Question

像年龄这样的对象数组有5条记录，基于sno我们想要第一个元素。

var ages=[
{name: "Jan", age: "90", sno: "1"},
{name: "Harry", age: "100", sno: "1"},
{name: "Tan", age: "10", sno: "2"},
{name: "Ron", age: "15", sno: "2"},
{name: "Roh", age: "19", sno: "2"},
]

输出：

[
{name: "Jan", age: "90", sno: "1"},
{name: "Tan", age: "10", sno: "2"},
]

我试图通过使用ES6 Javascript表达式解决这个问题，通过获取重复的sno，循环通过distinc sno并获取第一个元素。但我需要一个更好的方法。

是否可以在不使用任何内置函数的情况下实现它？

Answer 1

将Array.filter()与Set一起使用。在每次迭代时检查Set中是否存在sno。如果是，则返回false（删除项目），如果它没有添加到Set，并返回对Set的引用（将被转换为'true＆＃39;）。

＆＃13;

var ages = [{"name":"Jan","age":"90","sno":"1"},{"name":"Harry","age":"100","sno":"1"},{"name":"Tan","age":"10","sno":"2"},{"name":"Ron","age":"15","sno":"2"},{"name":"Roh","age":"19","sno":"2"}]

var result = ages.filter(function({ sno }) {
  return this.has(sno) ? false : this.add(sno)
}, new Set())

console.log(result)

＆＃13;

你可以使用for循环和辅助对象做同样的事情：

＆＃13;

var ages = [{"name":"Jan","age":"90","sno":"1"},{"name":"Harry","age":"100","sno":"1"},{"name":"Tan","age":"10","sno":"2"},{"name":"Ron","age":"15","sno":"2"},{"name":"Roh","age":"19","sno":"2"}]

var result = []
var helper = {}
var key

for(var i = 0; i < ages.length; i++) {
  item = ages[i]
  if(!helper[item.sno]) {
    helper[item.sno] = true
    result.push(item)
  }
}

console.log(result)

＆＃13;

Answer 2

您可以使用Set作为闭包，只过滤未知的sno。

var ages = [{ name: "Jan", age: "90", sno: "1" }, { name: "Harry", age: "100", sno: "1" }, { name: "Tan", age: "10", sno: "2" }, { name: "Ron", age: "15", sno: "2" }, { name: "Roh", age: "19", sno: "2" }],
    filtered = ages.filter((s => ({ sno }) => !s.has(sno) && s.add(sno))(new Set));
    
console.log(filtered);

Answer 3

通过reduce（由于some查找而不是非常高效）：

const ages =[
    {name: "Jan", age: "90", sno: "1"},
    {name: "Harry", age: "100", sno: "1"},
    {name: "Tan", age: "10", sno: "2"},
    {name: "Ron", age: "15", sno: "2"},
    {name: "Roh", age: "19", sno: "2"},
];

const result = ages.reduce((acc, x) => {
	if (acc.some(e => e.sno == x.sno) == false)
		acc.push(x);

	return acc;
}, []);

console.log(result);

Answer 4

晚会。

＆＃13;

var mem;
var result = [];
var ages=[
{name: "Jan", age: "90", sno: "1"}
, {name: "Harry", age: "100", sno: "1"}
, {name: "Tan", age: "10", sno: "2"}
, {name: "Ron", age: "15", sno: "2"}
, {name: "Roh", age: "19", sno: "2"}
];

mem = ages[0];
result.push(ages[0]);

for (var i=0; i<ages.length; i++) {
    
    if (mem.sno == ages[i].sno) {
        continue;
    } else {
        mem = ages[i];
        result.push(ages[i]);
    }
}

console.log(result);

＆＃13;

在javascript中从对象数组中删除重复属性对象的最佳方法是什么

4 个答案: