用不同集合中的元素替換陣列中的專案

我不確定以遞回方式進行此操作的正確方法。假設我有兩個陣列

array1 = [a, null, c, d, e]

array2 = [1, 2, 3]

我想創建表單的結果陣列

[
   [a, 1, c, d, e],
   [a, 2, c, d, e],
   [a, 3, c, d, e]
]

其中第二個陣列的每個元素都填充了 null 的位置。我可以很好地做到這一點，但是如果我有兩個 null 情況怎么辦？

array1 = [a, null, c, null, e]

array2 = [1, 2, 3]

我如何獲得以下內容？

[
   [a, 1, c, 1, e],
   [a, 1, c, 2, e],
   [a, 1, c, 3, e],
   [a, 2, c, 1, e],
   ...
   [a, 3, c, 3, e]
]

uj5u.com熱心網友回復：

Javascript代碼：

let array1 = ['a', null, 'c', null, 'e']
let array2 = [1, 2, 3]

let output = []
let cursorPosition = 0; // cursorPosition is a pointer for `array2` index
let nullPositions = []

// find nullable positions; in our case here will be 1, 4
for (let i = 0; i < array1.length; i  ) {
  if (array1[i] === null) {
    nullPositions.push(i)
  }
}

// find total expected numbers of arrays from given input and create that array
const totalArray = Math.pow(array2.length, nullPositions.length);
for (let i = 0; i < totalArray; i  ) {
  output.push([...array1])
}

// replacement logic for the `null` values
for (let i = 0; i < nullPositions.length; i  ) {
  cursorPosition = 0;
  // frequency: how often the cursor needs to change (depends on null position)
  let frequency = totalArray / Math.pow(array2.length, i   1);
  let resetCursorPositionOn = frequency;
  for (let j = 0; j < output.length; j  ) {
    if (j < resetCursorPositionOn) {
      output[j][nullPositions[i]] = array2[cursorPosition];
    } else {
      resetCursorPositionOn  = frequency;
      cursorPosition  ;
      if (cursorPosition === array2.length) {
        cursorPosition = 0;
      }
      output[j][nullPositions[i]] = array2[cursorPosition];
    }
  }
}
console.log(output);

輸出：

[
  ["a",1,"c",1,"e"],
  ["a",1,"c",2,"e"],
  ["a",1,"c",3,"e"],
  ["a",2,"c",1,"e"],
  ["a",2,"c",2,"e"],
  ["a",2,"c",3,"e"],
  ["a",3,"c",1,"e"],
  ["a",3,"c",2,"e"],
  ["a",3,"c",3,"e"]
]

它適用于任意數量的組合。

uj5u.com熱心網友回復：

您描述的第一種情況是map模式的實體，第二種情況是 -- flatMap。

換個角度看，map就像一個回圈，flatMap又像一個兩級嵌套回圈，從最深的一級回圈開始，一個一個地創建結果的元素：

地圖：

      for x in [1, 2, 3]:
         let res = { result of replacing
                     the first `null` with `x` 
                     in [a, null, c, d, e] }
         yield res

             -------------------------
               [a, null, c, d, e]
             ----------------------
   1           [a,  1,   c, d, e],
   2           [a,  2,   c, d, e],
   3           [a,  3,   c, d, e]

平面圖：

      for x in [1, 2, 3]:
         let res = { result of replacing
                     the first `null` with `x` 
                     in [a, null, c, null, e] }
         
         for x in [1, 2, 3]:
             let res2 = { result of replacing
                          the first `null` with `x` 
                          in res }                   ## res NB!
             yield res2

             -------------------------
             [  a, null, c, null, e  ]
             -------------------------
   1:        [  a,  1,   c, null, e  ],

        1      [a,  1,   c,   1 , e],
        2      [a,  1,   c,   2 , e],
        3      [a,  1,   c,   3 , e],

   2:        [  a,  2,   c, null, e  ],

        1      [a,  2,   c,   1 , e],
        2      [a,  2,   c,   2 , e],
        3      [a,  2,   c,   3 , e],

   3:        [  a,  3,   c, null, e  ]

        1      [a,  3,   c,   1 , e],
        2      [a,  3,   c,   2 , e],
        3      [a,  3,   c,   3 , e]

第一個就像用[1,2,3]替換結果替換每個元素；第二個就像用替換結果串列替換每個元素并將結果[1,2,3]串列附加在一起; 即將結果串列的元素拼接到位；即創建組合地圖，展平（因此得名，）。flatMap

標籤：数组 递归

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