基于javascript中的另一个数组过滤对象数组答案

【问题标题】：Filter array of objects based on another array in javascript基于javascript中的另一个数组过滤对象数组
【发布时间】：2018-04-04 07:06:43
【问题描述】：

给定一个对象数组：

people = [
    {id: "1", name: "abc", gender: "m", age:"15" },
    {id: "2", name: "a", gender: "m", age:"25" },
    {id: "3", name: "efg", gender: "f", age:"5" },
    {id: "4", name: "hjk", gender: "m", age:"35" },
    {id: "5", name: "ikly", gender: "m", age:"41" },
    {id: "6", name: "ert", gender: "f", age:" 30" },
    {id: "7", name: "qwe", gender: "f", age:" 31" },
    {id: "8", name: "bdd", gender: "m", age:" 78" },
]

还有一个想要的 id 数组：

id_filter = [1,4,5,8]

如何过滤数组people 以返回id_filter 和gender=m 中定义的目标ID？

【问题讨论】：

请使用const或let来声明变量，例如const people = []。

标签： javascript arrays filter

【解决方案1】：

在这种情况下，步行更有意义所以首先你需要遍历你的 people 数组然后你检查一个人的 id 是否等于你的过滤器列表

for(person in people) {
     for(id in id_filter) {
         if(person[id] == id && person[gender] == "m"){

         }
     }
}

【讨论】：

我想要一个过滤数组，而且 id 不是串行的。还需要添加另一个条件，即性别 = m

【解决方案2】：

最简单的过滤方法是使用数组的filter()函数，类似这样：

people.filter(person => id_filter.includes(person.id))

【讨论】：

如果我想添加另一个条件，即性别应该是 == m 怎么办？
people.filter((person) => person.gender == 'm')
真的是people.filter((person) => id_filter.includes(Number(person.id)) && person.gender === "m") 或people.filter(({ id, gender }) => id_filter.includes(Number(id)) && gender === "m")。

【解决方案3】：

    people = [
        {id : "1", name : "abc", gender : "m", age :"15" },
        {id : "2", name : "a", gender : "m", age :"25" },
        {id : "3", name : "efg", gender : "f", age :"5" },
        {id : "4", name : "hjk", gender : "m", age :"35" },
        {id : "5", name : "ikly", gender : "m", age :"41" },
        {id : "6", name : "ert", gender : "f", age :" 30" },
        {id : "7", name : "qwe", gender : "f", age :" 31" },
        {id : "8", name : "bdd", gender : "m", age :" 78" }
    ]
    var id_filter = ["1","4","5","8"], filteredPeople = []; 
    for( var i=people.length-1; i>=0; --i){ 
      if( id_filter.indexOf( people[i].id ) != -1 ){ 
        filteredPeople.push( people[i] ); 
      } 
    }
    console.log( filteredPeople );

【讨论】：

【解决方案4】：

带有Array.includes()功能：

var people = [
    {id : "1", name : "abc", gender : "m", age :"15" }, {id : "2", name : "a", gender : "m", age :"25" },
    {id : "3", name : "efg", gender : "f", age :"5" },  {id : "4", name : "hjk", gender : "m", age :"35" },
    {id : "5", name : "ikly", gender : "m", age :"41" }, {id : "6", name : "ert", gender : "f", age :" 30" },
    {id : "7", name : "qwe", gender : "f", age :" 31" }, {id : "8", name : "bdd", gender : "m", age :" 78" }
], 
    id_filter = [1,4,5,8],
    result = people.filter((o) => id_filter.includes(+o.id) && o.gender == "m");       
	
console.log(result);

+o.id - + 在这里用于将数字字符串转换为数字

【讨论】：

【解决方案5】：

您可以使用Array.prototype.filter 喜欢：

function filter(arr, ids, gender) {                              // takes an array of people arr, an array of indexes ids, and a gender and return the matched people objects from arr
  return arr.filter(function(obj) {                              // filtering each object...
    return ids.includes(obj.id) && obj.gender === gender;        // if this object is is included in the ids array and if its gender property is equal to the desired gender
  });
}

var people = [{id:"1",name:"abc",gender:"m",age:"15"},{id:"2",name:"a",gender:"m",age:"25"},{id:"3",name:"efg",gender:"f",age:"5"},{id:"4",name:"hjk",gender:"m",age:"35"},{id:"5",name:"ikly",gender:"m",age:"41"},{id:"6",name:"ert",gender:"f",age:"30"},{id:"7",name:"qwe",gender:"f",age:"31"},{id:"8",name:"bdd",gender:"m",age:"78"}];

console.log(filter(people, ["5", "7", "4"], "m"));               // filtering elements where id is one of ["5", "7", "4"] and the gender is "m".

注意：people 中对象的id 属性是字符串，因此您必须向filter 提供字符串ID 数组或将id 属性转换为传递给includes之前的编号。

【讨论】：

【解决方案6】：

const people = [
    {id: "1", name: "abc", gender: "m", age: "15" },
    {id: "2", name: "a", gender: "m", age: "25" },
    {id: "3", name: "efg", gender: "f", age: "5" },
    {id: "4", name: "hjk", gender: "m", age: "35" },
    {id: "5", name: "ikly", gender: "m", age: "41" },
    {id: "6", name: "ert", gender: "f", age: " 30" },
    {id: "7", name: "qwe", gender: "f", age: " 31" },
    {id: "8", name: "bdd", gender: "m", age: " 78" },
]

const idFilter = [1,4,5,8]

const idIsInList = id => idFilter.includes(+id) // "+id" to make sure it is a number, not a string
const genderIsMale = gender => gender === "m"
const result = people.filter(item => idIsInList(item.id) && genderIsMale(item.gender))

console.log(result)

【讨论】：

【解决方案7】：

您可以使用 array.filter() 在几个条件下获得您想要的输出。我也更正了你的 JSON。

var filtered = people.filter(function(item) {
        return id_filter.indexOf(item.id) !== -1 && item.gender==='m';
});

演示

var  people =[
  { "id": 1, "name": "abc", "gender": "m","age": "15" },
  { "id": 2, "name": "a", "gender": "m", "age": "25"  },
  { "id": 3,"name": "efg", "gender": "f","age": "5" },
  { "id": 4,"name": "hjk","gender": "m","age": "35" },
  {  "id": 5, "name": "ikly","gender": "m","age": "41" },
  { "id": 6, "name": "ert", "gender": "f", "age": "30" },
  { "id": 7, "name": "qwe", "gender": "f", "age": "31" },
  { "id":8, "name": "bdd",  "gender": "m", "age": " 8" }
];
var id_filter = [1,4,5,8];
var filtered = people.filter(function(item) {
    return id_filter.indexOf(item.id) !== -1 && item.gender==='m';
});
console.log(filtered);

【讨论】：

【解决方案8】：

您可以通过以下代码实现：

const filtered_people = people.filter(function(person){
    return id_filter.includes(person.id) && person.gender === 'm';
});

只要确保每个人的 id 是一个整数而不是一个字符串，就像你的例子一样。否则，includes() 函数将不匹配。此外，您的 people 数组存在内部语法问题。所以，最终的代码应该是这样的：

const people = [
    {id: 1, name: "abc", gender: "m", age:15},
    {id: 2, name: "a", gender: "m", age: 25},
    {id: 3, name: "efg", gender: "f", age: 5},
    {id: 4, name: "hjk", gender: "f", age: 35},
    {id: 5, name: "ikly", gender: "m", age: 41},
    {id: 6, name: "ert", gender: "f", age: 30},
    {id: 7, name: "qwe", gender: "f", age: 31},
    {id: 8, name: "bdd", gender: "m", age: 78}
]
const id_filter = [1,4,5,8]
const filtered_people = people.filter((person) => id_filter.includes(person.id) && person.gender === 'm')
console.log(filtered_people)

我希望这对你有帮助。祝你好运。

【讨论】：

【解决方案9】：

对于这种情况你可以使用filter和include函数，因为你的id是字符串，使用前需要解析。

var result = people.filter((person) => (id_filter.includes(parseInt(person.id)) && person.gender ==='m'))

【讨论】：

【解决方案10】：

如果您的id_filter 很大，您需要先将其转换为new Set()。这将允许恒定时间查找。然后，您可以使用 .filter() 迭代您的 people 数组，如果您在其中设置了 .has() id 并且性别等于 'm'，则返回 true。

const people = [ {id: "1", name: "abc", gender: "m", age:"15" }, {id: "2", name: "a", gender: "m", age:"25" }, {id: "3", name: "efg", gender: "f", age:"5" }, {id: "4", name: "hjk", gender: "m", age:"35" }, {id: "5", name: "ikly", gender: "m", age:"41" }, {id: "6", name: "ert", gender: "f", age:" 30" }, {id: "7", name: "qwe", gender: "f", age:" 31" }, {id: "8", name: "bdd", gender: "m", age:" 78" }, ];

const id_filter = new Set([1,4,5,8]);
const res = people.filter(({id, gender}) => id_filter.has(+id) && gender === 'm');
console.log(res);

总体而言，这种方法的时间复杂度为O(N + k)，而不是使用.includes() 或.indexOf() 方法时得到的O(Nk)，其中N 是@ 的长度987654337@数组，k是id_filter数组的长度

【讨论】：