Question

我有以下对象：

let res = {
   'name': {
      age: 10,
      density: 33,
      length: 2
    },
    'greet': {
       age: 33,
       density: 92,
       length: 3
    },
    'gyrt': {
        age: 91,
        density: 2,
        length: 47
     },
     .
     .
     .
}

其中有更多的键值对。我被要求按“排名”的降序返回键列表，该列表受以下约束控制：

年龄越大，排名越高
低密度单词的排名必须更高
长字必须排名较高

我很困惑如何实现这一目标。我正在尝试对对象进行排序，但这无济于事。我首先尝试使用age，然后使用density，然后使用length进行排序，但这显然不起作用，因为对象被一次又一次地重新排序，而忘记了先前的顺序。我该如何实现？

这是我正在尝试的：

let sortByAge = keys.sort((a, b) => Number(res[b].age) - Number(res[a].age));

let sortByDensity = keys.sort((a, b) => Number(res[b].density) - Number(res[a].density));

与length相同。我无法理解如何综合所有因素并对其进行排序。

Answer 1

您只需要在排序函数中放入一些条件即可检查年龄或密度是否相等，如果相等，则按下一组规则（年龄->密度->长度）进行排序：

let a={'name':{age:10,density:33,length:2},'greet':{age:33,density:92,length:3},'gyrt':{age:91,density:2,length:47},'foo':{age:91,density:3,length:47},'bar':{age:91,density:2,length:30},'baz':{age:10,density:2,length:47},'boo':{age:91,density:2,length:47},}

let sorted = Object.values(a);

sorted.sort((a,b) => {
  if(a.age === b.age) {
    if(a.density === b.density) {
      //sort by length
      return b.length - a.length;
    }
    //sort by density
    return a.density - b.density;
  }
  //sort by age
  return b.age - a.age;
});

console.log(sorted)

要对键进行排序（对值进行排序更好地显示其工作原理），只需将a和b替换为obj[a]和obj[b]

let obj={'name':{age:10,density:33,length:2},'greet':{age:33,density:92,length:3},'gyrt':{age:91,density:2,length:47},'foo':{age:91,density:3,length:47},'bar':{age:91,density:2,length:30},'baz':{age:10,density:2,length:47},'boo':{age:91,density:2,length:47},}

let sorted = Object.keys(obj);

sorted.sort((a,b) => {
  if(obj[a].age === obj[b].age) {
    if(obj[a].density === obj[b].density) {
      //sort by length
      return obj[b].length - obj[a].length;
    }
    //sort by density
    return obj[a].density - obj[b].density;
  }
  //sort by age
  return obj[b].age - obj[a].age;
});

console.log(sorted)

Answer 2

使用我介绍的here函数sortBy()，您可以创建一种将每个属性组合在一起的排序方法，如下所示：

const res = {name:{age:10,density:33,length:2},greet:{age:33,density:92,length:3},gyrt:{age:91,density:2,length:47},foo:{age:91,density:3,length:47},bar:{age:91,density:2,length:30},baz:{age:10,density:2,length:47},boo:{age:91,density:2,length:47}}

const sortBy = fn => (a, b) => -(fn(a) < fn(b)) || +(fn(a) > fn(b))
const pick = key => ({ [key]: value }) => value
const age = sortBy(pick('age'))
const density = sortBy(pick('density'))
const length = sortBy(pick('length'))
const sortFn = (a, b) =>
  -age(a[1], b[1]) ||
  density(a[1], b[1]) ||
  -length(a[1], b[1])

const keys = Object.entries(res).sort(sortFn).map(([key, value]) => key)

console.log(keys)

根据给定的标准对对象键进行排名

2 个答案: