Arrange a static array of objects by leveraging a separate array

Question

Arrange a static array of objects by leveraging a separate array

I have a variable that sets the order of objects in a specific way:

const order = ['large', 'medium', 'small'];

My goal is to arrange the data array objects according to the names specified in the order array:

const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}]

The desired outcome would be:

const data = [{name: "large", children: []}, {name: "medium", children: []}, {name: "small", children: []}]

Can someone help me achieve this sorting?

javascript arrays sorting

Answer 1

Answer №1

To achieve this, we utilize the index position of the name within the order array:

const order = ['large', 'medium', 'small'];
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}];

const sortArray = (arr, order) => {
  return data.sort((a, b) => order.indexOf(a.name) - order.indexOf(b.name)); 
}
console.log(sortArray(data, order));

Answer 2

To achieve this, we utilize the index position of the name within the order array:

const order = ['large', 'medium', 'small'];
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}];

const sortArray = (arr, order) => {
  return data.sort((a, b) => order.indexOf(a.name) - order.indexOf(b.name)); 
}
console.log(sortArray(data, order));

Answer 3

Answer №2

utilizing the lodash library

_(data).orderBy([(x) => {
    return sequence.findIndex(y => y.name === x.name);
}]).value();

Answer 4

utilizing the lodash library

_(data).orderBy([(x) => {
    return sequence.findIndex(y => y.name === x.name);
}]).value();

Answer 5

Answer №3

Here is my recommendation:

data  = data.sort(item => order.indexOf(item.name))

Alternatively, you could try:

data  = data.sort(item => -order.indexOf(item.name))

Answer 6

Here is my recommendation:

data  = data.sort(item => order.indexOf(item.name))

Alternatively, you could try:

data  = data.sort(item => -order.indexOf(item.name))

Answer 7

Answer №4

To achieve a custom sorting of elements based on a predefined order array, you can create a sort function that compares the index of the matched word.

Here's an example:

const order = ['red', 'green', 'blue'];
const data = [{color: "blue", value: 10}, {color: "red", value: 5}, {color: "green", value: 7}];

const sortedData = data.sort((a, b) => order.indexOf(a.color) - order.indexOf(b.color));

console.log(sortedData);

Answer 8

To achieve a custom sorting of elements based on a predefined order array, you can create a sort function that compares the index of the matched word.

Here's an example:

const order = ['red', 'green', 'blue'];
const data = [{color: "blue", value: 10}, {color: "red", value: 5}, {color: "green", value: 7}];

const sortedData = data.sort((a, b) => order.indexOf(a.color) - order.indexOf(b.color));

console.log(sortedData);

Answer 9

Answer №5

To improve efficiency when dealing with large arrays, it's best to avoid scanning in each call of the sorting callback function as this can lead to a time complexity of O(n²logn). Instead, consider converting the order information into a map (such as a plain object) and utilizing bucket sort for a linear time solution:

const order = ['large', 'medium', 'small'];
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}];


// Preprocessing step O(n)    
const map = Object.fromEntries(order.map((key, i) => [key, i]));
let result = order.map(() => []);
// Bucket sort step O(n)
for (let obj of data) result[map[obj.name]].push(obj);
// Flattening step O(n)
result = result.flat();
console.log(result);

Answer 10

To improve efficiency when dealing with large arrays, it's best to avoid scanning in each call of the sorting callback function as this can lead to a time complexity of O(n²logn). Instead, consider converting the order information into a map (such as a plain object) and utilizing bucket sort for a linear time solution:

const order = ['large', 'medium', 'small'];
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}];


// Preprocessing step O(n)    
const map = Object.fromEntries(order.map((key, i) => [key, i]));
let result = order.map(() => []);
// Bucket sort step O(n)
for (let obj of data) result[map[obj.name]].push(obj);
// Flattening step O(n)
result = result.flat();
console.log(result);

Arrange a static array of objects by leveraging a separate array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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