如何基于对象时间字段映射数组生成嵌套数组结构？

你想要的效果完全可以通过嵌套的map函数实现，不用局限于只处理单个字段。核心思路是先确定要分组的时间字段列表，然后针对每个字段，遍历原数组生成对应的城市-时间对象数组。

直接实现方案（硬编码字段） #

这是最直观的写法，适合字段固定的场景：

const list = [
  { city: "new york", current_time: "123", time1: "456", time2: "789" },
  { city: "london", current_time: "123", time1: "456", time2: "789" },
  { city: "tokyo", current_time: "123", time1: "456", time2: "789" }
];

// 定义需要分组的时间字段顺序
const timeFields = ['current_time', 'time1', 'time2'];

// 外层map遍历每个时间字段，内层map遍历每个城市生成目标对象
const result = timeFields.map(field => {
  return list.map(item => ({
    city: item.city,
    time: item[field]
  }));
});

console.log(result);

动态适配字段方案（无需硬编码） #

如果以后可能新增time3、time4这类字段，可以动态提取时间字段（排除city字段），让代码更灵活：

const list = [
  { city: "new york", current_time: "123", time1: "456", time2: "789" },
  { city: "london", current_time: "123", time1: "456", time2: "789" },
  { city: "tokyo", current_time: "123", time1: "456", time2: "789" }
];

// 自动提取所有非city的字段作为时间字段
const timeFields = Object.keys(list[0]).filter(key => key !== 'city');

const result = timeFields.map(field => 
  list.map(item => ({ city: item.city, time: item[field] }))
);

console.log(result);

关于forEach的替代写法 #

当然用forEach也能实现，但代码会稍显繁琐，因为需要手动创建数组并push元素：

const list = [
  { city: "new york", current_time: "123", time1: "456", time2: "789" },
  { city: "london", current_time: "123", time1: "456", time2: "789" },
  { city: "tokyo", current_time: "123", time1: "456", time2: "789" }
];

const result = [];
const timeFields = ['current_time', 'time1', 'time2'];

timeFields.forEach(field => {
  const group = [];
  list.forEach(item => {
    group.push({ city: item.city, time: item[field] });
  });
  result.push(group);
});

console.log(result);

对比下来，嵌套map的写法更简洁，也符合函数式编程的风格，推荐优先使用。

内容的提问来源于stack exchange，提问作者jando