JS中的递归挑战将所有可能的数组键组合为true |错误的版本，我附上输入和输出

Question

我已经找到了关于数组值之间所有可能组合的许多解决方案，但是我需要一些不同的东西，希望您能支持我。基本上是创建所有将数组键与true | false值组合的可能对象，如下所示：

输入：（应返回一个由32个对象组成的数组，2exp5、5个键中的两个可能值）

let properties = ['arm','lens','season','food','size'];

输出：

let combinations = [
  {"arm": "false","lens": "false","season": "false","food": "false","size": "false"}
  {"arm": "false","lens": "false","season": "false","food": "false","size": "true"}
  {"arm": "false","lens": "false","season": "false","food": "true","size": "true"}
  {"arm": "false","lens": "false","season": "true","food": "true","size": "true"}
  {"arm": "false","lens": "true","season": "true","food": "true","size": "true"}
  {"arm": "true","lens": "true","season": "true","food": "true","size": "true"}
  {"arm": "true","lens": "true","season": "true","food": "false","size": "true"}
  {"arm": "true","lens": "true","season": "false","food": "false","size": "true"}
  and so on...
]

非常感谢！

Answer 1

您可以为每个属性使用带有打开和关闭开关的2D矩阵。然后为每个键创建条目，并使用Object.fromEntries()

创建对象

0 0 0 0 0
0 0 0 0 1
0 0 0 1 0
0 0 0 1 1
etc

以下是代码段：

let keys = ['arm', 'lens', 'season', 'food', 'size'];

function combination(keys) {
  return Array.from({ length: 2 ** keys.length }, (_, i) =>
    Object.fromEntries(
      i.toString(2)
      .padStart(keys.length, 0)
      .split('')
      .map((binary, j) => [keys[j], String(Boolean(+binary))])
    )
  )
}

console.log(combination(keys))