Question

我有两个数组，第一个包含对象，第二个包含id。我想从第一个数组返回一个新数组，该数组具有第一个数组的ID。最佳和有效的方法是什么？

const firstArr = [{id: 1, city: London}, {id: 5, city: 'Berlin'}, {id: 10, city: 'Paris'}, {id: 2, city: 'Rome'}]

const secondArr = ['2', '5']

const wantedArr = [{id: 2, city: 'Rome'}, {id: 5, city: 'Berlin'}]

Answer 1

对于线性时间复杂度，将第二个数组转换为set，然后使用Set.prototype.has

const firstArr = [{id: 1, city: 'London'}, {id: 5, city: 'Berlin'}, {id: 10, city: 'Paris'}, {id: 2, city: 'Rome'}]
const secondArr = ['2', '5'];

let set = new Set(secondArr);
const res = firstArr.filter(x => set.has(String(x.id)));

console.log(res)

如果要根据secondArr保持结果数组的顺序，请首先从firstArr创建对象，然后在map()上使用secondArr

const firstArr = [{id: 1, city: 'London'}, {id: 5, city: 'Berlin'}, {id: 10, city: 'Paris'}, {id: 2, city: 'Rome'}]
const secondArr = ['2', '5'];

const obj = firstArr.reduce((ac,a) => (ac[a.id] = a,ac), {});

const res = secondArr.map(x => obj[x]);
console.log(res)

Answer 2

您可以使用.indexOf()或.includes()方法实现此目标

const firstArr = [{ id: 2, city: 'London' }, { id: 2, city: 'Tokyo' }, { id: 5, city: 'Berlin' }, { id: 10, city: 'Paris' }, { id: 6, city: 'Rome' }];

const secondArr = ['2', '5'];

const output = firstArr.filter((r) => {
  return secondArr.includes(`${r.id}`);
});

console.log(output);

const firstArr = [{ id: 2, city: 'London' }, { id: 2, city: 'Tokyo' }, { id: 5, city: 'Berlin' }, { id: 10, city: 'Paris' }, { id: 6, city: 'Rome' }];

const secondArr = ['2', '5'];

const output = firstArr.filter((r) => {
  return secondArr.indexOf(`${r.id}`) > -1;
});

console.log(output);

Answer 3

这应该可以解决问题：

secondArr.map(e => firstArr.filter(a => a.id == +e))

我正在使用+e将字符串转换为整数，这可能不是最好的方法，但肯定是最短的方法。

Answer 4

const wantedArr = firstArr.filter(({ id }) => secondArr.includes(`${id}`));

如何基于另一个数组中的值来过滤对象数组

4 个答案: