如何選擇2D矩陣的N個隨機元素的col/row或x/y坐標而沒有專案/坐標重復？

我有一個二維矩陣，這是一個例子 data = [["a", "b", "c", "d"], ["e", "g"], ["i", "j", "k"]]

我需要獲得N 個(x, y)沒有重復的隨機索引。

我已經在相同的背景關系中提出了不同的問題，這是選擇 2 x, y 組合的解決方案

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!');
}

uj5u.com熱心網友回復：

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; }

編輯由于...

謝謝它有效，但我需要索引 – Amine

但這不是問題的主題所陳述的...... “如何從 2D 矩陣中挑選 N 個隨機元素而不會在 javascript 中重復？” 盡管如此，我將提供第二個解決方案，它利用第一個給定的方法來完全滿足 OP 的要求。— 彼得·塞利格

為了滿足 OP 的要求，上述給定的方法略有更改為......

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; }

筆記

盡管甚至可能因為誤解，基礎方法本身并沒有改變這一點可能會變得更加清楚。這是...從沒有重復的陣列中選擇n隨機專案...和...

• Split the main task into more specialized sub tasks. (Like demonstrated with the implementation of pickNOf which solves a single problem perfectly, thus it does not need to be refactored.)

• The other task/s has/have to transform the input data in a way that it can be passed to the specialized task of picking n random items from an array without duplicates (which hopefully got proofed by the two different approaches each targeting a different requirement).

標籤：javascript 数组 矩阵 随机的 映射

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