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

Question

我找到了很多关于数组值之间所有可能组合的解决方案，但我需要一些不同的东西，我希望你能支持我。基本上是创建将数组键与 true|false 值组合在一起的所有可能对象，如下所示:

输入:(应该返回一个包含 32 个对象的数组，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开和关开关的 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

您总共需要2 ** keys.length数组中的对象。因此，使用 Array.from({ 2 ** keys.length }) 创建它

在 map 函数中，使用 row.toString(2) 为当前行创建一个二进制数

添加前导 0 直到字符串为 keys.length长:( "00001" )

拆分字符串以使其成为数组( ["0", "0", "0", "0", "1"] )

映射此数组并为相应的键创建一个条目数组[["arm","false"],["lens","false"],["season","false"],["food","false"],["size","false"]]

使用 Object.fromEntries() 从条目数组创建一个对象

这是一个片段:

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

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

console.log(combination(keys))