使用Javascript查找具有最低关键值的对象 [英] Finding object with lowest value for some key, in Javascript

问题描述

在我的Javascript程序中，我有一个Person对象的列表.

In my Javascript program, I have a list of Person objects.

例如

[
     "Michael": {
          "age": 45,
          "position": "manager",
          ...
     },
    "Dwight": {
          "age": 36,
          "position": "assistant manager",
          ...
     },
    ....
]

我想找到最小的Person.

我通过创建两个数组来实现此目的:创建所有Person中的一个，并创建其所有年龄中的一个，并获取最低年龄的索引并将其应用于第一个数组.喜欢:

I've accomplished this by creating two arrays: one of all the Persons and one of all their ages, and getting the index of the lowest age and applying it to the first array. Like:

var arrayOfPersons = [persons[0], persons[1], ....];
var arrayOfAges = [persons[0].age, persons[1].age, ....];
var min = arrayOfAges.indexOf(Math.max.apply(Math, arrayOfAges));
var youngestPerson = arrayOfPerson[min];

问题在于效率低下似乎不是最好的方法.同样，它也没有处理可能为最小的孩子打成平手的事实.

The problem with this is it is inefficient doesn't seem like the best way. Also it doesn't deal with the fact that there may be a tie for youngest.

有人知道一种更本地，更简单的方法吗?

Does anyone know of a more native, simpler way to do this?

推荐答案

您可以按age属性对person数组进行排序，然后选择第一个:

You can sort the persons array by age property and pick first one:

var persons = [
    { name: 'Michael', age: 45, position: 'manager' },
    { name: 'Dwight', age: 36, position: 'assistant manager' },
    { name: 'Foo', age: 99, position: 'foo' },
    { name: 'Bar', age: 37, position: 'bar' }
];

persons.sort(function(a, b) {
    return a.age > b.age;
});

console.log('Youngest person: ', persons[0]);

这篇关于使用Javascript查找具有最低关键值的对象的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！