Sort array of object based on another array

Question

So let's say I have an array:

const chunks = [
    {id: 0, names: ['app']}, 
    {id: 1, names: ['contact']}, 
    {id: 2, names: ['bootstrap']}
];

And I want it to be sorted based on the names property, so the order is like in this array:

const original = ['bootstrap', 'app', 'contact'];

What is the most efficient way to do this?

Why is names an array, and what should happen when it contains more or less than exactly one item? — Bergi
– Bergi, Commented Jun 19, 2017 at 16:46

Nina Scholz · Accepted Answer · 2019-09-23 14:22:47Z

1

You could use the delta of the indices of names in original.

const chunks = [{ id: 0, names: ['app'] }, { id: 1, names: ['contact'] }, { id: 2, names: ['bootstrap'] }],
    original = ['bootstrap', 'app', 'contact'];

chunks.sort((a, b) => original.indexOf(a.names[0]) - original.indexOf(b.names[0]));

console.log(chunks);

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

edited Sep 23, 2019 at 14:22

answered Jun 19, 2017 at 17:01

Nina Scholz

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2 Comments

Michael Over a year ago

Can someone explain to me how this works? What is the "delat of indices"?

Nina Scholz Over a year ago

@Michael, it should read delta, that is the difference between two values of a same group.

Vladimir Gordienko · Accepted Answer · 2017-06-19 16:48:47Z

1

Try this method:

const chunks = [{id: 0, names: ['app']}, {id: 1, names: ['contact']}, {id: 2, names: ['bootstrap']}];
const original = ['bootstrap', 'app', 'contact'];
let result = [];
for(let i = 0; i < original.length; i++) {
    for(let j = 0; j < chunks.length; j++) {
        if(chunks[j].names.indexOf(original[i]) !== -1) {
            result.push(chunks[j]);
        }
    }
}
console.log(result);

answered Jun 19, 2017 at 16:48

Vladimir Gordienko

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1 Comment

mdmb Over a year ago

This is what I came up with at the beginning, but @NinaScholz way is so much cleaner!

cheesenthusiast · Accepted Answer · 2017-06-19 17:14:17Z

1

Easy way: convert chunks into an object so you get the correct one with just the key, then map over the (already sorted) array to plug in the object in place.

cMap = chunks.reduce((p,c) => Object.assign( p, {[c.names[0]]: c} ), {});
const sorted = original.map(k => cMap[k]);

answered Jun 19, 2017 at 17:14

cheesenthusiast

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2 Comments

Bergi Over a year ago

Nice approach, but you'll want to mention that this only works when there's exactly one chunk per name.

cheesenthusiast Over a year ago

True, I didn't take that into consideration. Thankfully there are more discerning eyes on SO than myself :).

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