按值对对象数组进行排序

Question

I need to sort an array of objects by its values and I use this code to do this:

function compare(a,b){return dict[a]-dict[b]} Object.keys(dict).sort(compare)

it works if each value is different, but if two objects have the same value, it leaves them in the order they appear in the array but I want them to be sorted alphabetically. I couldn't figure out a way to do that.

dict = {a:1, d:4, c:2, b:4, e:5, f:3} should be:

{a:1, c:2, f:3, b:4, d:4, e:5 }

but I'm getting this:

{a:1, c:2, f:3, d:4, b:4, e:5 }

Answer 1

You can change the compareFunction to use localeCompare

dict[a] - dict[b] || a.localeCompare(b)

If dict[a] - dict[b] returns 0, it checks the next condition and sorts the keys alphabetically

Here's a snippet:

 const dict = {a:1, d:4, c:2, b:4, e:5, f:3} function compare(a, b) { return dict[a] - dict[b] || a.localeCompare(b) } const sorted = Object.keys(dict).sort(compare).reduce((acc, k) => (acc[k] = dict[k], acc), {}) console.log( sorted )

Answer 2

So compare the keys if the diff is equal to zero

 function compare(a, b) { var diff = dict[a] - dict[b] return diff === 0? a.localeCompare(b): diff } const dict = { a: 1, d: 4, c: 2, b: 4, e: 5, f: 3 } const result = Object.keys(dict).sort(compare).reduce((o, x) => ({...o, [x]: dict[x], }), {}) console.log(result)