1041

If I have a JavaScript object such as:

var list = { "you": 100, "me": 75, "foo": 116, "bar": 15 }; 

Is there a way to sort the properties based on value? So that I end up with

list = { "bar": 15, "me": 75, "you": 100, "foo": 116 }; 
9
  • 5
    Not only "sorting," but more importantly sorting numbers. Numbers are immune to Javascripts Array.sort() method, meaning you'll not just have to find a method for sorting properties, but you'll have to write your own function to compare the numerical values. Commented Jul 1, 2009 at 15:12
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    @GovindRai yet, in real world frontend applications we loop over object collections with IDs as the keys and the order is important if translated to HTML templates. You say they have no order, I say they have exactly the order that I see when console.logging them in the current browser. And that order can get reordered. As soon as you loop over them, they have an order. Commented Dec 19, 2016 at 15:27
  • 10
    @GovindRai: There is now a means of accessing properties in a specified order in the spec. Is it a good idea? Almost certainly not. :-) But it's there, as of ES2015. Commented Dec 31, 2016 at 16:22
  • 2
    @T.J.Crowder ah, you're absolutely right. Still, with all the caveats that come with this addition, I hope our fellow SO'ers will read the borderline before adopting of this new approach (as one always should). :D Commented Dec 31, 2016 at 20:17
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    2019 visitors: check this barely upvoted Object.entries-based answer which is the cleanest and most readable state of the art since ES2017: stackoverflow.com/a/37607084/245966 Commented May 3, 2019 at 22:33

46 Answers 46

1151

Move them to an array, sort that array, and then use that array for your purposes. Here's a solution:

let maxSpeed = { car: 300, bike: 60, motorbike: 200, airplane: 1000, helicopter: 400, rocket: 8 * 60 * 60 }; let sortable = []; for (var vehicle in maxSpeed) { sortable.push([vehicle, maxSpeed[vehicle]]); } sortable.sort(function(a, b) { return a[1] - b[1]; }); // [["bike", 60], ["motorbike", 200], ["car", 300], // ["helicopter", 400], ["airplane", 1000], ["rocket", 28800]] 

Once you have the array, you could rebuild the object from the array in the order you like, thus achieving exactly what you set out to do. That would work in all the browsers I know of, but it would be dependent on an implementation quirk, and could break at any time. You should never make assumptions about the order of elements in a JavaScript object.

let objSorted = {} sortable.forEach(function(item){ objSorted[item[0]]=item[1] }) 

In ES8, you can use Object.entries() to convert the object into an array:

const maxSpeed = { car: 300, bike: 60, motorbike: 200, airplane: 1000, helicopter: 400, rocket: 8 * 60 * 60 }; const sortable = Object.entries(maxSpeed) .sort(([,a],[,b]) => a-b) .reduce((r, [k, v]) => ({ ...r, [k]: v }), {}); console.log(sortable);


In ES10, you can use Object.fromEntries() to convert array to object. Then the code can be simplified to this:

const maxSpeed = { car: 300, bike: 60, motorbike: 200, airplane: 1000, helicopter: 400, rocket: 8 * 60 * 60 }; const sortable = Object.fromEntries( Object.entries(maxSpeed).sort(([,a],[,b]) => a-b) ); console.log(sortable);

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21 Comments

Can you please reformulate "you can rebuild" to "you can use array to maintain ordering of keys and pull values from object"? Not only it is non-standard, as you've yourself mentioned, this erroneous assumption is broken by more browsers than just Chrome today, so it's better not to encourage users to try it.
Here is a more compact version of your code. Object.keys(maxSpeed).sort(function(a, b) {return -(maxSpeed[a] - maxSpeed[b])});
@TheBrain: Just to add, keys() is only supported by IE9+ (and other modern browsers), if that is of concern. Also keys() excludes enumerable properties from the elements prototype chain (unlike for..in) - but that is usually more desirable.
_.pairs turns an object into [ [key1, value1], [key2, value2] ]. Then call sort on that. Then call _.object on it to turn it back.
Am I crazy or does the ES10 solution not work? When I try it, Object.fromEntries creates an object sorted by key, undoing all the work of sorting.
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517

We don't want to duplicate the entire data structure, or use an array where we need an associative array.

Here's another way to do the same thing as bonna:

var list = {"you": 100, "me": 75, "foo": 116, "bar": 15}; keysSorted = Object.keys(list).sort(function(a,b){return list[a]-list[b]}) console.log(keysSorted); // bar,me,you,foo

15 Comments

This seems to be sorting by key, not value, which is not what the question called for?
It's sorted by value, and displays keys -- but we lose the value count when it prints out, as it prints only keys.
Object property order is not guaranteed in JavaScript, so sorting should be done into an array, not an object (which is what you are referring to as an 'associative array').
Don't forget. keysSorted is an array! Not an object!
If you add .map(key => list[key]); to the end of the sort, it will return the whole object instead of just the key
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255

Your objects can have any amount of properties and you can choose to sort by whatever object property you want, number or string, if you put the objects in an array. Consider this array:

var arrayOfObjects = [ { name: 'Diana', born: 1373925600000, // Mon, Jul 15 2013 num: 4, sex: 'female' }, { name: 'Beyonce', born: 1366832953000, // Wed, Apr 24 2013 num: 2, sex: 'female' }, { name: 'Albert', born: 1370288700000, // Mon, Jun 3 2013 num: 3, sex: 'male' }, { name: 'Doris', born: 1354412087000, // Sat, Dec 1 2012 num: 1, sex: 'female' } ]; 

sort by date born, oldest first

// use slice() to copy the array and not just make a reference var byDate = arrayOfObjects.slice(0); byDate.sort(function(a,b) { return a.born - b.born; }); console.log('by date:'); console.log(byDate); 

sort by name

var byName = arrayOfObjects.slice(0); byName.sort(function(a,b) { var x = a.name.toLowerCase(); var y = b.name.toLowerCase(); return x < y ? -1 : x > y ? 1 : 0; }); console.log('by name:'); console.log(byName); 

http://jsfiddle.net/xsM5s/16/

6 Comments

Sort by name should not substr out the first character; else Diana and Debra have an undefined order. Also, your byDate sort actually uses num, not born.
This looks good, but note that to compare strings you can just use x.localeCompare(y)
@JemarJones localCompare looks like a very useful function! Just note that it won't be supported in every browser - IE 10 and less, Safari Mobile 9 and less.
@inorganik Yes very good note for those who need to support those browsers
This is an array of Objects, which is not the OP asked for (an Object with several properties)
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163

ECMAScript 2017 introduces Object.values / Object.entries. As the name suggests, the former aggregates all the values of an object into an array, and the latter does the whole object into an array of [key, value] arrays; Python's equivalent of dict.values() and dict.items().

The features make it pretty easier to sort any hash into an ordered object. As of now, only a small portion of JavaScript platforms support them, but you can try it on Firefox 47+.

EDIT: Now supported by all modern browsers!

let obj = {"you": 100, "me": 75, "foo": 116, "bar": 15}; let entries = Object.entries(obj); // [["you",100],["me",75],["foo",116],["bar",15]] let sorted = entries.sort((a, b) => a[1] - b[1]); // [["bar",15],["me",75],["you",100],["foo",116]] 

6 Comments

how does this possibly answers the question's title Sorting JavaScript Object by property value ? you misunderstood the question I think, since you are to change the original Object and not create a new Array from it.
@vsync This answer gives the same result as the accepted answer, but with less code and no temporary variable.
FWIW, this answer is clean af and is the only one that helped me.
Just note that this will not preserve keys.
For even further simplification, combine the process into one line: let sorted = Object.entries(obj).sort((a, b) => a[1] - b[1]);
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72

For completeness sake, this function returns sorted array of object properties:

function sortObject(obj) { var arr = []; for (var prop in obj) { if (obj.hasOwnProperty(prop)) { arr.push({ 'key': prop, 'value': obj[prop] }); } } arr.sort(function(a, b) { return a.value - b.value; }); //arr.sort(function(a, b) { return a.value.toLowerCase().localeCompare(b.value.toLowerCase()); }); //use this to sort as strings return arr; // returns array } var list = {"you": 100, "me": 75, "foo": 116, "bar": 15}; var arr = sortObject(list); console.log(arr); // [{key:"bar", value:15}, {key:"me", value:75}, {key:"you", value:100}, {key:"foo", value:116}] 

JSFiddle with the code above is here. This solution is based on this article.

Updated fiddle for sorting strings is here. You can remove both additional .toLowerCase() conversions from it for case sensitive string comparation.

4 Comments

What about an array of Objects? [ {name:Will}, {name:Bill}, {name:Ben} ]
Hi Will, you can use localeCompare function for this comparation. Added it to the answer above.
Great solution. The string sort needs a return
@Stano Correct and simple solution. Thanks so much.
63

An "arrowed" version of @marcusR 's answer for reference

var myObj = { you: 100, me: 75, foo: 116, bar: 15 }; keysSorted = Object.keys(myObj).sort((a, b) => myObj[a] - myObj[b]); alert(keysSorted); // bar,me,you,foo 

UPDATE: April 2017 This returns a sorted myObj object defined above.

const myObj = { you: 100, me: 75, foo: 116, bar: 15 }; const result = Object.keys(myObj) .sort((a, b) => myObj[a] - myObj[b]) .reduce( (_sortedObj, key) => ({ ..._sortedObj, [key]: myObj[key] }), {} ); document.write(JSON.stringify(result));

UPDATE: March 2021 - Object.entries with sort function (updated as per comments)

const myObj = { you: 100, me: 75, foo: 116, bar: 15 }; const result = Object .entries(myObj) .sort((a, b) => a[1] - b[1]) .reduce((_sortedObj, [k,v]) => ({ ..._sortedObj, [k]: v }), {}) document.write(JSON.stringify(result));

13 Comments

does not work for var myObj = {"1": {"Value": 40}, "2": {"Value": 10}, "3": {"Value": 30}, "4": {"Value": 20}};
The OP's question, along with this answer does not contain nested objects @Rohanil Maybe you'd want to ask another question instead of down-voting. Your nested object with various types obviously needs more than this solution provides
Just a note: your updates won't work actually. At least not everywhere and not because of entries. According to the standard, an object is an unordered collection of properties. Well, that means if you are trying to construct the new object after sorting it by property value or anything else, the property key ordering becomes again undefined. Chrome, for example, is by default ordering the property keys, thus every attempt to order them differently is useless. Your only chance is to get an "index" based on your ordering preferences and traverse the original object accordingly.
Yes @ZorgoZ, you are right and many people have mentioned it on this post. Most of the time, we use this function as a transformation before converting to another type (like JSON) or before another reduce function. If the objective is to mutate the object thereafter, then there might be unexpected results. I have had success with this function across engines.
Note: you don't want to use sort() on Object.entries()
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40

JavaScript objects are unordered by definition (see the ECMAScript Language Specification, section 8.6). The language specification doesn't even guarantee that, if you iterate over the properties of an object twice in succession, they'll come out in the same order the second time.

If you need things to be ordered, use an array and the Array.prototype.sort method.

7 Comments

Note that's there's been quite a bit of arguing about this. Most implementations keep the list in the order in which elements were added. IIRC, Chrome doesn't. There was argument about whether Chrome should fall in line with the other implementations. My belief is that a JavaScript object is a hash and no order should be assumed. I believe Python went through the same argument and a new ordered hash-like list was recently introduced. For most browsers, you CAN do what you want by recreating your object, adding elements by sorted value. But you shouldn't.
Edit. Chrome usually keeps order, but doesn't always. And here's the relevant Chromium bug: code.google.com/p/chromium/issues/detail?id=883
That bug report is unjustified, and those who relied on the undocumented behaviour are the ones with the bugs. Read ECMASCript section 8.6; it clearly states that "An Object is an unordered collection of properties". Anybody who found that it didn't seem that way in a few implementations, and then started to depend on that implementation-specific behaviour,made a big mistake, and they shouldn't be trying to shift the blame away from themselves. If I was on the Chrome team I'd mark that bug report as "Invalid, WontFix".
EcmaScript 5 actually defines the order of enumeration to be the order of insertion -- the absence of definition is considered a spec bug in ES3. It's worth noting that the EcmaScript spec defines behaviour that no one would consider sane -- for example spec behaviour is that syntax errors are in many cases thrown at runtime, incorrect use of continue, break, ++, --, const, etc according to the spec any engine that throws an exception before reaching that code is wrong
@olliej - I don't think this is correct. The spec says: "The mechanics and order of enumerating the properties (step 6.a in the first algorithm, step 7.a in the second) is not specified". This is on page 92 of the PDF of the final draft.
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OK, as you may know, javascript has sort() function, to sort arrays, but nothing for object...

So in that case, we need to somehow get array of the keys and sort them, thats the reason the apis gives you objects in an array most of the time, because Array has more native functions to play with them than object literal, anyway, the quick solotion is using Object.key which return an array of the object keys, I create the ES6 function below which does the job for you, it uses native sort() and reduce() functions in javascript:

function sortObject(obj) { return Object.keys(obj) .sort().reduce((a, v) => { a[v] = obj[v]; return a; }, {}); } 

And now you can use it like this:

let myObject = {a: 1, c: 3, e: 5, b: 2, d: 4}; let sortedMyObject = sortObject(myObject); 

Check the sortedMyObject and you can see the result sorted by keys like this:

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

Also this way, the main object won't be touched and we actually getting a new object.

I also create the image below, to make the function steps more clear, in case you need to change it a bit to work it your way:

Sorting a javascript object by property value

4 Comments

This sort by key, not by value.
@ROROROOROROR, it's just an indication how it works as they are not that different, but all good, I will add sorting by value too
Sorting by value is wrong and it's your dataset. Change c: 3 to c: 13 and you'll see it fall apart.
@VtoCorleone That's just a natural sorting error, 1 comes first, not a flaw in the algorithm as presented.
12

Update with ES6: If your concern is having a sorted object to iterate through (which is why i'd imagine you want your object properties sorted), you can use the Map object.

You can insert your (key, value) pairs in sorted order and then doing a for..of loop will guarantee having them loop in the order you inserted them

var myMap = new Map(); myMap.set(0, "zero"); myMap.set(1, "one"); for (var [key, value] of myMap) { console.log(key + " = " + value); } // 0 = zero // 1 = one 

3 Comments

Also in ES6 (ES2015): Object properties do have order (or there is a means of accessing them in a defined order, depending on your point of view). If the property names are strings and don't look like array indexes, the order is the order they were added to the object. This order is not specified to be respected by for-in or Object.keys, but it is defined to be respected by Object.getOwnPropertyNames and the other new methods for accessing property name arrays. So that's another option.
But where is `sort' operation? They are not sorted at all
@Green I think what I wanted to highlight here is that, with Map you actually have a data structure that can store in sorted order (sort your data, loop through it and store it in the map) where as you aren't guaranteed that with objects.
10
var list = { "you": 100, "me": 75, "foo": 116, "bar": 15 }; function sortAssocObject(list) { var sortable = []; for (var key in list) { sortable.push([key, list[key]]); } // [["you",100],["me",75],["foo",116],["bar",15]] sortable.sort(function(a, b) { return (a[1] < b[1] ? -1 : (a[1] > b[1] ? 1 : 0)); }); // [["bar",15],["me",75],["you",100],["foo",116]] var orderedList = {}; for (var idx in sortable) { orderedList[sortable[idx][0]] = sortable[idx][1]; } return orderedList; } sortAssocObject(list); // {bar: 15, me: 75, you: 100, foo: 116} 

3 Comments

Simple and Perfect! This answers the question. Thanks.
Key order is not guaranteed in javascript so this fails for me if keys are integers or the like
exact, ParoX. This is not working correctly when the key is integer type.
9

Sort values without multiple for-loops (to sort by the keys change index in the sort callback to "0")

const list = { "you": 100, "me": 75, "foo": 116, "bar": 15 }; let sorted = Object.fromEntries( Object.entries(list).sort( (a,b) => a[1] - b[1] ) ) console.log('Sorted object: ', sorted) 

Comments

9

Very short and simple!

var sortedList = {}; Object.keys(list).sort((a,b) => list[a]-list[b]).forEach((key) => { sortedList[key] = list[key]; }); 

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8

Underscore.js or Lodash.js for advanced array or object sorts

var data = { "models": { "LTI": [ "TX" ], "Carado": [ "A", "T", "A(пасс)", "A(груз)", "T(пасс)", "T(груз)", "A", "T" ], "SPARK": [ "SP110C 2", "sp150r 18" ], "Autobianchi": [ "A112" ] } }; var arr = [], obj = {}; for (var i in data.models) { arr.push([i, _.sortBy(data.models[i], function(el) { return el; })]); } arr = _.sortBy(arr, function(el) { return el[0]; }); _.map(arr, function(el) { return obj[el[0]] = el[1]; }); console.log(obj);
<script src="https://cdn.jsdelivr.net/npm/[email protected]/lodash.min.js" integrity="sha256-qXBd/EfAdjOA2FGrGAG+b3YBn2tn5A6bhz+LSgYD96k=" crossorigin="anonymous"></script>

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5

I am following the solution given by slebetman (go read it for all the details), but adjusted, since your object is non-nested.

// First create the array of keys/values so that we can sort it: var sort_array = []; for (var key in list) { sort_array.push({key:key,value:list[key]}); } // Now sort it: sort_array.sort(function(x,y){return x.value - y.value}); // Now process that object with it: for (var i=0;i<sort_array.length;i++) { var item = list[sort_array[i].key]; // now do stuff with each item } 

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5

Thanks to @orad for providing the answer in TypeScript. Now, We can use the below codesnippet in JavaScript.

function sort(obj,valSelector) { const sortedEntries = Object.entries(obj) .sort((a, b) => valSelector(a[1]) > valSelector(b[1]) ? 1 : valSelector(a[1]) < valSelector(b[1]) ? -1 : 0); return new Map(sortedEntries); } const Countries = { "AD": { "name": "Andorra", }, "AE": { "name": "United Arab Emirates", }, "IN": { "name": "India", }} // Sort the object inside object. var sortedMap = sort(Countries, val => val.name); // Convert to object. var sortedObj = {}; sortedMap.forEach((v,k) => { sortedObj[k] = v }); console.log(sortedObj); //Output: {"AD": {"name": "Andorra"},"IN": {"name": "India"},"AE": {"name": "United Arab Emirates"}}

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4
let toSort = {a:2323, b: 14, c: 799} let sorted = Object.entries(toSort ).sort((a,b)=> a[1]-b[1]) 

Output:

[ [ "b", 14 ], [ "c", 799 ], [ "a", 2323 ] ] 

Comments

4

There are many ways to do this, but since I didn't see any using reduce() I put it here. Maybe it seems utils to someone.

var list = { "you": 100, "me": 75, "foo": 116, "bar": 15 }; let result = Object.keys(list).sort((a,b)=>list[a]>list[b]?1:-1).reduce((a,b)=> {a[b]=list[b]; return a},{}); console.log(result);

Comments

4

Sorting object property by values

const obj = { you: 100, me: 75, foo: 116, bar: 15 }; const keysSorted = Object.keys(obj).sort((a, b) => obj[a] - obj[b]); const result = {}; keysSorted.forEach(key => { result[key] = obj[key]; }); document.write('Result: ' + JSON.stringify(result));

The desired output:

{"bar":15,"me":75,"you":100,"foo":116} 

References:

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4

Here's another way to achieve this

var list = { "you": 100, "me": 75, "foo": 116, "bar": 15 }; const sortedByValue = Object.fromEntries( Object.entries(list).sort((a, b) => { return a[1] - b[1]; }) ); console.log(sortedByValue);

Comments

3

This could be a simple way to handle it as a real ordered object. Not sure how slow it is. also might be better with a while loop.

Object.sortByKeys = function(myObj){ var keys = Object.keys(myObj) keys.sort() var sortedObject = Object() for(i in keys){ key = keys[i] sortedObject[key]=myObj[key] } return sortedObject } 

And then I found this invert function from: http://nelsonwells.net/2011/10/swap-object-key-and-values-in-javascript/

Object.invert = function (obj) { var new_obj = {}; for (var prop in obj) { if(obj.hasOwnProperty(prop)) { new_obj[obj[prop]] = prop; } } return new_obj; }; 

So

var list = {"you": 100, "me": 75, "foo": 116, "bar": 15}; var invertedList = Object.invert(list) var invertedOrderedList = Object.sortByKeys(invertedList) var orderedList = Object.invert(invertedOrderedList) 

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3

Just in case, someone is looking for keeping the object (with keys and values), using the code reference by @Markus R and @James Moran comment, just use:

var list = {"you": 100, "me": 75, "foo": 116, "bar": 15}; var newO = {}; Object.keys(list).sort(function(a,b){return list[a]-list[b]}) .map(key => newO[key] = list[key]); console.log(newO); // {bar: 15, me: 75, you: 100, foo: 116} 

1 Comment

You are returning an assignment in that map, forEach would be better
3

TypeScript

The following function sorts object by value or a property of the value. If you don't use TypeScript you can remove the type information to convert it to JavaScript.

/** * Represents an associative array of a same type. */ interface Dictionary<T> { [key: string]: T; } /** * Sorts an object (dictionary) by value or property of value and returns * the sorted result as a Map object to preserve the sort order. */ function sort<TValue>( obj: Dictionary<TValue>, valSelector: (val: TValue) => number | string, ) { const sortedEntries = Object.entries(obj) .sort((a, b) => valSelector(a[1]) > valSelector(b[1]) ? 1 : valSelector(a[1]) < valSelector(b[1]) ? -1 : 0); return new Map(sortedEntries); } 

Usage

var list = { "one": { height: 100, weight: 15 }, "two": { height: 75, weight: 12 }, "three": { height: 116, weight: 9 }, "four": { height: 15, weight: 10 }, }; var sortedMap = sort(list, val => val.height); 

The order of keys in a JavaScript object are not guaranteed, so I'm sorting and returning the result as a Map object which preserves the sort order.

If you want to convert it back to Object, you can do this:

var sortedObj = {} as any; sortedMap.forEach((v,k) => { sortedObj[k] = v }); 

2 Comments

can I have this in javascript, please.
Thank you very much, this solution worked for me. I have used the same thing in Javascript without Dictionary interface. const Countries = { "AD": { "name": "Andorra", }, "AE": { "name": "United Arab Emirates", }, "IN": { "name": "India", }, } // Sort the object inside object. var sortedMap = sort(Countries, val => val.name); // Convert to object. var sortedObj = {}; sortedMap.forEach((v,k) => { sortedObj[k] = v }); console.log(sortedObj); Output: {"AD": {"name": "Andorra"},"IN": {"name": "India"},"AE": {"name": "United Arab Emirates"}}
3
<pre> function sortObjectByVal(obj){ var keysSorted = Object.keys(obj).sort(function(a,b){return obj[b]-obj[a]}); var newObj = {}; for(var x of keysSorted){ newObj[x] = obj[x]; } return newObj; } var list = {"you": 100, "me": 75, "foo": 116, "bar": 15}; console.log(sortObjectByVal(list)); </pre> 

1 Comment

Welcome to stackoverflow. In addition to the answer you've provided, please consider providing a brief explanation of why and how this fixes the issue.
2

Object sorted by value (DESC)

function sortObject(list) { var sortable = []; for (var key in list) { sortable.push([key, list[key]]); } sortable.sort(function(a, b) { return (a[1] > b[1] ? -1 : (a[1] < b[1] ? 1 : 0)); }); var orderedList = {}; for (var i = 0; i < sortable.length; i++) { orderedList[sortable[i][0]] = sortable[i][1]; } return orderedList; } 

Comments

2
a = { b: 1, p: 8, c: 2, g: 1 } Object.keys(a) .sort((c,b) => { return a[b]-a[c] }) .reduce((acc, cur) => { let o = {} o[cur] = a[cur] acc.push(o) return acc } , []) 

output = [ { p: 8 }, { c: 2 }, { b: 1 }, { g: 1 } ]

Comments

2
 var list = { "you": 100, "me": 75, "foo": 116, "bar": 15 }; var tmpList = {}; while (Object.keys(list).length) { var key = Object.keys(list).reduce((a, b) => list[a] > list[b] ? a : b); tmpList[key] = list[key]; delete list[key]; } list = tmpList; console.log(list); // { foo: 116, you: 100, me: 75, bar: 15 } 

Comments

2
const arrayOfObjects = [ {name: 'test'}, {name: 'test2'} ] const order = ['test2', 'test'] const setOrder = (arrayOfObjects, order) => arrayOfObjects.sort((a, b) => { if (order.findIndex((i) => i === a.name) < order.findIndex((i) => i === b.name)) { return -1; } if (order.findIndex((i) => i === a.name) > order.findIndex((i) => i === b.name)) { return 1; } return 0; }); 

Comments

2

Another example with Object.values, sort() and the spread operator.

var paintings = { 0: { title: 'Oh my!', year: '2020', price: '3000' }, 1: { title: 'Portrait V', year: '2021', price: '2000' }, 2: { title: 'The last leaf', year: '2005', price: '600' } } 

We transform the object into an array of objects with Object.values:

var toArray = Object.values(paintings) 

Then we sort the array (by year and by price), using the spread operator to make the original array inmutable and the sort() method to sort the array:

var sortedByYear = [...toArray].sort((a, b) => a.year - b.year) var sortedByPrice = [...toArray].sort((a, b) => a.price - b.price) 

Finally, we generate the new sorted objects (again, with the spread operator to keep the original form of object of objects with a [x: number] as key):

var paintingsSortedByYear = { ...sortedByYear } var paintingsSortedByPrice = { ...sortedByPrice } 

Hope this could be helpful!

Comments

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A follow up answer to a long outdated question. I wrote two functions, one in which it sorts by keys, and the other by values, and returns the object in its sorted form in both functions. It should also work on strings as that is the reason why I am posting this (was having difficulty with some of the above on sorting by values if the values weren't numeric).

const a = { absolutely: "works", entirely: 'zen', best: 'player', average: 'joe' } const prop_sort = obj => { return Object.keys(obj) .sort() .reduce((a, v) => { a[v] = obj[v]; return a; }, {}); } const value_sort = obj => { const ret = {} Object.values(obj) .sort() .forEach(val => { const key = Object.keys(obj).find(key => obj[key] == val) ret[key] = val }) return ret } console.log(prop_sort(a)) console.log(value_sort(a))

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many similar and useful functions: https://github.com/shimondoodkin/groupbyfunctions/

function sortobj(obj) { var keys=Object.keys(obj); var kva= keys.map(function(k,i) { return [k,obj[k]]; }); kva.sort(function(a,b){ if(a[1]>b[1]) return -1;if(a[1]<b[1]) return 1; return 0 }); var o={} kva.forEach(function(a){ o[a[0]]=a[1]}) return o; } function sortobjkey(obj,key) { var keys=Object.keys(obj); var kva= keys.map(function(k,i) { return [k,obj[k]]; }); kva.sort(function(a,b){ k=key; if(a[1][k]>b[1][k]) return -1;if(a[1][k]<b[1][k]) return 1; return 0 }); var o={} kva.forEach(function(a){ o[a[0]]=a[1]}) return o; } 

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