使用ES6 / lodash進行迭代和過濾

Question

我們有一系列對象，如：

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

以及對象中對象的屬性值數組

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

我們如何使用es6在names數組中找到名字的人，如：

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

Answer 1

ES6

使用帶謂詞的filter函數，並在其中檢查names數組中是否存在names 。

 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()來獲取所需的結果。

DEMO

 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))) 

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

Answer 3

我建議使用indexOf因為includes在IE瀏覽器中不起作用。 此外，使用{name}作為Destructuring賦值，它將保存對象的name屬性的值。

 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)) 

Answer 4

lodash

你可以嘗試以下

 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

如果你想在執行它n復雜，這里是一個辦法做到這一點：

1. 使用鍵創建一個地圖，將人的姓名和值作為人物的對象。
2. 映射條件數組並從步驟1中的地圖創建中提取人物對象。

這是一個有效的演示：

 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); 

注意：此解決方案假定源數組中只有唯一的人名。 需要調整它來處理重復項。

Answer 6

有關詳細信息，請參閱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)