Two arrays with varying sizes, utilizing one array repeatedly within a loop once its maximum index is reached

Question

Two arrays with varying sizes, utilizing one array repeatedly within a loop once its maximum index is reached

I have two arrays set up, and my goal is to iterate through the larger array and assign a property to it using randomly selected IDs from the smaller array.

//The actual length of the users array is 140
const users = [
  {
    name: "John Roberts",
    uid: "49ikds_dm3idmssmmi9sz"
  },
  {
    name: "Peter Jones",
    uid: "fmi33_sm39imsz9z9nb"
  }
]

//The actual length of the cars array is 424
const cars = [
  {
    manufacturer: "BMW",
    model: "320d",
    year: "2010",
    user: null
  },
  {
    manufacturer: "BMW",
    model: "530d",
    year: "2018",
    user: null
  },
  {
    manufacturer: "AUDI",
    model: "RS6",
    year: "2014",
    user: null
  }
]

for(let i = 0; i < cars.length; i++){
  //In this example, if the index is 2 or higher, users[2] will be undefined
  cars[i].user = users[i].uid;
}

Essentially, I am looking for a way to efficiently utilize the small users array. Once the variable i reaches 2 or more in the example above, then accessing users[2] will return undefined.

Does anyone have a clever solution to suggest that could help me overcome this challenge?

javascript arrays for-loop

Answer 1

Answer №1

Utilize the % operator to ensure accurate indexing and reusing of users.

const users = [
  {
    name: 'John Roberts',
    uid: '49ikds_dm3idmssmmi9sz',
  },
  {
    name: 'Peter Jones',
    uid: 'fmi33_sm39imsz9z9nb',
  }
];

const cars = [
  {
    manufacturer: 'BMW',
    model: '320d',
    year: '2010',
    user: null,
  },
  {
    manufacturer: 'BMW',
    model: '530d',
    year: '2018',
    user: null,
  },
  {
    manufacturer: 'AUDI',
    model: 'RS6',
    year: '2014',
    user: null,
  },
];

for (let i = 0; i < cars.length; i += 1){
  cars[i].user = users[i % users.length].uid;
}

console.log(cars);

A jaw-dropping ES6 solution! (must see)

const users = [
  {
    name: 'John Roberts',
    uid: '49ikds_dm3idmssmmi9sz',
  },
  {
    name: 'Peter Jones',
    uid: 'fmi33_sm39imsz9z9nb',
  }
];

const cars = [
  {
    manufacturer: 'BMW',
    model: '320d',
    year: '2010',
    user: null,
  },
  {
    manufacturer: 'BMW',
    model: '530d',
    year: '2018',
    user: null,
  },
  {
    manufacturer: 'AUDI',
    model: 'RS6',
    year: '2014',
    user: null,
  },
];

const result = cars.map((car, index) => ({
  ...car,
  
  user: users[index % users.length].uid,
}));

console.log(result);

Answer 2

Utilize the % operator to ensure accurate indexing and reusing of users.

const users = [
  {
    name: 'John Roberts',
    uid: '49ikds_dm3idmssmmi9sz',
  },
  {
    name: 'Peter Jones',
    uid: 'fmi33_sm39imsz9z9nb',
  }
];

const cars = [
  {
    manufacturer: 'BMW',
    model: '320d',
    year: '2010',
    user: null,
  },
  {
    manufacturer: 'BMW',
    model: '530d',
    year: '2018',
    user: null,
  },
  {
    manufacturer: 'AUDI',
    model: 'RS6',
    year: '2014',
    user: null,
  },
];

for (let i = 0; i < cars.length; i += 1){
  cars[i].user = users[i % users.length].uid;
}

console.log(cars);

A jaw-dropping ES6 solution! (must see)

const users = [
  {
    name: 'John Roberts',
    uid: '49ikds_dm3idmssmmi9sz',
  },
  {
    name: 'Peter Jones',
    uid: 'fmi33_sm39imsz9z9nb',
  }
];

const cars = [
  {
    manufacturer: 'BMW',
    model: '320d',
    year: '2010',
    user: null,
  },
  {
    manufacturer: 'BMW',
    model: '530d',
    year: '2018',
    user: null,
  },
  {
    manufacturer: 'AUDI',
    model: 'RS6',
    year: '2014',
    user: null,
  },
];

const result = cars.map((car, index) => ({
  ...car,
  
  user: users[index % users.length].uid,
}));

console.log(result);

Answer 3

Answer №2

To properly handle the user[i] variable, you can check for its existence and assign a default value:

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[i] && users[i].uid || /*Default value:*/ null;
}

An alternative approach is using an if statement:

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined 
  if (users[i]) {
    cars[i].user = users[i].uid;
  } else {
    cars[i].user = null; // Or another default
  }
}

------------------ Update to take account of comment -----------------

If you want to loop through users and circle back to the beginning of the array when you reach the end, you can utilize modulo %

let user_idx = null
for(let i = 0; i < cars.length; i++){

  user_idx = i % users.length

  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[user_idx].uid || null;
}

Answer 4

To properly handle the user[i] variable, you can check for its existence and assign a default value:

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[i] && users[i].uid || /*Default value:*/ null;
}

An alternative approach is using an if statement:

for(let i = 0; i < cars.length; i++){
  //if index is 2 or greater in this example. users[2] will be undefined 
  if (users[i]) {
    cars[i].user = users[i].uid;
  } else {
    cars[i].user = null; // Or another default
  }
}

------------------ Update to take account of comment -----------------

If you want to loop through users and circle back to the beginning of the array when you reach the end, you can utilize modulo %

let user_idx = null
for(let i = 0; i < cars.length; i++){

  user_idx = i % users.length

  //if index is 2 or greater in this example. users[2] will be undefined
  cars[i].user = users[user_idx].uid || null;
}

Answer 5

Answer №3

element, you will find a straightforward solution. All you need to do is implement the following logic:

$.grep(users, function(user, i ) {  cars[i].user = user.uid;});

Answer 6

element, you will find a straightforward solution. All you need to do is implement the following logic:

$.grep(users, function(user, i ) {  cars[i].user = user.uid;});

Answer 7

Answer №4

var persons = [
{
name: "Emily Johnson",
uid: "34ids_fa2iiemasl123v"
},
{
name: "Michael Smith",
uid: "fmi33_sm39imsz9z9nb"
}
]

//Actual length of the array is 424
var vehicles = [
{
make: "Toyota",
model: "Camry",
year: "2015",
owner: null
},
{
make: "Honda",
model: "Civic",
year: "2019",
owner: null
},
{
make: "Ford",
model: "Mustang",
year: "2021",
owner: null
}
]

vehicles.map(function(vehicle){
var randomIndex = Math.floor(Math.random() * Math.floor(persons.length))
vehicle.owner = persons[randomIndex].uid;
})
console.log(vehicles)

Answer 8

var persons = [
{
name: "Emily Johnson",
uid: "34ids_fa2iiemasl123v"
},
{
name: "Michael Smith",
uid: "fmi33_sm39imsz9z9nb"
}
]

//Actual length of the array is 424
var vehicles = [
{
make: "Toyota",
model: "Camry",
year: "2015",
owner: null
},
{
make: "Honda",
model: "Civic",
year: "2019",
owner: null
},
{
make: "Ford",
model: "Mustang",
year: "2021",
owner: null
}
]

vehicles.map(function(vehicle){
var randomIndex = Math.floor(Math.random() * Math.floor(persons.length))
vehicle.owner = persons[randomIndex].uid;
})
console.log(vehicles)

Answer 9

Answer №5

choice A:

for(let index = 0; index < vehicles.length; index++){
   vehicles[index].driver = drivers[index % drivers.length].license;
}

choice B:

for(let index = 0; index < vehicles.length; index++){
   vehicles[index].driver = drivers[Math.round(Math.random() * (drivers.length - 1))].license;
}

Answer 10

choice A:

for(let index = 0; index < vehicles.length; index++){
   vehicles[index].driver = drivers[index % drivers.length].license;
}

choice B:

for(let index = 0; index < vehicles.length; index++){
   vehicles[index].driver = drivers[Math.round(Math.random() * (drivers.length - 1))].license;
}

Two arrays with varying sizes, utilizing one array repeatedly within a loop once its maximum index is reached

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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