Javascript 用javascript中的矩阵的列（转置）交换行

Swap rows with columns (transposition) of a matrix in javascript

提问by Bakhtiyor

For instance I have a matrix like this:

例如我有一个这样的矩阵：

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

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

我需要将其转换为这样的矩阵：

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

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

实现这一目标的最佳和最优方法是什么？

采纳答案by troynt

See article: Transpose An Array In JavaScript and jQuery

请参阅文章：在 JavaScript 和 jQuery 中转置数组

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by hobs

DuckDuckingturned up thisby 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().

DuckDucking打开了这个由肯。令人惊讶的是，它比Nikita的回答更简洁和完整。它在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by Nikita Rybak

Just like in any other language:

就像在任何其他语言中一样：

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:

你只需要在 JS 中以不同的方式构造二维数组。像这样：

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by KIT-Inwi

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

我没有足够的声誉来发表评论（wtf.），所以我需要将Ken 的更新版本作为单独的答案发布：

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

回答by Anton Iokov

Compact version of Hobs' answerusing arrow functions from ES6:

使用 ES6 中的箭头函数的 Hobs答案的紧凑版本：

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

回答by Pedro Justo

You can use Object.keysand Array.prototype.map:

您可以使用Object.keys和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]
]));