Javascript：更新矩阵中的列答案

【问题标题】：Javascript: update a column in matrixJavascript：更新矩阵中的列
【发布时间】：2020-06-08 07:05:33
【问题描述】：

我想更新矩阵的整个列。

我首先初始化矩阵，如：

    m =({length:10}, (_,i) => Array.from({length:2}, (_,j) => i+'x'+j));

然后，我有一个包含 2 个元素（例如 a2）的数组，我想通过操作将其放入矩阵的第 j 列。如果我使用 matlab，我会执行以下操作：m(:,j)=a2+ m(:,j)

如何在 javascript 中做到这一点？

【问题讨论】：

向我们展示您的尝试。

标签： javascript arrays matrix

【解决方案1】：

你的意思是这样的吗？：

const m = Array.from({length:10}, (_,i) => Array.from({length:2}, (_,j) => i+'x'+j));

console.log("before", m);

const j = 2;
const a2 = "z"
const n = m.map(row => {row.splice(j, 0, a2); return row});

console.log("after", n);

操作前：

before [
  [    "0x0",    "0x1"  ],
  [    "1x0",    "1x1"  ],
  [    "2x0",    "2x1"  ],
  [    "3x0",    "3x1"  ],
  [    "4x0",    "4x1"  ],
  [    "5x0",    "5x1"  ],
  [    "6x0",    "6x1"  ],
  [    "7x0",    "7x1"  ],
  [    "8x0",    "8x1"  ],
  [    "9x0",    "9x1"  ]
]

在操作之后（即在第 2 列添加“z”）：

after [
  [    "0x0",    "0x1",    "z"  ],
  [    "1x0",    "1x1",    "z"  ],
  [    "2x0",    "2x1",    "z"  ],
  [    "3x0",    "3x1",    "z"  ],
  [    "4x0",    "4x1",    "z"  ],
  [    "5x0",    "5x1",    "z"  ],
  [    "6x0",    "6x1",    "z"  ],
  [    "7x0",    "7x1",    "z"  ],
  [    "8x0",    "8x1",    "z"  ],
  [    "9x0",    "9x1",    "z"  ]
]

或者如果a2 本身应该是一个数组？然后，它是一维数组的简单情况：

const m = Array.from({length:10}, (_,i) => Array.from({length:2}, (_,j) => i+'x'+j));

console.log("before", m);

const j = 2;
const a2 = Array.from({length:10}, (_,j) => 'z'+j)
const n = m.map((row, i) => {row.splice(j, 0, a2[i]); return row});

console.log("after", n);

操作后（即在第 2 列添加“z{ i }”）：

after [
  [    "0x0",    "0x1",    "z0"  ],
  [    "1x0",    "1x1",    "z1"  ],
  [    "2x0",    "2x1",    "z2"  ],
  [    "3x0",    "3x1",    "z3"  ],
  [    "4x0",    "4x1",    "z4"  ],
  [    "5x0",    "5x1",    "z5"  ],
  [    "6x0",    "6x1",    "z6"  ],
  [    "7x0",    "7x1",    "z7"  ],
  [    "8x0",    "8x1",    "z8"  ],
  [    "9x0",    "9x1",    "z9"  ]
]

或者当a2是一个二维数组时更复杂的情况：

const m = Array.from({length:10}, (_,i) => Array.from({length:2}, (_,j) => i+'x'+j));

console.log("before", m);

const j = 2;
const a2 = Array.from({length:10}, (_,i) => Array.from({length:2}, (_,j) => i+'z'+j));
const n = m.map((row, i) => {row.splice(j, 0, a2[i]); return row.flat()});

console.log("after", n);

操作后（即在第 2 列添加“z{ i }”）：

after [
  [    "0x0",    "0x1",    "0z0",    "0z1"  ],
  [    "1x0",    "1x1",    "1z0",    "1z1"  ],
  [    "2x0",    "2x1",    "2z0",    "2z1"  ],
  [    "3x0",    "3x1",    "3z0",    "3z1"  ],
  [    "4x0",    "4x1",    "4z0",    "4z1"  ],
  [    "5x0",    "5x1",    "5z0",    "5z1"  ],
  [    "6x0",    "6x1",    "6z0",    "6z1"  ],
  [    "7x0",    "7x1",    "7z0",    "7z1"  ],
  [    "8x0",    "8x1",    "8z0",    "8z1"  ],
  [    "9x0",    "9x1",    "9z0",    "9z1"  ]
]

【讨论】：

【解决方案2】：

我认为使用Array.prototype.forEach 和Array.prototype.splice 是实现此目的的好方法：

// Setup for demo
const Row = (w) =>Array.from({length:w}).fill('').map((c,i)=>c=i);
const Matrix = (h, w) => Array.from({length:h}).fill('').map(r=>r=Row(w));
const print = (mtx) => mtx.forEach(r=>console.log([...r].join(", ")));

// Original state of matrix
const m1 = Matrix(2,10);
print(m1);
console.log('----------');

// Insertion function
const insertCol = (mtx, cInd, newCol) => {
  mtx.forEach( (row, rInd) => { row.splice(cInd, 0, newCol[rInd]); }
  );
}

// Matrix with inserted column
const someCol = ['X','Y'];
insertCol(m1, 3, someCol);
print(m1);

【讨论】：