javascript – Sorting object property by values

The Question :

756 people think this question is useful

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
};

The Question Comments :
  • 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.
  • Before you read the answers: The answer is No. The ordering of object properties is non-standard in ECMAScript. You should never make assumptions about the order of elements in a JavaScript object. An Object is an unordered collection of properties. The answers below show you how to “use” sorted properties, using the help of arrays, but never actually alter the order of properties of objects themselves. So, no, it’s not possible. Even if you build an object with presorted properties, it is not guaranteed that they will display in the same order in the future. Read on :).
  • @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.
  • @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.
  • 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

The Answer 1

771 people think this answer is useful

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

var maxSpeed = {
    car: 300, 
    bike: 60, 
    motorbike: 200, 
    airplane: 1000,
    helicopter: 400, 
    rocket: 8 * 60 * 60
};
var 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.

var 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);

The Answer 2

443 people think this answer is useful

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

The Answer 3

199 people think this answer is useful

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/

The Answer 4

72 people think this answer is useful

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]]

The Answer 5

64 people think this answer is useful

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) { 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.

The Answer 6

46 people think this answer is useful

UPDATE: August 2020 Testing revealed that this no longer works in chrome. It appears chrome forces objects ordered by keys, even if an alternate order was used to construct a new object. See the last example.

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.
Object.keys(myObj)
  .sort((a, b) => myObj[a] - myObj[b])
  .reduce(
    (_sortedObj, key) => ({
      ..._sortedObj,
      [key]: myObj[key],
    }),
    {}
  );

Try it here!

UPDATE: October 2018 – Object.entries version

Object
 .entries(myObj)
 .sort()
 .reduce((_sortedObj, [k,v]) => ({
   ..._sortedObj, 
   [k]: v
 }), {})

Try it here!

UPDATE: July 2020 – Object.entries with sort function (updated as per comments)

Object
 .entries(myObj)
 .sort(([k1, v1], [k2, v2]) => {
    if(k1 < k2) { return -1; }
    if(k1 > k2) { return 1; }
    return 0;
 })
 .reduce((_sortedObj, [k,v]) => ({
   ..._sortedObj, 
   [k]: v
 }), {})

Try it here!

The Answer 7

37 people think this answer is useful

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.

The Answer 8

30 people think this answer is useful

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

The Answer 9

11 people think this answer is useful
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}

The Answer 10

9 people think this answer is useful

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

The Answer 11

7 people think this answer is useful

Very short and simple!

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

The Answer 12

6 people think this answer is useful

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);

demo

The Answer 13

5 people think this answer is useful

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
}

The Answer 14

5 people think this answer is useful

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) 

The Answer 15

3 people think this answer is useful
<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>

The Answer 16

2 people think this answer is useful

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)

The Answer 17

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

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

The Answer 18

2 people think this answer is useful

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}

The Answer 19

2 people think this answer is useful

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 });

The Answer 20

2 people think this answer is useful
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;
    });

The Answer 21

2 people think this answer is useful
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 ] ]

The Answer 22

1 people think this answer is useful

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;
}

The Answer 23

1 people think this answer is useful

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;
}

The Answer 24

1 people think this answer is useful

Here is one more example:

function sortObject(obj) {
  var arr = [];
  var prop;
  for (prop in obj) {
    if (obj.hasOwnProperty(prop)) {
      arr.push({
        'key': prop,
        'value': obj[prop]
      });
    }
  }
  arr.sort(function(a, b) {
    return a.value - b.value;
  });
  return arr; // returns array
}
var list = {
  car: 300,
  bike: 60,
  motorbike: 200,
  airplane: 1000,
  helicopter: 400,
  rocket: 8 * 60 * 60
};
var arr = sortObject(list);
console.log(arr);

The Answer 25

1 people think this answer is useful

here is the way to get sort the object and get sorted object in return

let sortedObject = {}
sortedObject = Object.keys(yourObject).sort((a, b) => {
                        return yourObject[a] - yourObject[b] 
                    }).reduce((prev, curr, i) => {
                        prev[i] = yourObject[curr]
                        return prev
                    }, {});

you can customise your sorting function as per your requirement

The Answer 26

1 people think this answer is useful
    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 }

The Answer 27

1 people think this answer is useful

input is object, output is object, using lodash & js built-in lib, with descending or ascending option, and does not mutate input object

eg input & output

{
  "a": 1,
  "b": 4,
  "c": 0,
  "d": 2
}
{
  "b": 4,
  "d": 2,
  "a": 1,
  "c": 0
}

The implementation

const _ = require('lodash');

const o = { a: 1, b: 4, c: 0, d: 2 };


function sortByValue(object, descending = true) {
  const { max, min } = Math;
  const selector = descending ? max : min;

  const objects = [];
  const cloned = _.clone(object);

  while (!_.isEmpty(cloned)) {
    const selectedValue = selector(...Object.values(cloned));
    const [key, value] = Object.entries(cloned).find(([, value]) => value === selectedValue);

    objects.push({ [key]: value });
    delete cloned[key];
  }

  return _.merge(...objects);
}

const o2 = sortByValue(o);
console.log(JSON.stringify(o2, null, 2));

The Answer 28

1 people think this answer is useful

my solution with sort :

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

let sorted = Object.entries(list).sort((a,b) => a[1] - b[1]);

for(let element of sorted) {
    console.log(element[0]+ ": " + element[1]);
}

The Answer 29

1 people think this answer is useful

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);

The Answer 30

1 people think this answer is useful

To find frequency of each element and sort it by frequency/values.

let response = ["apple", "orange", "apple", "banana", "orange", "banana", "banana"];
let frequency = {};
response.forEach(function(item) {
  frequency[item] = frequency[item] ? frequency[item] + 1 : 1;
});
console.log(frequency);
let intents = Object.entries(frequency)
  .sort((a, b) => b[1] - a[1])
  .map(function(x) {
    return x[0];
  });
console.log(intents);

Outputs:

{ apple: 2, orange: 2, banana: 3 }
[ 'banana', 'apple', 'orange' ]

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