Lodash sorting object by values, without losing the key - sorting

Question

Let's say I have an object:
{Derp: 17, Herp: 2, Asd: 5, Foo: 8, Qwe: 12}
And I need to sort it by value. What I'm looking to get is:
{Derp: 17, Qwe: 12, Foo: 8, Asd: 5, Herp: 2}
I'd like to use lodash for it. When I use _.sortBy it doesn't retain the keys how ever:
_.sortBy({Derp: 17, Herp: 2, Asd: 5, Foo: 8, Qwe: 12}).reverse();
// [17, 12, 8, 5, 2]
Hell, I'd even settle for just the array of keys, but still sorted by the value in the input:
['Derp', 'Herp', 'Foo', 'Asd', 'Qwe']

Answer 1

This worked for me
o = _.fromPairs(_.sortBy(_.toPairs(o), 1).reverse())
Here's an example:
var o = {
a: 2,
c: 3,
b: 1
};
o = _.fromPairs(_.sortBy(_.toPairs(o), 1).reverse())
console.log(o);
<script src="https://cdn.jsdelivr.net/lodash/4.17.4/lodash.min.js"></script>

Answer 2

I was struggling with a similar problem and I was able to solve it doing some transforms with lodash. For your problem it would be:
let doo = {Derp: 17, Herp: 2, Asd: 5, Foo: 8, Qwe: 12};
let foo = _.chain(doo)
.map((val, key) => {
return { name: key, count: val }
})
.sortBy('count')
.reverse()
.keyBy('name')
.mapValues('count')
.value();
console.log(foo);
// Derp: 17, Qwe: 12, Foo: 8, Asd: 5, Herp: 2 }

Answer 3

You could try like this,
_.mapValues(_.invert(_.invert(obj)),parseInt);
Object {Herp: 2, Asd: 5, Foo: 8, Qwe: 12, Derp: 17}
or
var obj = {Derp: 17, Herp: 2, Asd: 5, Foo: 8, Qwe: 12}
var result = _.reduceRight(_.invert(_.invert(obj)), function(current, val, key){
current[key] = parseInt(val);
return current;
},{});
Object {Derp: 17, Qwe: 12, Foo: 8, Asd: 5, Herp: 2}
or
Using Chain methods:
_.chain(obj).invert().invert().reduceRight(function(current, val, key){
current[key] = parseInt(val);
return current;
},{}).value()
Object {Derp: 17, Qwe: 12, Foo: 8, Asd: 5, Herp: 2}
Note: It depends on browser usally object properties order is not gurrantee in most case.

Answer 4

Vanilla ES6 version
const obj = {
a: 2,
c: 3,
b: 1
}
const sorted = Object.entries(obj)
.sort((a, b) => a[1] <= b[1] ? -1 : 1)
.reduce((acc, pair) => {
acc[pair[0]] = pair[1]
return acc
}, {})
// Output: { b: 1, a: 2, c: 3 }
This'll also let you access the sort callback, allowing for more granular sorting on deeper object properties.

Answer 5

Simple Vanila javascript approch
Object.fromEntries(Object.entries(obj).sort((a, b) => a[1] - b[1]));