如何将对象数组转换为按部门合并名称的短数组？

原始数组 #

[
 { 
  department:"Reviewer",
  name:"John"
 },
 { 
  department:"Reviewer",
  name:"Carol"
 },
 { 
  department:"Reviewer",
  name:"June"
 },
 { 
  department:"Assistant Reviewer",
  name:"Mac"
 },
 { 
  department:"Assistant Reviewer",
  name:"Issac"
 }
]

期望转换结果 #

[
     { 
      department:"Reviewer",
      name:"John;Carol;June;"
     },
     { 
      department:"Assistant Reviewer",
      name:"Mac;Issac;"
     },
]

你的代码问题分析 #

你之前的写法存在两处关键问题：

1. new Set(obj.department) 语法错误，obj是数组，无法直接取department属性，应该先遍历数组提取每个元素的department值
2. map回调里的逻辑没有关联部门与对应名称的关系，也未实现同部门名称的累加拼接

正确实现方法 #

方法一：使用reduce高效分组（推荐）

const arr = [
 { department:"Reviewer", name:"John" },
 { department:"Reviewer", name:"Carol" },
 { department:"Reviewer", name:"June" },
 { department:"Assistant Reviewer", name:"Mac" },
 { department:"Assistant Reviewer", name:"Issac" }
];

// 按部门分组，收集对应名称数组
const groupedData = arr.reduce((accumulator, current) => {
  // 若当前部门未在分组中，初始化空数组
  if (!accumulator[current.department]) {
    accumulator[current.department] = [];
  }
  // 将当前名称加入对应部门的数组
  accumulator[current.department].push(current.name);
  return accumulator;
}, {});

// 将分组对象转换为目标数组格式
const mergedResult = Object.entries(groupedData).map(([dept, names]) => ({
  department: dept,
  name: names.join(';') + ';' // 拼接名称并添加末尾分号
}));

console.log(mergedResult);

方法二：先取唯一部门再拼接名称（逻辑直观）

const arr = [/* 原始数组 */];

// 获取所有唯一部门
const uniqueDepartments = [...new Set(arr.map(item => item.department))];

// 遍历每个部门，筛选并拼接对应名称
const mergedResult = uniqueDepartments.map(dept => ({
  department: dept,
  name: arr
           .filter(item => item.department === dept)
           .map(item => item.name)
           .join(';') + ';'
}));

两种方法都能实现需求，方法一的reduce只遍历数组一次，性能更优；方法二逻辑更易懂，适合数据量较小的场景。

内容的提问来源于stack exchange，提问作者C Sharper