Sort an array by object property value same as string in second array

Question

Sorry about the title, I'm really not sure how to word this. I'm stuck, I need to sort an array by its object's title property value, into the same order of a second array of strings.

Basically, I have two arrays...

var titles = [
  'title1',
  'title2',
  'title3',
  'title4',
  'title5'
]

var objects = [
  {
    title: 'title5',
    ...//other values
  }, {
    title: 'title3',
    ...//other values
  }
  , {
    title: 'title4',
    ...//other values
  }, {
    title: 'title1',
    ...//other values
  }, {
    title: 'title2',
    ...//other values
  } 
]

I want to order the objects array in the same order as the strings in titles array. Is this possible? My code base has access to _underscore, or I can use plain ES6.

no, whatever the order of the titles array is, those values could be anything as they are strings. — Stuart
– Stuart, Commented Feb 11, 2019 at 11:30
think of the titles array as a configuration for displaying the objects array. We use this currently to filter the objects array to contain only the match values, but i also now need to sort it in the same order. — Stuart
– Stuart, Commented Feb 11, 2019 at 11:34

Nenad Vracar · Accepted Answer · 2019-02-11 11:30:00Z

2

You could create object from titles array and use it for ordering your array of objects.

var objects = [{"title":"title5"},{"title":"title3"},{"title":"title4"},{"title":"title1"},{"title":"title2"}]
var titles = ['title1','title2','title3','title4','title5'].reduce((r, e, i) => Object.assign(r, {[e]: i}), {})

objects.sort((a, b) => titles[a.title] - titles[b.title])
console.log(objects)

edited Feb 11, 2019 at 11:30

answered Feb 11, 2019 at 11:28

Nenad Vracar

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3 Comments

Ulysse BN Over a year ago

This works only because you assume titles are in alphabetical order.

Mark Over a year ago

That's not true @UlysseBN -- it's sorting by the index in the array as it should. This is a good, efficient way to do this.

Ulysse BN Over a year ago

Oh I didn't see the index, maybe you should use more readable variables :)

TKoL · Accepted Answer · 2019-02-11 11:30:43Z

1

Yes, it's very possible. I personally would follow the 'decorate-sort-undecorate' pattern:

var titles = [
  'title1',
  'title2',
  'title3',
  'title4',
  'title5'
]

var objects = [
  {
    title: 'title5',
    //other values
  }, {
    title: 'title3',
    //other values
  }
  , {
    title: 'title4',
    //other values
  }, {
    title: 'title1',
    //other values
  }, {
    title: 'title2',
    //other values
  } 
]

var rankings = objects.map( o => ({
    object: o,
    ranking: titles.indexOf(o.title)
}))

rankings.sort((a, b) => a.ranking-b.ranking)

var sortedObjects = rankings.map(r => r.object);

console.log(sortedObjects);

answered Feb 11, 2019 at 11:30

TKoL

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Comments

Treble Snake · Accepted Answer · 2019-02-11 11:37:02Z

0

You could use _.indexBy to create an Object from objects variable and then map titles:

const objectsByTitle = _.indexBy(objects, 'title');
const result = titles.map(it => objectsByTitle[it]);

Titles should be unique ofc.

answered Feb 11, 2019 at 11:37

Treble Snake

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Comments

Marcelo Macedo · Accepted Answer · 2019-02-11 11:29:22Z

-1

Can you check if this resolve what you are needing for?

var titles = [
  'title1',
  'title2',
  'title3',
  'title4',
  'title5'
]

var objects = [
  {
    title: 'title5',
  }, {
    title: 'title3',
  }
  , {
    title: 'title4',
  }, {
    title: 'title1',
  }, {
    title: 'title2',
  } 
]

function compare(a,b) {
  if (a.title < b.title)
    return -1;
  if (a.title > b.title)
    return 1;
  return 0;
}

console.log(objects.sort(compare));

https://jsfiddle.net/sxtn57gj/

answered Feb 11, 2019 at 11:29

Marcelo Macedo

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1 Comment

Mark Over a year ago

The op doesn't want to sort by title, they want to sort by where the title apprears in titles array, which you are ignoring.

Ulysse BN · Accepted Answer · 2019-02-12 14:59:15Z

-1

There is a simple (but costful) solution:

let titles = ['foo', 'bar', 'fizz', 'buzz']

let objects = [{title: 'fizz'}, {title: 'bar'}, {title: 'buzz'}, {title: 'foo'}]

console.log(titles.map(title => objects.find(o => o.title === title)))

However this is a o(n²) solution, I'm pretty sure your needs may be a little bit different and you could user an object, with title as keys.

edited Feb 12, 2019 at 14:59

answered Feb 11, 2019 at 11:27

Ulysse BN

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2 Comments

Mark Over a year ago

I didn't downvote this — it works and might be good enough, but I suspect the downvote is because this is a quadratic time solution when a O(nlogn) solution exists.

Ulysse BN Over a year ago

@MarkMeyer I'd love to agree, but I wrote it in my answer already... I'll precise though, thanks for the comment.

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