Question

我有这样的数组：

people = [
{id:1,name:'abc',age:18},
{id:2,name:'abcd',age:20},
{id:3,name:'abce',age:18},
{id:4,name:'abcf',age:18},
{id:5,name:'abcg',age:21},
{id:6,name:'abch',age:20},
{id:7,name:'abci',age:17},
]

现在，我想按JavaScript中的出现频率（年龄）对数组顺序进行排序。输出如下：

people = [
{id:1,name:'abc',age:18},
{id:3,name:'abce',age:18},
{id:4,name:'abcf',age:18},
{id:2,name:'abcd',age:20},
{id:6,name:'abch',age:20},
{id:5,name:'abcg',age:21},
{id:7,name:'abci',age:17},
]

那我该怎么做？

我尝试按照Nina Scholz的回答进行操作，但是我遇到了这样的错误。

var people = [{ id: 1, name: 'abc', age: 18 }, { id: 2, name: 'abcd', age: 20 }, { id: 3, name: 'abce', age: 18 }, { id: 4, name: 'abcf', age: 18 }, { id: 5, name: 'abcg', age: 21 }, { id: 6, name: 'abch', age: 20 }, { id: 7, name: 'abcg', age: 21 }, { id: 8, name: 'abci', age: 17 }],
count = people.reduce((r, { age }) => (r[age] = (r[age] || 0) + 1, r), {});

people.sort((a, b) => count[b.age] - count[a.age]);

console.log(people);

输出为：

people = [
{id:1,name:'abc',age:18},
{id:3,name:'abce',age:18},
{id:4,name:'abcf',age:18},
{id:2,name:'abcd',age:20},
{id:5,name:'abcg',age:21},
{id:6,name:'abch',age:20},
{id:5,name:'abcg',age:21},
{id:7,name:'abci',age:17},
]

Answer 1

只需计数年龄，然后按计数降序即可。

var people = [{ id: 1, name: 'abc', age: 18 }, { id: 2, name: 'abcd', age: 20 }, { id: 3, name: 'abce', age: 18 }, { id: 4, name: 'abcf', age: 18 }, { id: 5, name: 'abcg', age: 21 }, { id: 6, name: 'abch', age: 20 }, { id: 7, name: 'abci', age: 17 }],
    count = people.reduce((r, { age }) => (r[age] = (r[age] || 0) + 1, r), {});

people.sort((a, b) => count[b.age] - count[a.age]);

console.log(people);

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

要按count 和 age进行排序，请添加logical OR ||以及年龄段。

   var people = [{ id: 1, name: 'abc', age: 18 }, { id: 2, name: 'abcd', age: 20 }, { id: 3, name: 'abce', age: 18 }, { id: 4, name: 'abcf', age: 18 }, { id: 5, name: 'abcg', age: 21 }, { id: 6, name: 'abch', age: 20 }, { id: 7, name: 'abcg', age: 21 }, { id: 8, name: 'abci', age: 17 }],
    count = people.reduce((r, { age }) => (r[age] = (r[age] || 0) + 1, r), {});

people.sort((a, b) =>
    count[b.age] - count[a.age] ||
    b.age - a.age
);

console.log(people);

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

Answer 2

尝试一下：

let people = [{ id: 1, name: 'abc', age: 18 }, { id: 2, name: 'abcd', age: 20 }, { id: 3, name: 'abce', age: 18 }, { id: 4, name: 'abcf', age: 18 }, { id: 5, name: 'abcg', age: 21 }, { id: 6, name: 'abch', age: 20 }, { id: 7, name: 'abci', age: 17 }];
let tmpAge = {};
for (let i = 0; i < people.length; i++) {
  if (!tmpAge[people[i].age]) {
    tmpAge[people[i].age] = 1;
  } else {
    tmpAge[people[i].age]++;
  }
}
people.sort((a, b) => tmpAge[b.age] - tmpAge[a.age]);
console.log(people);

Answer 3

const people = [
  { id: 1, name: 'abc', age: 18 },
  { id: 2, name: 'abcd', age: 20 },
  { id: 3, name: 'abce', age: 18 },
  { id: 4, name: 'abcf', age: 18 },
  { id: 5, name: 'abcg', age: 21 },
  { id: 6, name: 'abch', age: 20 },
  { id: 7, name: 'abci', age: 17 },
];

const agesFrequency = people.reduce((previousValue, currentValue) => {
  const { age } = currentValue;

  return {
    ...previousValue,
    [age]: (
      previousValue[age] || 0
    ) + 1,
  };
}, {});

people.sort((
  (a, b) =>
    agesFrequency[b.age] - agesFrequency[a.age]
));

console.log(people);

Answer 4

我解决了。将功能排序更改为

java.util.Supplier

_{代表OP发布}

如何按出现频率对数组排序？

4 个答案: