Question

我有以下源数组：

const list = [
    {
        students: [ 'peter', 'bob', 'john']
    },
    {
        students: [ 'thomas', 'sarah', 'john']
    },
    {
        students: [ 'john', 'sarah', 'jack']
    }
];

我想获得唯一的学生姓名和人数，最终结果应该是：

{
    'john': 3,
    'sarah': 2,
    'thomas': 1,
    'jack': 1,
    'peter': 1,
    'bob': 1
}

这是我的尝试：

const unique = list.reduce(function(total, curr){
  const students = curr.students;
  for (c of students) {
    if (!total[c]) {
      total[c] = 1
    } else {
      total[c] += 1;
    }
  }
  return total;

}, {});

有更好的方法吗？还是更快更清洁的方式？谢谢

Answer 1

我先将数组展平，然后用reduce进行计数：

const list = [
    {
        students: [ 'peter', 'bob', 'john']
    },
    {
        students: [ 'thomas', 'sarah', 'john']
    },
    {
        students: [ 'john', 'sarah', 'jack']
    }
];


const allStudents = list.flatMap(({ students }) => students);
const count = allStudents.reduce((a, name) => {
  a[name] = (a[name] || 0) + 1;
  return a;
}, {});
console.log(count);

如果您还希望对属性进行排序，则将对象的Object.entries进行排序，然后使用Object.fromEntries将其变回对象：

const list = [
    {
        students: [ 'peter', 'bob', 'john']
    },
    {
        students: [ 'thomas', 'sarah', 'john']
    },
    {
        students: [ 'john', 'sarah', 'jack']
    }
];


const allStudents = list.flatMap(({ students }) => students);
const count = allStudents.reduce((a, name) => {
  a[name] = (a[name] || 0) + 1;
  return a;
}, {});
const sorted = Object.fromEntries(
  Object.entries(count).sort((a, b) => b[1] - a[1])
);
console.log(sorted);

如果您的环境不支持flatMap或fromEntries，请使用polyfill或通过其他方法展平/分组：

const list = [
    {
        students: [ 'peter', 'bob', 'john']
    },
    {
        students: [ 'thomas', 'sarah', 'john']
    },
    {
        students: [ 'john', 'sarah', 'jack']
    }
];


const allStudents = [].concat(...list.map(({ students }) => students));
const count = allStudents.reduce((a, name) => {
  a[name] = (a[name] || 0) + 1;
  return a;
}, {});
const sortedEntries = Object.entries(count).sort((a, b) => b[1] - a[1]);
const sortedObj = sortedEntries.reduce((a, [prop, val]) => {
  a[prop] = val;
  return a;
}, {});
console.log(sortedObj);

请记住，对象属性顺序仅在ES6 +环境中指定。尽管规范未保证Object.fromEntries会按照与条目相同的顺序创建对象，但幸运的是，在我遇到的任何实现中，它仍然会做。（如果您仍然担心它，可以使用老式的reduce方法来创建对象，就像在第三段中一样）

Answer 2

尝试使用函数式编程：map和2 reduce方法的组合。

const listMapped = list.map(it=> it.students)

const listReduced = listMapped.reduce((acc, rec) => {
return [...acc.concat(rec)]
}, [])

const listCounted = listReduced.reduce((acc, rec) => {
acc[rec]
? acc[rec] += 1
: acc[rec] = 1
return acc 
}, {})

console.log(listCounted)

创建对象数组中数组项的唯一列表和计数

2 个答案: