基于公共键合并二维对象数组并求和basecount的优化方法

问题描述 #

原始二维对象数组：

[
  [ 
    { basecount: 343, campaign_name: '1.Stay_Connected_Plus:SP_JCA' },
    { basecount: 342, campaign_name: '1.Stay_Connected_Plus:JP_JCA' }
  ],
  [ 
    { basecount: 102, campaign_name: '1.Stay_Connected_Plus:SP_JCA' },
    { basecount: 102, campaign_name: '1.Stay_Connected_Plus:PP_JCA' }   
  ]
]

期望输出：

[ 
  { basecount: 445, campaign_name: '1.Stay_Connected_Plus:SP_JCA' },
  { basecount: 342, campaign_name: '1.Stay_Connected_Plus:JP_JCA' },
  { basecount: 102, campaign_name: '1.Stay_Connected_Plus:PP_JCA' }     
]

核心需求：当campaign_name相同时，累加对应的basecount；不同则保留原对象。已通过嵌套遍历、匹配campaign_name求和的方式实现，想寻求更高效的优化方案。

优化实现方案 #

可以利用对象映射+数组扁平化的方式实现，时间复杂度为O(n)（n为所有对象的总数），比嵌套遍历的O(n²)更高效：

const originalArray = [
  [ 
    { basecount: 343, campaign_name: '1.Stay_Connected_Plus:SP_JCA' },
    { basecount: 342, campaign_name: '1.Stay_Connected_Plus:JP_JCA' }
  ],
  [ 
    { basecount: 102, campaign_name: '1.Stay_Connected_Plus:SP_JCA' },
    { basecount: 102, campaign_name: '1.Stay_Connected_Plus:PP_JCA' }   
  ]
];

// 1. 扁平化二维数组为一维
const flatArray = originalArray.flat();

// 2. 用对象做映射，key为campaign_name，value为累加后的对象
const map = {};
flatArray.forEach(item => {
  if (map[item.campaign_name]) {
    map[item.campaign_name].basecount += item.basecount;
  } else {
    map[item.campaign_name] = {...item}; // 浅拷贝原对象，避免修改原始数据
  }
});

// 3. 将映射对象的值转为数组，得到最终结果
const result = Object.values(map);

console.log(result);

方案优势 #

• 扁平化数组后只需一次遍历即可完成累加，数据量越大，效率优势越明显；
• 用campaign_name作为键值，匹配过程是O(1)的哈希查找，避免嵌套遍历中的重复匹配；
• 浅拷贝原对象，不会修改原始数组的数据，保证数据独立性。

也可以用Array.reduce()简化写法，逻辑和上面一致：

const result = originalArray.flat().reduce((acc, curr) => {
  acc[curr.campaign_name] = acc[curr.campaign_name] 
    ? { ...curr, basecount: acc[curr.campaign_name].basecount + curr.basecount }
    : { ...curr };
  return acc;
}, {});

console.log(Object.values(result));

内容的提问来源于stack exchange，提问作者Atul Kumar