如何从对象集合中提取最详细的唯一层级对象？

我有一个存储公司层级选择的对象集合，示例如下：

{
  1:   { division: "division1" },
  2:   { division: "division2" },
  11:  { division: "division2", branch: "branch2" },
  41:  { division: "division2", branch: "branch2", department: "department2" },
  100: { division: "division2", branch: "branch2", department: "department10" },
  102: { division: "division2", branch: "branch3" },
  130: { division: "division17" },
  144: { division: "division17" , branch: "branch22" },
  200: { division: "division50" }
}

需要从中提取唯一且最详细的对象，预期结果如下（无需保留原键）：

{
  1:   { division: "division1" },
  41:  { division: "division2", branch: "branch2", department: "department2" },
  100: { division: "division2", branch: "branch2", department: "department10" },
  102: { division: "division2", branch: "branch3" },
  144: { division: "division17" , branch: "branch22" },
  200: { division: "division50" }
}

这些数据来自三个下拉框的选择，层级关系为division > branch > department。比如用户先选division2再选branch2，会同时添加{ division: "division2" }和{ division: "division2", branch: "branch2" }，但只需要最详细的后者。

可以用JavaScript实现这个筛选逻辑，核心思路是过滤掉所有被其他层级更深的对象完全包含的项，具体代码如下：

const input = {
  1:   { division: "division1" },
  2:   { division: "division2" },
  11:  { division: "division2", branch: "branch2" },
  41:  { division: "division2", branch: "branch2", department: "department2" },
  100: { division: "division2", branch: "branch2", department: "department10" },
  102: { division: "division2", branch: "branch3" },
  130: { division: "division17" },
  144: { division: "division17" , branch: "branch22" },
  200: { division: "division50" }
};

// 把输入对象转成值数组（不需要原键）
const items = Object.values(input);

// 筛选出最详细的项：排除所有被其他层级更深的对象包含的项
const detailedItems = items.filter(item => {
  return !items.some(other => {
    if (item === other) return false;
    // 检查当前项的所有字段值都和other匹配
    const allFieldsMatch = Object.keys(item).every(key => other[key] === item[key]);
    // 检查other的字段数比当前项多（层级更深）
    const hasMoreDetails = Object.keys(other).length > Object.keys(item).length;
    return allFieldsMatch && hasMoreDetails;
  });
});

// 可选：把结果转回对象（键可以自定义，这里用索引）
const resultObj = detailedItems.reduce((acc, item, idx) => {
  acc[idx + 1] = item;
  return acc;
}, {});

console.log(resultObj);

逻辑说明 #

1. 转数组：先用Object.values把输入对象转成只包含层级数据的数组，抛弃原键。
2. 筛选核心：
   • 对每个项，用some检查是否存在另一个对象：既完全匹配当前项的所有字段值，又有更多的字段（意味着层级更深、信息更详细）。
   • 如果存在这样的对象，说明当前项是冗余的，直接过滤掉；否则保留。
3. 转回对象（可选）：如果需要对象格式，用reduce把筛选后的数组转回对象，键可以根据需求自定义（比如索引、层级字段拼接的唯一标识）。

内容的提问来源于stack exchange，提问作者pileup