javascript - 如何使用函数式编程重构 javascript 代码

Question

我正在尝试通过一本教科书来学习使用 JavaScript 的函数式编程。我必须使用函数式编程重构一些代码，这很困难。如何使用 map()、filter() 和 reduce() 重写此代码。

我必须以更实用的方式重写以下代码。我已经尝试过，但我不断收到空数组。

const students = [{
    name: "Anna",
    sex: "f",
    grades: [4.5, 3.5, 4]
  },
  {
    name: "Dennis",
    sex: "m",
    country: "Germany",
    grades: [5, 1.5, 4]
  },
  {
    name: "Martha",
    sex: "f",
    grades: [5, 4, 2.5, 3]
  },
  {
    name: "Brock",
    sex: "m",
    grades: [4, 3, 2]
  }
];

// Compute female student results
const femaleStudentsResults = [];
for (const student of students) {
  if (student.sex === "f") {
    let gradesSum = 0;
    for (const grade of student.grades) {
      gradesSum += grade;
    }
    const averageGrade = gradesSum / student.grades.length;
    femaleStudentsResults.push({
      name: student.name,
      avgGrade: averageGrade
    });
  }
}

console.log(femaleStudentsResults);

结果必须与此相同:

[对象{avgGrade: 4，名称:'安娜'}，对象{avgGrade: 3.625，名称:'玛莎'}]

我首先将所有女学生分开

const Females = Students.filter(gender => sex.sex === 'f')

然后，返回我的成绩数组

constgrade=females.map(grade=>grade.grades);

但是当我尝试使用 reduce() 求成绩平均值时:

 const grade = females
   .map(grade => grade.grades)
   .reduce((accum,curr)=>accum + curr)

我得到以下结果4.5,3.5,45,4,2.5,3。我怎样才能防止这种情况并只获得每个学生的平均成绩平均值？

Answer 1

使用filter(获取女学生)，然后使用.map(将学生对象转换为具有avgGrade的对象) :

const students = [{
    name: "Anna",
    sex: "f",
    grades: [4.5, 3.5, 4]
  },
  {
    name: "Dennis",
    sex: "m",
    country: "Germany",
    grades: [5, 1.5, 4]
  },
  {
    name: "Martha",
    sex: "f",
    grades: [5, 4, 2.5, 3]
  },
  {
    name: "Brock",
    sex: "m",
    grades: [4, 3, 2]
  }
];

const sum = (a, b) => a + b;
const femaleStudentsResults = students
  .filter(({ sex }) => sex === 'f')
  .map(({ name, grades }) => ({ name, avgGrade: grades.reduce(sum) / grades.length }));
console.log(femaleStudentsResults);