Sort each Object inside Array of Objects by its values

Question

I have this function below to sort the keys of an object accordly their values.

var obj = {"A": 50, "B": 10, "C": 150 };

var ordered = Object.keys(obj).sort(function(a,b) {
            return obj[a] > obj[b] ? -1 :
                   obj[b] > obj[a] ? 1 : 0;
        });

console.log(ordered); // => [C, A, B];

However, when I have an Array of Objects, this function returns nothing.

For an Array of objects like this below:

var objs = [
    {"A": 50, "B": 10, "C": 150 },
    {"A": 60, "B": 100, "C": 10 },
    {"A": 150, "B": 100, "C": 30 } 
]

I have tried something like this:

let op =[];
objs.forEach((obj) => {
    let ordered = {};
     Object.keys(obj).sort().forEach((a,b) => {
      ordered = obj[a] > obj[b] ? -1 : obj[b] > obj[a] ? 1 : 0;
    });
  op.push(ordered);
});

console.log(op);

But this function returns an error: Identifier 'op' has already been declared.

I need to return something similar to the example above, but applied to each object. The result would be:

console.log(ordered); // => [[C, A, B], [B, A, C], [A, B, C]]

Identifier 'op' has already been declared. You've only declared op once in the code you posted, though... — CertainPerformance
– CertainPerformance, Commented Dec 28, 2018 at 1:29
If you’re executing this code in your console by copy-pasting it, you shouldn’t paste the let op line. You could also wrap it in a block. — Sebastian Simon
– Sebastian Simon, Commented Dec 28, 2018 at 1:32
@CertainPerformance this is the entire function. It seemed weird to me as well... Maybe there is another issue in this code. — Atoyansk
– Atoyansk, Commented Dec 28, 2018 at 1:33
The code you posted does not produce the error you describe, though... — CertainPerformance
– CertainPerformance, Commented Dec 28, 2018 at 1:36

Robby Cornelissen · Accepted Answer · 2018-12-28 01:42:45Z

1

Cannot reproduce the error you describe, but here's another way to do it, using Object.entries() with some destructuring:

const objs = [
    {"A": 50, "B": 10, "C": 150 },
    {"A": 60, "B": 100, "C": 10 },
    {"A": 150, "B": 100, "C": 30 } 
];

const ordered = objs.map(o => Object.entries(o)
                                    .sort(([k1, v1], [k2, v2]) => v2 - v1)
                                    .map(([k]) => k));

console.log(ordered);

answered Dec 28, 2018 at 1:42

Robby Cornelissen

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Comments

Mark · Accepted Answer · 2018-12-28 01:41:46Z

0

In your code you are putting the sort logic in the forEach which isn't going to work. Since you already have a sort that works for you, to get your desired output, you can just make it a function and map it over the array:

function ordered(obj) {
  return Object.keys(obj).sort(function(a, b) {
    return obj[a] > obj[b] ? -1 :
      obj[b] > obj[a] ? 1 : 0;
  });
}


var objs = [
    {"A": 50, "B": 10, "C": 150 },
    {"A": 60, "B": 100, "C": 10 },
    {"A": 150, "B": 100, "C": 30 } 
]


console.log(objs.map(ordered))

If you prefer the forEach code, you just need a couple changes:

var objs = [
    {"A": 50, "B": 10, "C": 150 },
    {"A": 60, "B": 100, "C": 10 },
    {"A": 150, "B": 100, "C": 30 } 
]

let op =[];

objs.forEach((obj) => {                                // just on forEach
    let ordered = Object.keys(obj).sort((a, b) => 
        obj[a] > obj[b] ? -1 : obj[b] > obj[a] ? 1 : 0 // sort 
    )
    op.push(ordered);                                  // push sorted array
});

console.log(op);

edited Dec 28, 2018 at 1:41

answered Dec 28, 2018 at 1:33

Mark

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1 Comment

Atoyansk Over a year ago

Great, both solutions returned the expected values

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