Question

我们有一系列对象，如：

const persons = [{
    name: "john",
    age: 23
}, {
    name: "lisa",
    age: 43
}, {
    name: "jim",
    age: 101
}, {
    name: "bob",
    age: 67
}];

对象

中对象的属性值数组

const names = ["lisa", "bob"]

我们如何使用es6在names数组中找到名字的人，比如：

const filteredPersons = [{
    name: "lisa",
    age: 43
}, {
    name: "bob",
    age: 67
}];

Answer 1

ES6

将filter函数与谓词一起使用，并在其中检查names数组中是否存在名称。

const persons = [
    {name: "john", age:23},
    {name: "lisa", age:43},
    {name: "jim", age:101},
    {name: "bob", age:67}
];

const names = ["lisa", "bob"];

const filtered = persons.filter(person => names.includes(person.name));

console.log(filtered);

Answer 2

您可以使用filter()和inlcudes()来获取所需的结果。

<强>样本

＆＃13;

const persons = [{
  name: "john",
  age: 23
}, {
  name: "lisa",
  age: 43
}, {
  name: "jim",
  age: 101
}, {
  name: "bob",
  age: 67
}];
const names = ["lisa", "bob"];

console.log(persons.filter(({
  name
}) => names.includes(name)))

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

Answer 3

我建议使用indexOf，因为includes在IE浏览器中不起作用。此外，使用{name}作为解构赋值，它将保存对象的name属性值。

＆＃13;

const persons = [{
  name: "john",
  age: 23
}, {
  name: "lisa",
  age: 43
}, {
  name: "jim",
  age: 101
}, {
  name: "bob",
  age: 67
}];
const names = ["lisa", "bob"];

console.log(persons.filter(({name}) => names.indexOf(name) !== -1))

＆＃13;

Answer 4

lodash

您可以尝试以下

const persons = [{name: "john", age: 23},
    {name: "lisa",age: 43}, 
    {name: "jim", age: 101}, 
    {name: "bob",age: 67}];


const names = ["lisa", "bob"]


const filteredPersons = _.filter(persons, function(person) {
    return _.indexOf(names, person.name) !== -1;
});

console.log(filteredPersons);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.1/lodash.js"></script>

Answer 5

如果您想以n复杂度执行此操作，可以使用以下方法：

使用关键字创建一个地图作为人物的名称和值作为人物的对象。
映射条件数组并从步骤1中创建的地图中提取人物对象。

这是一个有效的演示：

＆＃13;

const persons = [{
  name: "john",
  age: 23
}, {
  name: "lisa",
  age: 43
}, {
  name: "jim",
  age: 101
}, {
  name: "bob",
  age: 67
}];

const names = ["lisa", "bob"];
const map = persons.reduce((acc, item) => {
  acc[item.name] = item;
  return acc;
}, {});
const result = names.map(name => map[name]);
console.log(result);

＆＃13;

注意：此解决方案假定源数组中只有唯一的人名。需要调整它来处理重复项。

Answer 6

有关详细信息，请参阅Closures，Set和Array.prototype.filter()。

// Input.
const persons = [{name: "john",age: 23}, {name: "lisa",age: 43}, {name: "jim",age: 101}, {name: "bob",age: 67}]
const names = ["lisa", "bob"]

// Filter.
const filter = (A, B) => (s => A.filter(x => s.has(x.name)))(new Set(B))

// Output.
const output = filter(persons, names)

// Proof.
console.log(output)

使用ES6 / lodash进行迭代和过滤

6 个答案: