Is there any function so that i can calculate the length of each users skill's at once and then find which user have the maximum skills?

Question

Find the person who has many skills in the users' object.

const users = {
  Alex: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript'],
    age: 20,
    isLoggedIn: false,
    points: 30
  },
  Asab: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript', 'Redux', 'MongoDB', 'Express', 'React', 'Node'],
    age: 25,
    isLoggedIn: false,
    points: 50
  },
  Brook: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript', 'React', 'Redux'],
    age: 30,
    isLoggedIn: true,
    points: 50
  },
  Daniel: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript', 'Python'],
    age: 20,
    isLoggedIn: false,
    points: 40
  },
}

Is there any function so that I can calculate the length of each users skill's at once and then find which user have the maximum skills?

Answer 1

you can do something like this

const userWithMaxSkills = users => Object.entries(users)
.reduce((res, [username, data]) => {
    if(data.skills.length > res.maxSkills){
      return {
        maxSkills: data.skills.length,
        user: {
          [username]: data
        }
      }
    }
    return res
}, 
 { 
  maxSkills: -1,
  user: undefined
 }
).user

const users = {
  Alex: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript'],
    age: 20,
    isLoggedIn: false,
    points: 30
  },
  Asab: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript', 'Redux', 'MongoDB', 'Express', 'React', 'Node'],
    age: 25,
    isLoggedIn: false,
    points: 50
  },
  Brook: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript', 'React', 'Redux'],
    age: 30,
    isLoggedIn: true,
    points: 50
  },
  Daniel: {
    email: '[email protected]',
    skills: ['HTML', 'CSS', 'JavaScript', 'Python'],
    age: 20,
    isLoggedIn: false,
    points: 40
  },
}

console.log(userWithMaxSkills(users))