When provided with two arrays, generate an object by pairing each element of the first array with the corresponding key from the second array

Question

When provided with two arrays, generate an object by pairing each element of the first array with the corresponding key from the second array

Despite trying countless times, I can't seem to figure out where I'm going wrong.

function userChecker(arr1,arr2) {

  let map = {};

  arr1.filter((user) => {

    arr2.filter((obj) => {


      if (user == obj.user) {
        map[user] = obj;
      } 
    
      else {
        map['untracked'] = [user];
      }


    });

  });

console.log(map)

}


userChecker([39471379, 44471379, 25471379, 35471379, 29471379,55471379],[{user: 39471379, salary: 250000}, {user: 44471379, salary: 260000}, {user: 35471379, salary: 148700},{user: 29471379, salary: 270500}]);

A function is needed that takes in two arrays: one containing user numbers and the other containing objects with employee data (user number and salary). The function should create an object where the user number is the key and the corresponding value is the object with all the data. // If an employee with a specific user number is not found, it means they do not work for the company. In such cases, a new entry must be created with the key being "untracked" and the value as an array of all user numbers that are not found.

Expected Output:

{39471379: {ID: 39471379, salary: 250000},
44471379: {ID: 44471379, salary: 260000},
35471379: {ID: 35471379, salary: 148700},
29471379: {ID: 29471379, salary: 270500},
"untracked": [25471379,55471379]}

javascript arrays object

Answer 1

Answer №1

To create a user data object with reduced user ids and retrieve specific data by cross-referencing the object.

function getFilteredUserData(users, data) {
    const ids = Object.fromEntries(data.map(obj => [obj.user, obj]));
    return users.reduce((result, user) => {
        if (ids[user]) result[user] = ids[user];
        else result.untracked.push(user);
        return result;
    }, { untracked: []});
}

const
    users = [39471379, 44471379, 25471379, 35471379, 29471379, 55471379],
    data = [{ user: 39471379,  salary: 250000 }, { user: 44471379, salary: 260000 }, { user: 35471379, salary: 148700 }, { user: 29471379, salary: 270500 }],
    filteredResult = getFilteredUserData(users, data);

console.log(filteredResult);

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

Answer 2

To create a user data object with reduced user ids and retrieve specific data by cross-referencing the object.

function getFilteredUserData(users, data) {
    const ids = Object.fromEntries(data.map(obj => [obj.user, obj]));
    return users.reduce((result, user) => {
        if (ids[user]) result[user] = ids[user];
        else result.untracked.push(user);
        return result;
    }, { untracked: []});
}

const
    users = [39471379, 44471379, 25471379, 35471379, 29471379, 55471379],
    data = [{ user: 39471379,  salary: 250000 }, { user: 44471379, salary: 260000 }, { user: 35471379, salary: 148700 }, { user: 29471379, salary: 270500 }],
    filteredResult = getFilteredUserData(users, data);

console.log(filteredResult);

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

Answer 3

Answer №2

Implementing a strategy using a single loop, along with the usage of find.

const ids = [39471379, 44471379, 25471379, 35471379, 29471379,55471379];
const users = [{user: 39471379, salary: 250000}, {user: 44471379, salary: 260000}, {user: 35471379, salary: 148700},{user: 29471379, salary: 270500}];

function userCheck(ids, users) {
  
  // Initializing two arrays, one for tracked users and another for untracked users
  const output = [];
  const untracked = [];
  
  // Looping through the ids array
  for (const id of ids) {
    
    // Checking if there is a matching user based on id
    const foundUser = users.find(user => {
      return user.user === id;
    });
    
    // Adding the user object to output array if found
    if (foundUser) {
      output.push({ [id]: foundUser });
    
    // Otherwise adding the id to the untracked array
    } else {
      untracked.push(id);
    }
  }
  
  // Combining the tracked and untracked arrays and returning them 
  return [...output, { untracked }];
}

console.log(userCheck(ids, users));

For more information, refer to:

Spread syntax

Answer 4

Implementing a strategy using a single loop, along with the usage of find.

const ids = [39471379, 44471379, 25471379, 35471379, 29471379,55471379];
const users = [{user: 39471379, salary: 250000}, {user: 44471379, salary: 260000}, {user: 35471379, salary: 148700},{user: 29471379, salary: 270500}];

function userCheck(ids, users) {
  
  // Initializing two arrays, one for tracked users and another for untracked users
  const output = [];
  const untracked = [];
  
  // Looping through the ids array
  for (const id of ids) {
    
    // Checking if there is a matching user based on id
    const foundUser = users.find(user => {
      return user.user === id;
    });
    
    // Adding the user object to output array if found
    if (foundUser) {
      output.push({ [id]: foundUser });
    
    // Otherwise adding the id to the untracked array
    } else {
      untracked.push(id);
    }
  }
  
  // Combining the tracked and untracked arrays and returning them 
  return [...output, { untracked }];
}

console.log(userCheck(ids, users));

For more information, refer to:

Spread syntax

Answer 5

Answer №3

My approach involved utilizing a forEach loop in conjunction with the find() method.

const ids = [39471379, 44471379, 25471379, 35471379, 29471379,55471379];
const users = [{user: 39471379, salary: 250000}, {user: 44471379, salary: 260000}, {user: 35471379, salary: 148700},{user: 29471379, salary: 270500}];
    function checkUsers (ids, users) {
        let resultObject = {};
        resultObject["untracked"] = [];
        console.log(resultObject);
    
        ids.forEach(element => {
            const id = element;
            const foundUser = users.find(ele => ele.user === element);
            if(foundUser) {
                resultObject[id] = {
                    [id] : foundUser
                }
            } else {
                resultObject["untracked"].push(id);
            }
        });
        return resultObject;
    
    const finalResult = checkUsers(ids, users);
    console.log(finalResult);

Answer 6

My approach involved utilizing a forEach loop in conjunction with the find() method.

const ids = [39471379, 44471379, 25471379, 35471379, 29471379,55471379];
const users = [{user: 39471379, salary: 250000}, {user: 44471379, salary: 260000}, {user: 35471379, salary: 148700},{user: 29471379, salary: 270500}];
    function checkUsers (ids, users) {
        let resultObject = {};
        resultObject["untracked"] = [];
        console.log(resultObject);
    
        ids.forEach(element => {
            const id = element;
            const foundUser = users.find(ele => ele.user === element);
            if(foundUser) {
                resultObject[id] = {
                    [id] : foundUser
                }
            } else {
                resultObject["untracked"].push(id);
            }
        });
        return resultObject;
    
    const finalResult = checkUsers(ids, users);
    console.log(finalResult);

Answer 7

Answer №4

To generate an o object from the data array with a specific format for all users, you can utilize the Array.prototype.reduce() method to populate properties that will be spread across the result object using the Spread syntax (...). Lastly, for handling untracked, you only need to employ the Array.prototype.filter() function once to filter out the ids from the users array that are absent in the properties of the o object.

Snippet:

const users = [39471379, 44471379, 25471379, 35471379, 29471379, 55471379]
const data = [{ user: 39471379, salary: 250000 },{ user: 44471379, salary: 260000 },{ user: 35471379, salary: 148700 },{ user: 29471379, salary: 270500 }]

const userCheck = (users, data) => {
  const o = data.reduce((a, c) => (a[c.user] = c, a), {})
  return {
    ...o,
    untracked: users.filter(user => !o[user])
  }
}

console.log(userCheck(users, data))

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

Answer 8

To generate an o object from the data array with a specific format for all users, you can utilize the Array.prototype.reduce() method to populate properties that will be spread across the result object using the Spread syntax (...). Lastly, for handling untracked, you only need to employ the Array.prototype.filter() function once to filter out the ids from the users array that are absent in the properties of the o object.

Snippet:

const users = [39471379, 44471379, 25471379, 35471379, 29471379, 55471379]
const data = [{ user: 39471379, salary: 250000 },{ user: 44471379, salary: 260000 },{ user: 35471379, salary: 148700 },{ user: 29471379, salary: 270500 }]

const userCheck = (users, data) => {
  const o = data.reduce((a, c) => (a[c.user] = c, a), {})
  return {
    ...o,
    untracked: users.filter(user => !o[user])
  }
}

console.log(userCheck(users, data))

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

When provided with two arrays, generate an object by pairing each element of the first array with the corresponding key from the second array

Answer №1

Answer №2

Answer №3

Answer №4

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