JavaScript中如何过滤数组获取每个用户的最高百分比值

实现思路 #

• 遍历数组时以userId为唯一标识建立临时映射
• 对同一用户的多条记录，仅保留percent数值最大的条目
• 遍历结束后将映射转换为结果数组

代码实现 #

方案1：保留完整原对象结构 #

let data = [ 
  {
    "userId": "1",
    "percent": 97.58,
  },
  {
    "userId": "1",
    "percent": 92.01,
  },
  {
    "userId": "2",
    "percent": 91.64,
  },
  {
    "userId": "2",
    "percent": 91.64,
  },
  {
    "userId": "3",
    "percent": 91.64,
  }
];

const maxPerUser = [];
const userMaxMap = new Map();

data.forEach(item => {
  const currentMax = userMaxMap.get(item.userId);
  if (!currentMax || item.percent > currentMax.percent) {
    userMaxMap.set(item.userId, item);
  }
});

maxPerUser.push(...userMaxMap.values());
console.log(maxPerUser);

输出结果

[
  { userId: '1', percent: 97.58 },
  { userId: '2', percent: 91.64 },
  { userId: '3', percent: 91.64 }
]

方案2：仅获取用户ID和对应最高percent值 #

const userPercentMap = data.reduce((acc, curr) => {
  acc[curr.userId] = Math.max(acc[curr.userId] || -Infinity, curr.percent);
  return acc;
}, {});

console.log(userPercentMap);

输出结果

{ '1': 97.58, '2': 91.64, '3': 91.64 }

内容的提问来源于stack exchange，提问作者farzanmanafi