如何在没有 item/coordinate 重复项的情况下选择二维矩阵的 N 个随机元素的 col/row 或 x/y 坐标？

Question

我有一个二维矩阵，下面是一个例子data = [["a", "b", "c", "d"], ["e", "g"], ["i", "j", "k"]]

我需要得到 N 个没有重复的随机 (x, y) 索引。

我已经问了一个不同的问题相同的上下文，这是选择 2 x，y 组合的解决方案

const data = [["a", "b", "c", "d"], ["e", "g"], ["i", "j", "k"]];

function combinations(data) {
  const i11 = Math.floor(Math.random() * data.length);
  const i12 = Math.floor(Math.random() * data[i11].length);

  const dataLength = data[i11].length > 1 ? data.length : data.length - 1;

  let i21 = Math.floor(Math.random() * dataLength);
  if (i21 >= i11 && data[i11].length === 1) ++i21;

  const innerDataLength = i21 === i11 ? data[i21].length - 1 : data[i21].length;
  let i22 = Math.floor(Math.random() * innerDataLength);
  if (i21 === i11 && i22 >= i12) ++i22;
  
  return [[i11, i12], [i21, i22]];
}

console.log(combinations(data));

for (let i = 0; i < 10000; ++i) {
    const [[i11, i12], [i21, i22]] = combinations(data);
    if (i11 === i21 && i12 == i22) console.log('Test failed!');
}

Answer 1

function pickNOf(list, n) {
  // - create a new shallow array copy.
  // - does decouple the original reference, thus it
  //   prevents its further mutation by e.g. `splice`.
  list = Array.from(list);

  // - creates and returns an array of the desired length
  //   by a mapping task which constantly slices random
  //   items from the shallow array copy.
  return Array.from({ length: n }, () => list.splice(
    // - `splice` does mutate the list by removing a
    //   single item from its randomly chosen index.
    Math.floor(Math.random() * list.length), 1
  )[0]);
}

function pickNUniqueItemsFromNestedArray(arr, n) {
  // - flatten nested array structure,
  // - channel it through a `Set` in order
  //   to keep just unique values,
  // - return the result of the forwarded
  //   `pickNOf` call.
  return pickNOf([...new Set(
    arr.flat(Infinity)
  )], n);
}

const data = [
  ["a", "b", "c", "d"],
  ["e", "g"],
  ["i", "j", "k"]
];
console.log(
  "pickNUniqueItemsFromNestedArray(data, 9) ...",
  pickNUniqueItemsFromNestedArray(data, 9)
);
console.log(
  "pickNUniqueItemsFromNestedArray(data, 5) ...",
  pickNUniqueItemsFromNestedArray(data, 5)
);
console.log(
  "pickNUniqueItemsFromNestedArray(data, 3) ...",
  pickNUniqueItemsFromNestedArray(data, 3)
);

.as-console-wrapper { min-height: 100%!important; top: 0; }

编辑由于...

Thank you it works, but i need indexes – Amine

But that's not what the question's topic doe state ... "How to pick N random elements from a 2D matrix without duplicates in javascript?" Nevertheless I will provide a second solution which utilizes the first given approach in order to fully meet the OP's requirements. – Peter Seliger

为了满足 OP 的要求，上面给出的方法略微更改为 ...

function pickNOf(list, n) {
  // - create a new shallow array copy.
  // - does decouple the original reference, thus it
  //   prevents its further mutation by e.g. `splice`.
  list = Array.from(list);

  // - creates and returns an array of the desired length
  //   by a mapping task which constantly slices random
  //   items from the shallow array copy.
  return Array.from({ length: n }, () => list.splice(
    // - `splice` does mutate the list by removing a
    //   single item from its randomly chosen index.
    Math.floor(Math.random() * list.length), 1
  )[0]);
}

function pickNUniqueIndexTuplesFrom2dMatrix(matrix, n) {
  // - create a flat array of any of a
  //   matrix' index coordinates/tuples.
  return pickNOf(
    matrix.flatMap((arr, rowIdx) =>
      arr.map((_, colIdx) => [rowIdx, colIdx])
    ), n
  );
}

const data = [
  ["a", "b", "c", "d"],
  ["e", "g"],
  ["i", "j", "k"]
];
console.log(
  "pickNUniqueIndexTuplesFrom2dMatrix(data, 9) ...",
  pickNUniqueIndexTuplesFrom2dMatrix(data, 9)
);
console.log(
  "pickNUniqueIndexTuplesFrom2dMatrix(data, 5) ...",
  pickNUniqueIndexTuplesFrom2dMatrix(data, 5)
);
console.log(
  "pickNUniqueIndexTuplesFrom2dMatrix(data, 3) ...",
  pickNUniqueIndexTuplesFrom2dMatrix(data, 3)
);

.as-console-wrapper { min-height: 100%!important; top: 0; }

备注

尽管或什至可能因为误解，基本方法本身并没有改变这一点可能会变得更加清楚。这是... 从数组中挑选 n 个没有重复项的随机项目 ... 和 ...

将主要任务拆分为更专业的子任务。 （如pickNOf的实现演示，完美解决单个问题，无需重构。）
另一个 task/s has/have 以一种可以将输入数据传递给从没有重复的数组（希望通过两种不同的方法得到证明，每种方法针对不同的要求）。

如何在没有 item/coordinate 重复项的情况下选择二维矩阵的 N 个随机元素的 col/row 或 x/y 坐标？

How does one pick the col/row or x/y coordinates of N random elements of a 2D matrix without item/coordinate duplicates?

javascript

arrays

matrix

random

mapping