Sort array by specific order [duplicate]

Question

I have an array

peeps = [{name: "Mike", age: 31},{name: "Dave", age: 31},{name: "Delta", age: 31},{name: "Clint", age: 31}]

And I need to sort them by an arbitrary order of

newOrder = ['Delta', 'Dave', 'Mike', 'Clint']

How would I reference newOrder to sort peeps, also bear in mind the data in peeps is not guaranteed to have all the names in newOrder

The name will be known, newOrder will be static but peeps might have all the names or only 1 of the names in peeps. I've tried to figure out a way to do it by sorting alphabetically and then reversing the order but I cant figure it out. — UXCODA
– UXCODA, Commented Nov 26, 2020 at 9:03

michael · Accepted Answer · 2020-11-26 09:04:53Z

4

const peeps = [
  { name: 'Mike', age: 31 },
  { name: 'Dave', age: 31 },
  { name: 'Delta', age: 31 },
  { name: 'Clint', age: 31 }
];
const newOrder = ['Delta', 'Dave', 'Mike', 'Clint'];

peeps.sort((a, b) => newOrder.indexOf(a.name) - newOrder.indexOf(b.name));

console.log(peeps);

answered Nov 26, 2020 at 9:04

michael

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Ran Marciano · Accepted Answer · 2020-11-26 09:08:22Z

2

This is one way:

const peeps = [{name: "Mike", age: 31},{name: "Dave", age: 31},{name: "Delta", age: 31},{name: "Clint", age: 31}];
const newOrder = ['Delta', 'Dave', 'Mike', 'Clint'];

const notInNewOrder = peeps.filter(i => !newOrder.includes(i.name));

const response = newOrder.map(n => ({...peeps.find(peep=> peep.name === n)}));

console.log(response.concat(notInNewOrder));

answered Nov 26, 2020 at 9:08

Ran Marciano

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