I have variable that defines the order of objects like this:
const order = ['large', 'medium', 'small'];
I want to match that order of data array objects using names from order array:
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}]
so the effect should be following:
const data = [{name: "large", children: []}, {name: "medium", children: []}, {name: "small", children: []}]
How can I achieve this ?
We can do this by using the index of the name in 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));using lodash
_(data).sortBy([(x) => {
return order.findIndex(y => y.name === x.name);
}]).value();
I suggess :
data = data.sort(item => order.indexOf(item.name))
or
data = data.sort(item => -order.indexOf(item.name))
You could use a custom sort function that compares the index of matched word in the order array.
Something like
const order = ['large', 'medium', 'small'];
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}]
const result = data.sort((a, b) => order.indexOf(a) < order.indexOf(b) ? 1 : -1);
console.log(result)If your actual arrays are big, then try to avoid a scan in each call of the sort callback function, as that will mean the algorithm has a time complexity of O(n²logn). It can be done in linear time if you first convert the order information into a map (potentially in the form of a plain object) and use bucket sort:
const order = ['large', 'medium', 'small'];
const data = [{name: "small", children: []}, {name: "medium", children: []}, {name: "large", children: []}];
// preprocessing O(n)
const map = Object.fromEntries(order.map((key, i) => [key, i]));
let result = order.map(() => []);
// bucket sort O(n)
for (let obj of data) result[map[obj.name]].push(obj);
// flatten O(n)
result = result.flat();
console.log(result);