如何将嵌套对象数组中的ID替换为匹配对象（10k+数据场景）

现有包含10000+条记录的对象数组obj1，其中每个对象的technologies字段为ID字符串数组。需要遍历该数组，将每个ID与对象数组obj2的id字段匹配，用匹配到的obj2对象替换原ID，生成目标数组。

示例数据 #

obj1 #

obj1 = [
  {
    "control_id": "red1234",
    "security_domain": "astrem",
    "control_statement": "testing",
    "descriptio": "test",
    "label": "minimal",
    "technologies": ["180"],
    "reference": {"string": "string"},
    "evaluation": "",
    "category": null
  },
  {
    "control_id": "red1234",
    "security_domain": "astrem",
    "control_statement": "testing",
    "descriptio": "test",
    "label": "minimal",
    "technologies": ["180", "320","3213"],
    "reference": {"string": "string"},
    "evaluation": "",
    "category": null
  }
]

obj2 #

obj2 = [
  {"id": 94,"name": "SUSE Linux Enterprise 12.x"},
  {"id": 174,"name": "Ubuntu 18.x"},
  {"id": 106,"name": "Windows 2016 Server"},
  {"id": 180,"name": "Windows 2019 Server"},
  {"id": 53,"name": "Windows 2012 Server"},
  {"id": 217,"name": "Red Hat Enterprise Linux 8.x"},
  {"id": 81,"name": "Red Hat Enterprise Linux 7.x"},
  {"id": 109,"name": "Amazon Linux AMI"},
  {"id": 126,"name": "Amazon Linux 2 AMI"},
  {"id": 0,"name": "All OSs"},
  {"id": 1,"name": "Windows Platforms"},
  {"id": 2,"name": "Linux Platforms"},
  {"id": 320,"name": "Windows 2012 Server"},
  {"id": 3213,"name": "Windows 1999 Server"}
]

目标输出 #

obj1 = [
  {
    "control_id": "red1234",
    "security_domain": "astrem",
    "control_statement": "testing",
    "descriptio": "test",
    "label": "minimal",
    "technologies": [{"id": 180,"name": "Windows 2019 Server"}],
    "reference": {"string": "string"},
    "evaluation": "",
    "category": null
  },
  {
    "control_id": "red1234",
    "security_domain": "astrem",
    "control_statement": "testing",
    "descriptio": "test",
    "label": "minimal",
    "technologies": [
      {"id": 180,"name": "Windows 2019 Server"},
      {"id": 320,"name": "Windows 2012 Server"},
      {"id": 3213,"name": "Windows 1999 Server"}
    ],
    "reference": {"string": "string"},
    "evaluation": "",
    "category": null
  }
]

尝试的错误代码 #

obj1.map(element => element.technologies.forEach((techArrayList, index) => 
     this.operatingSystem.find(o => {
      if (techArrayList == o.id) {
         obj1[element.technologies].technologies[index].replace(o);
      }
     })));

错误分析 #

1. forEach无返回值，在map中使用会导致最终结果数组全是undefined
2. obj1[element.technologies]写法错误：element.technologies是数组，不能作为索引访问obj1
3. 数组元素不能用replace方法替换，需直接赋值
4. 每次用find遍历obj2查找匹配项，在10000+数据场景下效率极低

高效实现方案 #

先将obj2转换为映射表（Map），把id转为字符串作为键（因为obj1中的ID是字符串类型），对应对象作为值，这样查找匹配项的时间复杂度为O(1)，大幅提升处理效率。之后遍历obj1，替换每个对象的technologies字段即可。

// 构建技术ID到对象的映射表
const techMap = new Map(obj2.map(item => [String(item.id), item]));

// 生成目标数组，不修改原obj1
const targetArray = obj1.map(item => ({
  // 复制原对象的所有属性
  ...item,
  // 将technologies的ID字符串替换为匹配的obj2对象
  technologies: item.technologies
    .map(techId => techMap.get(techId))
    // 可选：过滤掉找不到匹配项的元素，若不需要可删除此句
    .filter(Boolean)
}));

代码说明 #

• 使用Map构建映射表：仅遍历obj2一次，后续查找无需重复遍历
• 用扩展运算符...item复制原对象，避免修改原数组obj1
• technologies.map将每个ID字符串转换为对应的对象
• filter(Boolean)用于过滤找不到匹配项的ID（如果存在无效ID的情况）

内容的提问来源于stack exchange，提问作者Lanka