How to sort an object based on its value's length

Question

I have an object of key (str): value (arr) pair.

const obj = {
   '1': []
   '2': ['a', 'b'],
   '3': [],
   '4': ['c']
}

I would like to sort this by the number of elements that each key has in the value.

So that it would be like,

const obj = {
   '2': ['a', 'b'],
   '4': ['c']
   '1': []
   '3': [],
}

After some searching, I tried this syntax

const sortable = Object.fromEntries(
    Object.entries(obj).sort(([,a],[,b]) => b.length)
);

But this gives me undefined.

Any help?

EDIT

Sorry the keys are String

More detailed example

const obj = {
    '1.2.3': [1,2,3,4],
    'ABC12': [],
    'CAA11': [3,5],
    '4.4.1': [1,2,3],
}

to

const obj = {
    '1.2.3': [1,2,3,4],
    '4.4.1': [1,2,3],
    'CAA11': [3,5],
    'ABC12': [],
}

You can't sort your object in that way if you want to preserve the keys as they are: integer-like object keys will always be sorted first AND numerically: https://www.stefanjudis.com/today-i-learned/property-order-is-predictable-in-javascript-objects-since-es2015/#1-integer-indices — Terry, Sep 27 '20 at 22:01
You're talking about changing the order of keys in an object, which isn't a trivial thing; what's worse, those keys are _numeric_ (and order of those is always incremental). What you can do is either create an array from that object (where each element is essentially { key: value } pair), or maintain another array of keys ordered properly and base your view on that. — raina77ow, Sep 27 '20 at 22:01
Just as a note, you are missing some commas in the first two code snippets. — shreyasm-dev, Sep 27 '20 at 22:15
Would it work for you to transform the object into an array of objects? Like this [{key: "1.2.3", values: [...]}, {key: "4.4.1", values: [...]}] etc... that would mantain a strict order as you want it and it's pretty easy to do — alotropico, Sep 27 '20 at 22:42

score 2 · Answer 1 · answered Sep 27 '20 at 22:03

What you are looking for is most probably:

Object.entries(obj).sort(([, a], [, b]) => b.length - a.length);

However, the output of the code above is an array consisting of enumerable property [key, value] pairs of the object again. Sorting the object's keys inside of the object itself can't be achieved.

Aleks · Accepted Answer · 2020-09-27T22:46:26.893

1

Assuming that all keys are integers

Object.keys(obj).sort((a,b)=>a-b).reduce((acc, key)=>((acc[key]=obj[key]), acc),{});

Returns new object with sorted keys

Edit: Sorted based on value length (array length);

Object.keys(obj).sort((a,b)=>obj[b].length - obj[a].length)
    .reduce((acc, key)=>((acc[key]=obj[key]), acc),{});

const obj = {
    '1.2.3': [1,2,3,4],
    'ABC12': [],
    'CAA11': [3,5],
    '4.4.1': [1,2,3],
}
const res = Object.keys(obj)
                  .sort((a,b)=>obj[b].length - obj[a].length)
                  .reduce((acc, key)=>((acc[key]=obj[key]), acc),{});
console.log(res);

edited Sep 27 '20 at 22:46

answered Sep 27 '20 at 22:10

Aleks

894
10
14

What if the keys are string? – Dawn17 Sep 27 '20 at 22:14
if it is a number string ie. `'3'` it will work just fine – Aleks Sep 27 '20 at 22:20
But I am trying to sort based on the value's length, not the key – Dawn17 Sep 27 '20 at 22:20
Every key in every object is a string. – kind user Sep 27 '20 at 22:21
@kind user i think it's a common knowledge – Aleks Sep 27 '20 at 22:26
I added a new example if that helps. – Dawn17 Sep 27 '20 at 22:28
yeah, my edited version should work, test it and see how it goes – Aleks Sep 27 '20 at 22:29
The function itself works, but somehow the object re-orders the object to its original order. Do you know why? – Dawn17 Sep 27 '20 at 22:49
Are you sure you are not logging old object? The function returns new object, you have to discard the old one or overwrite it. I need to know what do you do with new object to get an idea what is happening. – Aleks Sep 27 '20 at 22:51
Also i have to say that some browsers are inconsistent with preserving of an object key order. Ideally, as some other people have said, you would have a separate ordered array with keys.That's the only sure way to do it with non numeric keys ('3' , etc.) – Aleks Sep 27 '20 at 22:59
https://stackoverflow.com/questions/5525795/does-javascript-guarantee-object-property-order – Aleks Sep 27 '20 at 23:05
You could use `Map` if that's an option. It does preserve the order of insertion. – Aleks Sep 27 '20 at 23:13

Osman Durdag · Answer 3 · 2020-09-27T22:36:14.890

0

let obj = {
   '1.2.3': [1,2,3,4],
  'ABC12': [],
  'CAA11': [3,5],
  '4.4.1': [1,2,3],
}
console.log(obj);
let entries = Object.entries(obj).sort((a,b) => { return b[1].length - a[1].length;});
let tempObj = {};
for(let i = 0; i < entries.length; i++){
  tempObj[entries[i][0]] = entries[i][1]; 
}
obj = tempObj;
console.log(obj);

edited Sep 27 '20 at 22:36

answered Sep 27 '20 at 22:21

Osman Durdag

955
1
7
18

I added a new example if that helps – Dawn17 Sep 27 '20 at 22:28
Thank you, this is helpfull – Osman Durdag Sep 27 '20 at 22:35

How to sort an object based on its value's length

3 Answers3