下面，函数redux1删除与未列出的键相对应的条目keys_to_keep from data object.
鉴于我有一个要保留的对象键列表，我该如何重写redux1以更清洁的方式，最好利用map, filter or 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)
我现在的redux2将返回对象不会删除age.
您可以使用以下组合Array#map https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map and Array#reduce https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/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));
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