefined values to preserve sparse array
            if (original[i][j] === undefined) continue;
            // create row if it doesn't exist yet
            if (copy[j] === undefined) copy[j] = [];
            // swap the x and y coords for the copy
            copy[j][i] = original[i][j];
        }
    }
    return copy;
}

console.log(transpose([
    [1,2,3],
    [4,5,6],
    [7,8,9]
]));

回答4:

I don't have enough reputation to comment (wtf.), so I need to post Ken's updated version as a separate answer:

function transpose(a) {
    return a[0].map(function (_, c) { return a.map(function (r) { return r[c]; }); });
}

回答5:

Compact version of Hobs' answer using arrow functions from ES6:

function transpose(matrix) {
    return Object.keys(matrix[0])
        .map(colNumber => matrix.map(rowNumber => rowNumber[colNumber]));
}

回答6:

You can use Object.keys and Array.prototype.map:

function transpose(arr) {
  return Object.keys(arr[0]).map(function (c) {
    return arr.map(function (r) {
      return r[c];
    });
  });
}

console.log(transpose([
    [1,2,3],
    [4,5,6],
    [7,8,9]
]));