Sorting object property by values
JavascriptSortingPropertiesObjectJavascript Problem Overview
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
};
Javascript Solutions
Solution 1 - Javascript
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);
Solution 2 - Javascript
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
Solution 3 - Javascript
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);
Solution 4 - Javascript
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]]
Solution 5 - Javascript
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.
Solution 6 - Javascript
An "arrowed" version of @marcusR 's [answer][1] 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));
[1]: https://stackoverflow.com/a/16794116/382536 "Marcus R's answer"
Solution 7 - Javascript
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.
Solution 8 - Javascript
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:
Solution 9 - Javascript
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}
Solution 10 - Javascript
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][1] 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
[1]: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map "Map"
Solution 11 - Javascript
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)
Solution 12 - Javascript
Very short and simple!
var sortedList = {};
Object.keys(list).sort((a,b) => list[a]-list[b]).forEach((key) => {
sortedList[key] = list[key]; });
Solution 13 - Javascript
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>
Solution 14 - Javascript
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
}
Solution 15 - Javascript
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 ] ]
Solution 16 - Javascript
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}
Solution 17 - Javascript
<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>
Solution 18 - Javascript
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);
Solution 19 - Javascript
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"}}
Solution 20 - Javascript
> ### 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:
Solution 21 - Javascript
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)
Solution 22 - Javascript
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;
}
Solution 23 - Javascript
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 }
Solution 24 - Javascript
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 } ]
Solution 25 - Javascript
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 });
Solution 26 - Javascript
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;
});
Solution 27 - Javascript
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]);
}
Solution 28 - Javascript
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))
Solution 29 - Javascript
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;
}
Solution 30 - Javascript
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);
Solution 31 - Javascript
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
Solution 32 - Javascript
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));
Solution 33 - Javascript
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' ]
Solution 34 - Javascript
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!
Solution 35 - Javascript
Another way to solve this:-
var res = [{"s1":5},{"s2":3},{"s3":8}].sort(function(obj1,obj2){
var prop1;
var prop2;
for(prop in obj1) {
prop1=prop;
}
for(prop in obj2) {
prop2=prop;
}
//the above two for loops will iterate only once because we use it to find the key
return obj1[prop1]-obj2[prop2];
});
//res will have the result array
Solution 36 - Javascript
Thank you and continue answer @Nosredna
Now that we understand object need to be converted to array then sort the array. this is useful for sorting array (or converted object to array) by string:
Object {6: Object, 7: Object, 8: Object, 9: Object, 10: Object, 11: Object, 12: Object}
6: Object
id: "6"
name: "PhD"
obe_service_type_id: "2"
__proto__: Object
7: Object
id: "7"
name: "BVC (BPTC)"
obe_service_type_id: "2"
__proto__: Object
//Sort options
var sortable = [];
for (var vehicle in options)
sortable.push([vehicle, options[vehicle]]);
sortable.sort(function(a, b) {
return a[1].name < b[1].name ? -1 : 1;
});
//sortable => prints
[Array[2], Array[2], Array[2], Array[2], Array[2], Array[2], Array[2]]
0: Array[2]
0: "11"
1: Object
id: "11"
name: "AS/A2"
obe_service_type_id: "2"
__proto__: Object
length: 2
__proto__: Array[0]
1: Array[2]
0: "7"
1: Object
id: "7"
name: "BVC (BPTC)"
obe_service_type_id: "2"
__proto__: Object
length: 2
Solution 37 - Javascript
Try this. Even your object is not having the property based on which you are trying to sort also will get handled.
Just call it by sending property with object.
var sortObjectByProperty = function(property,object){
console.time("Sorting");
var sortedList = [];
emptyProperty = [];
tempObject = [];
nullProperty = [];
$.each(object,function(index,entry){
if(entry.hasOwnProperty(property)){
var propertyValue = entry[property];
if(propertyValue!="" && propertyValue!=null){
sortedList.push({key:propertyValue.toLowerCase().trim(),value:entry});
}else{
emptyProperty.push(entry);
}
}else{
nullProperty.push(entry);
}
});
sortedList.sort(function(a,b){
return a.key < b.key ? -1 : 1;
//return a.key < b.key?-1:1; // Asc
//return a.key < b.key?1:-1; // Desc
});
$.each(sortedList,function(key,entry){
tempObject[tempObject.length] = entry.value;
});
if(emptyProperty.length>0){
tempObject.concat(emptyProperty);
}
if(nullProperty.length>0){
tempObject.concat(nullProperty);
}
console.timeEnd("Sorting");
return tempObject;
}
Solution 38 - Javascript
Using query-js you can do it like this
list.keys().select(function(k){
return {
key: k,
value : list[k]
}
}).orderBy(function(e){ return e.value;});
You can find an introductory article on query-js here
Solution 39 - Javascript
I made a plugin just for this, it accepts 1 arg which is an unsorted object, and returns an object which has been sorted by prop value. This will work on all 2 dimensional objects such as {"Nick": 28, "Bob": 52}...
var sloppyObj = {
'C': 78,
'A': 3,
'B': 4
};
// Extend object to support sort method
function sortObj(obj) {
"use strict";
function Obj2Array(obj) {
var newObj = [];
for (var key in obj) {
if (!obj.hasOwnProperty(key)) return;
var value = [key, obj[key]];
newObj.push(value);
}
return newObj;
}
var sortedArray = Obj2Array(obj).sort(function(a, b) {
if (a[1] < b[1]) return -1;
if (a[1] > b[1]) return 1;
return 0;
});
function recreateSortedObject(targ) {
var sortedObj = {};
for (var i = 0; i < targ.length; i++) {
sortedObj[targ[i][0]] = targ[i][1];
}
return sortedObj;
}
return recreateSortedObject(sortedArray);
}
var sortedObj = sortObj(sloppyObj);
alert(JSON.stringify(sortedObj));
Here is a demo of the function working as expected http://codepen.io/nicholasabrams/pen/RWRqve?editors=001
Solution 40 - Javascript
Couln't find answer above that would both work and be SMALL, and would support nested objects (not arrays), so I wrote my own one :) Works both with strings and ints.
function sortObjectProperties(obj, sortValue){
var keysSorted = Object.keys(obj).sort(function(a,b){return obj[a][sortValue]-obj[b][sortValue]});
var objSorted = {};
for(var i = 0; i < keysSorted.length; i++){
objSorted[keysSorted[i]] = obj[keysSorted[i]];
}
return objSorted;
}
Usage:
/* sample object with unsorder properties, that we want to sort by
their "customValue" property */
var objUnsorted = {
prop1 : {
customValue : 'ZZ'
},
prop2 : {
customValue : 'AA'
}
}
// call the function, passing object and property with it should be sorted out
var objSorted = sortObjectProperties(objUnsorted, 'customValue');
// now console.log(objSorted) will return:
{
prop2 : {
customValue : 'AA'
},
prop1 : {
customValue : 'ZZ'
}
}
Solution 41 - Javascript
function sortObjByValue(list){
var sortedObj = {}
Object.keys(list)
.map(key => [key, list[key]])
.sort((a,b) => a[1] > b[1] ? 1 : a[1] < b[1] ? -1 : 0)
.forEach(data => sortedObj[data[0]] = data[1]);
return sortedObj;
}
sortObjByValue(list);
Github Gist Link
Solution 42 - Javascript
I have tried in my own way
var maxSpeed = {
car: 300,
bike: 60,
motorbike: 200,
airplane: 1000,
helicopter: 400,
rocket: 8 * 60 * 60
};
var sorted = {}
Object.keys(maxSpeed).sort ((a,b) => maxSpeed[a] - maxSpeed[b]).map(item => sorted[item] = maxSpeed[item]);
console.log(sorted)
Solution 43 - Javascript
I have an array of objects and I have sorted them in descending order based on value attribute
var objs_1 = [{ "label": "allsight", "value": 0, "series": 0, "key": "Secondary Sources" }, { "label": "customertransaction", "value": 84, "series": 0, "key": "Secondary Sources" }, { "label": "emailfrom", "value": 20, "series": 0, "key": "Secondary Sources" }, { "label": "webchat", "value": 20, "series": 0, "key": "Secondary Sources" }, { "label": "deterministicmatch", "value": 0, "series": 0, "key": "Secondary Sources" }];
objs_1.sort(function(a, b) {
return b.value - a.value;
});
console.log(objs_1);
Solution 44 - Javascript
If I am having a Object like this ,
var dayObj = {
"Friday":["5:00pm to 12:00am"] ,
"Wednesday":["5:00pm to 11:00pm"],
"Sunday":["11:00am to 11:00pm"],
"Thursday":["5:00pm to 11:00pm"],
"Saturday":["11:00am to 12:00am"]
}
want to sort it by day order,
we should have the daySorterMap first,
var daySorterMap = {
// "sunday": 0, // << if sunday is first day of week
"Monday": 1,
"Tuesday": 2,
"Wednesday": 3,
"Thursday": 4,
"Friday": 5,
"Saturday": 6,
"Sunday": 7
}
Initiate a separate Object sortedDayObj,
var sortedDayObj={};
Object.keys(dayObj)
.sort((a,b) => daySorterMap[a] - daySorterMap[b])
.forEach(value=>sortedDayObj[value]= dayObj[value])
You can return the sortedDayObj