使用 reduce 转换数组（属性未定义）答案

【问题标题】：Transform array using reduce (property undefined)使用 reduce 转换数组（属性未定义）
【发布时间】：2021-03-30 02:56:51
【问题描述】：

我想转换一个数组以重新组合同一国家/地区的所有城市。我有一个对象数组

locations = [
{
cities: ["Paris"]
country: "France"
},
{
cities: ["Marseille"]
country: "France"
},
{
cities: ["Kiev"]
country: "Ukraine"
},
{
cities: ["Odessa"]
country: "Ukraine"
}
...

我想要这个：

locations = [
{
cities: ["Paris", "Marseille"]
country: "France"
},
{
cities: ["Kiev, Odessa"]
country: "Ukraine"
},
...

所以，我的想法是：

locations.reduce((prev, curr) => {
    if (prev.country === curr.country) {
      prev.cities.push(curr.cities[0])
      // and delete curr Object
    }
  })

对于第一对 equals 国家，curr.cities[0] 被推送到 prev.cities，但不用于下一次迭代： TypeError：无法读取未定义的属性“国家/地区”

[![console.log(locations) - TypeError: Cannot read property 'country' of undefined][1]][1]

【问题讨论】：

您能否分享每个cities 数组中的值以及您在转换对象后尝试获得的结果？
你不会从 reduce 内部函数返回 prev。所以在下一次迭代中 prev 不存在。

标签： javascript arrays filter functional-programming reduce

【解决方案1】：

我认为要实现所需的输出格式，您可以执行以下操作：

locations = [
  { cities: ["Paris"], country: "France" },
  { cities: ["Marseille"], country: "France" },
  { cities: ["Kiev"], country: "Ukraine" },
  { cities: ["Odesa"], country: "Ukraine" }
];

var grouped = Object.entries(
  locations.reduce((a, {country, cities}) => ({
    ...a, [country]: [...(a[country] || []), ...cities]
  }), {})
).map(([country, cities]) => ({ country: country, cities: cities }), []);
  
console.log(grouped);

【讨论】：

【解决方案2】：

使用reduce时，需要返回prev以供下一次迭代使用，同时初始化为{}。此外，您需要注意每个国家/地区的初始添加，其中应存储具有 cities 数组的对象：

const locations = [
  { country: "France", cities: ['city1','city2'] },
  { country: "France", cities: ['city3','city4'] },
  { country: "Ukraine", cities: ['city5'] },
  { country: "Ukraine", cities: ['city6'] }
];

const res = Object.values(locations.reduce((prev, curr) => {
  const { country, cities = [] } = curr;
  const [city] = cities;
  if (prev[country] && city) {
    prev[country].cities.push(city);
  } else {
    prev[country] = { country, cities: city ? [city] : [] };
  }
  return prev;
}, {}));

console.log(res);

【讨论】：

【解决方案3】：

您可以创建一个以国家/地区为键的地图，并以最初为空的数组作为对应值。然后填充这些数组，最后从该 Map 中提取条目。这具有线性时间复杂度：

let locations = [{cities: ["Paris"],country: "France"},{cities: ["Marseille"],country: "France"},{cities: ["Kiev"],country: "Ukraine"},{cities: ["Odesa"], country: "Ukraine"}];

let map = new Map(locations.map(({country}) => [country, {country, cities: []}]));
for (let loc of locations) map.get(loc.country).cities.push(...loc.cities);
let res = Array.from(map.values());

console.log(res);

【讨论】：