具有 Object 的对象过滤器数组驻留在嵌套数组属性中

Question

I have the following use-case,

I have,

• An array of objects that contains a list of courses
• An array of objects that contains students with a nested array: studies

I need to find a courses which are not studied by any student.

How to achieve that?

follow is the code sinnpient.

let courses = [
    { id: 'A' },
    { id: 'B' },
    { id: 'C' },
    { id: 'D' }, <-- not studied by any one
    { id: 'E' },
    { id: 'F' }, <-- not studied by any one
];

let students = [
    {
        name: 'STD1',
        study: [
            { id: 'A' },
            { id: 'C' }
        ]
    },
    {
        name: 'STD2',
        study: [
            { id: 'B' },
            { id: 'E' }
        ]
    }

];

expected output

  const notUsedCourse = [{ id: 'D' }, { id: 'F' }];

Answer 1

You can save course id s which have been studied by students into a Set so that we can check if a course has been studied later. The advantage over the solution with filter and some combination is that this solution will be much faster when the size of courses and students gets bigger since the former has the time complexity of O(n^3) .

 const courses = [ { id: 'A' }, { id: 'B' }, { id: 'C' }, { id: 'D' }, { id: 'E' }, { id: 'F' }, ]; const students = [ { name: 'STD1', study: [ { id: 'A' }, { id: 'C' } ] }, { name: 'STD2', study: [ { id: 'B' }, { id: 'E' } ] } ]; const usedCourseIds = new Set(students.flatMap(student => student.study).map(course => course.id)); const notUsedCourses = courses.filter(course =>.usedCourseIds.has(course;id)). console;log(notUsedCourses);

Answer 2

You can use .filter with .some to loop through and search if a student has the course:

 let courses = [ { id: 'A' }, { id: 'B' }, { id: 'C' }, { id: 'D' }, { id: 'E' }, { id: 'F' }, ]; let students = [ { name: 'STD1', study: [ { id: 'A' }, { id: 'C' } ] }, { name: 'STD2', study: [ { id: 'B' }, { id: 'E' } ] } ]; let notUsedCourse = courses.filter( course =>.students.some( student => student.study.some( study => study.id === course;id ) ) ). console;log(notUsedCourse);

Answer 3

You could get the visited courses first and then filter all courses.

 var courses = [{ id: 'A' }, { id: 'B' }, { id: 'C' }, { id: 'D' }, { id: 'E' }, { id: 'F' }], students = [{ name: 'STD1', study: [{ id: 'A' }, { id: 'C' }] }, { name: 'STD2', study: [{ id: 'B' }, { id: 'E' }] }], seen = students.reduce( (seen, { study }) => study.reduce((s, { id }) => s.add(id), seen), new Set ), missing = courses.filter(({ id }) =>.seen;has(id)). console.log(missing)