仅保留数组中每个对象中的选定键

Question

编辑：不是重复的，因为我的问题不要求删除特定键，而是在数组中找不到所有键。

在下面，函数从对象中redux1删除与未列出的键对应的条目。keys_to_keepdata

鉴于我有一个要保留的对象键列表，我如何redux1以更简洁的方式重写，最好使用map,filter或reduce?

var data = [
	{name: 'John', city: 'London', age: 42},
	{name: 'Mike', city: 'Warsaw', age: 18},
	{name: 'Jim', city: 'New York', age: 22},
	{name: 'Celine', city: 'Tokyo', age: 54},
]

var keys_to_keep = ['name', 'city']

function redux1(data) {
	data.forEach((person) => {
		Object.keys(person).forEach((key) => {
			if (!keys_to_keep.includes(key)) {
				delete (person[key])
			}
		})
	})
	console.log(data)
}

function redux2(data) {	
	var reduced = data.filter(person => Object.keys(person).filter(key => keys_to_keep.includes(key)))
	console.log(reduced)
}

redux1(data)
//redux2(data)

My current redux2 will return objects will not remove age.

Answer 1

You could use a combination of Array#map and Array#reduce:

const data = [
	{name: 'John', city: 'London', age: 42},
	{name: 'Mike', city: 'Warsaw', age: 18},
	{name: 'Jim', city: 'New York', age: 22},
	{name: 'Celine', city: 'Tokyo', age: 54},
]

const keys_to_keep = ['name', 'city'];

const redux = array => array.map(o => keys_to_keep.reduce((acc, curr) => {
  acc[curr] = o[curr];
  return acc;
}, {}));

console.log(redux(data));

Answer 2

Array#map使用and的版本比接受的答案稍短Object.fromEntries()：

const data = [
    {name: 'John', city: 'London', age: 42},
    {name: 'Mike', city: 'Warsaw', age: 18},
    {name: 'Jim', city: 'New York', age: 22},
    {name: 'Celine', city: 'Tokyo', age: 54},
]

const keys_to_keep = ['name', 'city'];

const redux1 = list => list.map(o => Object.fromEntries(
    keys_to_keep.map(k => [k, o[k]])
));

console.log(redux1(data));