I'm looking to sort an array of about 200-300 objects, sorting on a specific key and a given order (asc/desc). The order of results must be consistent and stable.
What would be the best algorithm to use, and could you provide an example of it's implementation in javascript?
Thanks!
It is possible to get a stable sorting from a non-stable sort function.
Before sorting you get the position of all the elements.
In your sort condition, if both elements are equal, then you sort by the position.
Tada! You've got a stable sort.
I've written an article about it on my blog if you want to know more about this technique and how to implement it: http://blog.vjeux.com/2010/javascript/javascript-sorting-table.html
Since you are looking for something stable, the merge sort should do.
http://www.stoimen.com/blog/2010/07/02/friday-algorithms-javascript-merge-sort/
The code can be found at the above website:
function mergeSort(arr)
{
if (arr.length < 2)
return arr;
var middle = parseInt(arr.length / 2);
var left = arr.slice(0, middle);
var right = arr.slice(middle, arr.length);
return merge(mergeSort(left), mergeSort(right));
}
function merge(left, right)
{
var result = [];
while (left.length && right.length) {
if (left[0] <= right[0]) {
result.push(left.shift());
} else {
result.push(right.shift());
}
}
while (left.length)
result.push(left.shift());
while (right.length)
result.push(right.shift());
return result;
}
EDIT:
According to this post, it looks like Array.Sort in some implementations uses a merge sort.
Somewhat shorter version of the same thing using ES2017 features like arrow functions and destructuring:
Function
var stableSort = (arr, compare) => arr
.map((item, index) => ({item, index}))
.sort((a, b) => compare(a.item, b.item) || a.index - b.index)
.map(({item}) => item)
It accepts input array and compare function:
stableSort([5,6,3,2,1], (a, b) => a - b)
It also returns new array instead of making in-place sort like the built-in Array.sort() function.
Test
If we take the following input array, initially sorted by weight:
// sorted by weight
var input = [
{ height: 100, weight: 80 },
{ height: 90, weight: 90 },
{ height: 70, weight: 95 },
{ height: 100, weight: 100 },
{ height: 80, weight: 110 },
{ height: 110, weight: 115 },
{ height: 100, weight: 120 },
{ height: 70, weight: 125 },
{ height: 70, weight: 130 },
{ height: 100, weight: 135 },
{ height: 75, weight: 140 },
{ height: 70, weight: 140 }
]
Then sort it by height using stableSort:
stableSort(input, (a, b) => a.height - b.height)
Results in:
// Items with the same height are still sorted by weight
// which means they preserved their relative order.
var stable = [
{ height: 70, weight: 95 },
{ height: 70, weight: 125 },
{ height: 70, weight: 130 },
{ height: 70, weight: 140 },
{ height: 75, weight: 140 },
{ height: 80, weight: 110 },
{ height: 90, weight: 90 },
{ height: 100, weight: 80 },
{ height: 100, weight: 100 },
{ height: 100, weight: 120 },
{ height: 100, weight: 135 },
{ height: 110, weight: 115 }
]
However sorting the same input array using the built-in Array.sort() (in Chrome/NodeJS):
input.sort((a, b) => a.height - b.height)
Returns:
var unstable = [
{ height: 70, weight: 140 },
{ height: 70, weight: 95 },
{ height: 70, weight: 125 },
{ height: 70, weight: 130 },
{ height: 75, weight: 140 },
{ height: 80, weight: 110 },
{ height: 90, weight: 90 },
{ height: 100, weight: 100 },
{ height: 100, weight: 80 },
{ height: 100, weight: 135 },
{ height: 100, weight: 120 },
{ height: 110, weight: 115 }
]
Resources
Wikipedia
MDN
JSFiddle
Update
Array.prototype.sort is now stable in V8 v7.0 / Chrome 70!
Previously, V8 used an unstable QuickSort for arrays with more than 10 elements. Now, we use the stable TimSort algorithm.
source
I know that this question has been answered for some time, but I happen to have a good stable merge sort implementation for Array and jQuery in my clipboard, so I'll share it in the hopes that some future searchers might find it useful.
It allows you to specify your own comparison function just like the normal Array.sort implementation.
Implementation
// Add stable merge sort to Array and jQuery prototypes
// Note: We wrap it in a closure so it doesn't pollute the global
// namespace, but we don't put it in $(document).ready, since it's
// not dependent on the DOM
(function() {
// expose to Array and jQuery
Array.prototype.mergeSort = jQuery.fn.mergeSort = mergeSort;
function mergeSort(compare) {
var length = this.length,
middle = Math.floor(length / 2);
if (!compare) {
compare = function(left, right) {
if (left < right)
return -1;
if (left == right)
return 0;
else
return 1;
};
}
if (length < 2)
return this;
return merge(
this.slice(0, middle).mergeSort(compare),
this.slice(middle, length).mergeSort(compare),
compare
);
}
function merge(left, right, compare) {
var result = [];
while (left.length > 0 || right.length > 0) {
if (left.length > 0 && right.length > 0) {
if (compare(left[0], right[0]) <= 0) {
result.push(left[0]);
left = left.slice(1);
}
else {
result.push(right[0]);
right = right.slice(1);
}
}
else if (left.length > 0) {
result.push(left[0]);
left = left.slice(1);
}
else if (right.length > 0) {
result.push(right[0]);
right = right.slice(1);
}
}
return result;
}
})();
Example Usage
var sorted = [
'Finger',
'Sandwich',
'sandwich',
'5 pork rinds',
'a guy named Steve',
'some noodles',
'mops and brooms',
'Potato Chip Brand® chips'
].mergeSort(function(left, right) {
lval = left.toLowerCase();
rval = right.toLowerCase();
console.log(lval, rval);
if (lval < rval)
return -1;
else if (lval == rval)
return 0;
else
return 1;
});
sorted == ["5 pork rinds", "a guy named Steve", "Finger", "mops and brooms", "Potato Chip Brand® chips", "Sandwich", "sandwich", "some noodles"];
You can use the following function to perform a stable sort regardless of the native implementation, based on the assertion made in this answer.
Do note that as of ECMAScript 2019, the specification requires that the builtin sort() method perform a stable sort. With that in mind, an explicit stable sort function like the one below is still relevant if you are required to support older browsers that are not specification compliant.
// ECMAScript 5 implementation
function stableSort(array, compareFunction) {
'use strict';
var length = array.length;
var indices = new Uint32Array(length);
var i;
var slice;
// reference values by indices
for (i = 0; i < length; ++i) {
indices[i] = i;
}
// sort with fallback based on indices
indices.sort(function stableCompareFunction(compareFunction, a, b) {
var order = Number(compareFunction(this[a], this[b]));
return order || a - b;
}.bind(array, compareFunction));
slice = array.slice();
// re-order original array to stable sorted values
for (i = 0; i < length; ++i) {
array[i] = slice[indices[i]];
}
return array;
}
// usage
const array = Array(500000).fill().map(() => Number(Math.random().toFixed(4)));
const alwaysEqual = () => 0;
const isUnmoved = (value, index) => value === array[index];
// not guaranteed to be stable before ES2019
console.log(
'sort() stable?',
array.slice().sort(alwaysEqual).every(isUnmoved)
);
// guaranteed to be stable
console.log(
'stableSort() stable?',
stableSort(array.slice(), alwaysEqual).every(isUnmoved)
);
// performance using realistic scenario with unsorted big data
function time(arraySlice, algorithm, compare) {
var start;
var stop;
start = performance.now();
algorithm(arraySlice, compare);
stop = performance.now();
return stop - start;
}
const ascending = (a, b) => a - b;
const msSort = time(array.slice(), (array, compare) => array.sort(compare), ascending);
const msStableSort = time(array.slice(), (array, compare) => stableSort(array, compare), ascending);
console.log('sort()', msSort.toFixed(3), 'ms');
console.log('stableSort()', msStableSort.toFixed(3), 'ms');
console.log('sort() / stableSort()', (100 * msSort / msStableSort).toFixed(3) + '%');
Running the performance tests implemented above, stableSort() appears to run at about 72% of the speed of sort() on version 88 of Google Chrome and Microsoft Edge.
Using .bind() on the inline function within stableSort() used to boost relative performance significantly by avoiding unneeded scoped references on each call.
In practice, this no longer makes a difference since modern engines automatically perform this optimization now, but it is left in the implementation anyway in order to continue improving performance in older browsers which don't ship with this optimization.
The following sorts the supplied array, by applying the supplied compare function, returning the original index comparison when the compare function returns 0:
function stableSort(arr, compare) {
var original = arr.slice(0);
arr.sort(function(a, b){
var result = compare(a, b);
return result === 0 ? original.indexOf(a) - original.indexOf(b) : result;
});
return arr;
}
The example below sorts an array of names by surname, retaining the order of equal surnames:
var names = [
{ surname: "Williams", firstname: "Mary" },
{ surname: "Doe", firstname: "Mary" },
{ surname: "Johnson", firstname: "Alan" },
{ surname: "Doe", firstname: "John" },
{ surname: "White", firstname: "John" },
{ surname: "Doe", firstname: "Sam" }
]
function stableSort(arr, compare) {
var original = arr.slice(0);
arr.sort(function(a, b){
var result = compare(a, b);
return result === 0 ? original.indexOf(a) - original.indexOf(b) : result;
});
return arr;
}
stableSort(names, function(a, b) {
return a.surname > b.surname ? 1 : a.surname < b.surname ? -1 : 0;
})
names.forEach(function(name) {
console.log(name.surname + ', ' + name.firstname);
});
Here's a stable implementation. It works by using the native sort, but in cases where elements compare as equal, you break ties using the original index position.
function stableSort(arr, cmpFunc) {
//wrap the arr elements in wrapper objects, so we can associate them with their origional starting index position
var arrOfWrapper = arr.map(function(elem, idx){
return {elem: elem, idx: idx};
});
//sort the wrappers, breaking sorting ties by using their elements orig index position
arrOfWrapper.sort(function(wrapperA, wrapperB){
var cmpDiff = cmpFunc(wrapperA.elem, wrapperB.elem);
return cmpDiff === 0
? wrapperA.idx - wrapperB.idx
: cmpDiff;
});
//unwrap and return the elements
return arrOfWrapper.map(function(wrapper){
return wrapper.elem;
});
}
a non-thorough test
var res = stableSort([{a:1, b:4}, {a:1, b:5}], function(a, b){
return a.a - b.a;
});
console.log(res);
another answer alluded to this, but didn't post teh codez.
but, its not fast according to my benchmark. I modified a merge sort impl to accept a custom comparator function, and it was much faster.
You can also use Timsort. This is a really complicated algorithm (400+ lines, hence no source code here), so see Wikipedia's description or use one of the existing JavaScript implementations:
GPL 3 implementation. Packaged as Array.prototype.timsort. Appears to be an exact rewrite of Java code.
Public domain implementation Meant as a tutorial, the sample code only shows its use with integers.
Timsort is a highly optimized hybrid of mergesort and shuffle sort and is the default sorting algorithm in Python and in Java (1.7+). It is a complicated algorithm, since it uses different algorithms for many special cases. But as a result it's extremely fast under a wide variety of circumstances.
A simple one mergeSort from http://www.stoimen.com/blog/2010/07/02/friday-algorithms-javascript-merge-sort/
var a = [34, 203, 3, 746, 200, 984, 198, 764, 9];
function mergeSort(arr)
{
if (arr.length < 2)
return arr;
var middle = parseInt(arr.length / 2);
var left = arr.slice(0, middle);
var right = arr.slice(middle, arr.length);
return merge(mergeSort(left), mergeSort(right));
}
function merge(left, right)
{
var result = [];
while (left.length && right.length) {
if (left[0] <= right[0]) {
result.push(left.shift());
} else {
result.push(right.shift());
}
}
while (left.length)
result.push(left.shift());
while (right.length)
result.push(right.shift());
return result;
}
console.log(mergeSort(a));
I have to sort multidimensional arrays by an arbitrary column, and then by another. I use this function to sort:
function sortMDArrayByColumn(ary, sortColumn){
//Adds a sequential number to each row of the array
//This is the part that adds stability to the sort
for(var x=0; x<ary.length; x++){ary[x].index = x;}
ary.sort(function(a,b){
if(a[sortColumn]>b[sortColumn]){return 1;}
if(a[sortColumn]<b[sortColumn]){return -1;}
if(a.index>b.index){
return 1;
}
return -1;
});
}
Notice that ary.sort never returns zero, which is where some implementations of the "sort" function make decisions that might not be right.
This is pretty darn fast, too.
Here's how you could extend JS default Array object with a prototype method utilizing MERGE SORT. This method allows for sorting on a specific key (first parameter) and a given order ('asc'/'desc' as second parameter)
Array.prototype.mergeSort = function(sortKey, direction){
var unsortedArray = this;
if(unsortedArray.length < 2) return unsortedArray;
var middle = Math.floor(unsortedArray.length/2);
var leftSubArray = unsortedArray.slice(0,middle).mergeSort(sortKey, direction);
var rightSubArray = unsortedArray.slice(middle).mergeSort(sortKey, direction);
var sortedArray = merge(leftSubArray, rightSubArray);
return sortedArray;
function merge(left, right) {
var combined = [];
while(left.length>0 && right.length>0){
var leftValue = (sortKey ? left[0][sortKey] : left[0]);
var rightValue = (sortKey ? right[0][sortKey] : right[0]);
combined.push((direction === 'desc' ? leftValue > rightValue : leftValue < rightValue) ? left.shift() : right.shift())
}
return combined.concat(left.length ? left : right)
}
}
You can test this out yourself by dropping the above snippet into your browser console, then trying:
var x = [2,76,23,545,67,-9,12];
x.mergeSort(); //[-9, 2, 12, 23, 67, 76, 545]
x.mergeSort(undefined, 'desc'); //[545, 76, 67, 23, 12, 2, -9]
Or order based on a specific field in an array of objects:
var y = [
{startTime: 100, value: 'cat'},
{startTime: 5, value: 'dog'},
{startTime: 23, value: 'fish'},
{startTime: 288, value: 'pikachu'}
]
y.mergeSort('startTime');
y.mergeSort('startTime', 'desc');
So I needed a stable sort for my React+Redux app, and Vjeux's answer here helped me. However, my (generic) solution seems different than the others I see here so far, so I'm sharing it in case anyone else has a matching use-case:
I really just want to have something similar to the sort() API, where I can pass a comparator function.
Sometimes I can sort in-place, and sometimes my data is immutable (because Redux) and I need a sorted copy instead. So I need a stable sorting function for each use-case.
ES2015.
My solution is to create a typed array of indices, then use a comparison function to sort these indices based on the to-be-sorted array. Then we can use the sorted indices to either sort the original array or create a sorted copy in a single pass. If that's confusing, think of it this way: where you would normally pass a comparison function like:
(a, b) => {
/* some way to compare a and b, returning -1, 0, or 1 */
};
You now instead use:
(i, j) => {
let a = arrayToBeSorted[i], b = arrayToBeSorted[j];
/* some way to compare a and b, returning -1 or 1 */
return i - j; // fallback when a == b
}
Speed is good; it is basically the built-in sorting algorithm is, plus two linear passes in the end, and one extra layer of pointer indirection overhead.
Happy to receive feedback on this approach. Here is my full implementation of it it:
/**
* - `array`: array to be sorted
* - `comparator`: closure that expects indices `i` and `j`, and then
* compares `array[i]` to `array[j]` in some way. To force stability,
* end with `i - j` as the last "comparison".
*
* Example:
* ```
* let array = [{n: 1, s: "b"}, {n: 1, s: "a"}, {n:0, s: "a"}];
* const comparator = (i, j) => {
* const ni = array[i].n, nj = array[j].n;
* return ni < nj ? -1 :
* ni > nj ? 1 :
* i - j;
* };
* stableSortInPlace(array, comparator);
* // ==> [{n:0, s: "a"}, {n:1, s: "b"}, {n:1, s: "a"}]
* ```
*/
function stableSortInPlace(array, comparator) {
return sortFromIndices(array, findIndices(array, comparator));
}
function stableSortedCopy(array, comparator){
let indices = findIndices(array, comparator);
let sortedArray = [];
for (let i = 0; i < array.length; i++){
sortedArray.push(array[indices[i]]);
}
return sortedArray;
}
function findIndices(array, comparator){
// Assumes we don't have to worry about sorting more than
// 4 billion elements; if you know the upper bounds of your
// input you could replace it with a smaller typed array
let indices = new Uint32Array(array.length);
for (let i = 0; i < indices.length; i++) {
indices[i] = i;
}
// after sorting, `indices[i]` gives the index from where
// `array[i]` should take the value from, so to sort
// move the value at at `array[indices[i]]` to `array[i]`
return indices.sort(comparator);
}
// If I'm not mistaken this is O(2n) - each value is moved
// only once (not counting the vacancy temporaries), and
// we also walk through the whole array once more to check
// for each cycle.
function sortFromIndices(array, indices) {
// there might be multiple cycles, so we must
// walk through the whole array to check.
for (let k = 0; k < array.length; k++) {
// advance until we find a value in
// the "wrong" position
if (k !== indices[k]) {
// create vacancy to use "half-swaps" trick,
// props to Andrei Alexandrescu
let v0 = array[k];
let i = k;
let j = indices[k];
while (j !== k) {
// half-swap next value
array[i] = array[j];
// array[i] now contains the value it should have,
// so we update indices[i] to reflect this
indices[i] = i;
// go to next index
i = j;
j = indices[j];
}
// put original array[k] back in
// and update indices
array[i] = v0;
indices[i] = i;
}
}
return array;
}
I know this has been plenty answered. I just wanted to go ahead an post a quick TS implementation for anyone that landed here looking for that.
export function stableSort<T>( array: T[], compareFn: ( a: T, b: T ) => number ): T[] {
const indices = array.map( ( x: T, i: number ) => ( { element: x, index: i } ) );
return indices.sort( ( a, b ) => {
const order = compareFn( a.element, b.element );
return order === 0 ? a.index - b.index : order;
} ).map( x => x.element );
}
The method does no longer run in-place, as the native sort does. I also want to point out that it is not the most efficient. It adds two loops of the order O(n). though sort itself is most likely O(n log(n)) so it's less than that.
Some of the solutions mentioned are more performant, thought this might be less code, also using internal Array.prototype.sort.
(For a Javascript solution, just remove all the types)
According to the v8 dev blog and caniuse.com Array.sort is already stable as required by the spec in modern browsers, so you don't need to roll your own solution.
The only exception I can see is Edge, which should soon move over to chromium and support it as well.
function sort(data){
var result=[];
var array = data;
const array2=data;
const len=array2.length;
for(var i=0;i<=len-1;i++){
var min = Math.min.apply(Math,array)
result.push(min);
var index=array.indexOf(min)
array.splice(index,1);
}
return result;
}
sort([9,8,5,7,9,3,9,243,4,5,6,3,4,2,4,7,4,9,55,66,33,66]);
Counting Sort is faster than merge sort (it performs in O(n) time) and it is intended for use on integers.
Math.counting_sort = function (m) {
var i
var j
var k
var step
var start
var Output
var hash
k = m.length
Output = new Array ()
hash = new Array ()
// start at lowest possible value of m
start = 0
step = 1
// hash all values
i = 0
while ( i < k ) {
var _m = m[i]
hash [_m] = _m
i = i + 1
}
i = 0
j = start
// find all elements within x
while ( i < k ) {
while ( j != hash[j] ) {
j = j + step
}
Output [i] = j
i = i + 1
j = j + step
}
return Output
}
Example:
var uArray = new Array ()<br/>
var sArray = new Array ()<br/><br/>
uArray = [ 10,1,9,2,8,3,7,4,6,5 ]<br/>
sArray = Math.counting_sort ( uArray ) // returns a sorted array
I recently went through an interview where they asked me to come up with a sorting algorithm that really thew me off. Does anyone know the solution to this if it were to be done in javascript?
Problem 1: Double Sort
Please write a method which accepts an array of strings. Each element
can either be a number ("165") or a word ("dog"). Your method should
sort and print the array such that (1) The words are printed in
alphabetical order and the numbers in numerical order, and (2) the
order of words and numbers within the array is the same.
Examples (input => output):
sort(['5', '4', 'dog', '1', 'cat'])
=> ['1', '4', 'cat', '5', 'dog']
sort(['dog', 'cat'])
=> ['cat', 'dog']
sort('5', '3')
=> ['3', '5']
You can use standard library sort functions, and should assume that
all inputs will be valid. If you make any other assumptions, please
document those as well. You can use any programming language that
you'd like.
Additionally, you may assume that you'll be given a utility method
that returns whether a given String is a valid number (e.g.
isNumber(), where isNumber('dog') returns false, and isNumber('15')
returns true).
Here's a simple approach - filter the input array into separate arrays of only strings and only numbers. Then sort the homogeneously typed arrays per their natural ordering. Then produce a final sorted array populated by the type of the indices in the original array.
For example:
function doubleSort(arr) {
// Separate the values by type.
var numbers=[], strings=[];
arr.forEach(function(x) {
if (isNumber(x)) {
numbers.push(Number(x));
} else {
strings.push(x);
}
});
// Sort strings and numbers separately.
strings.sort();
numbers.sort(function(a, b) { return a - b; });
// Merge the sorted arrays by type from the input array.
var sorted=[], nextNumber=0, nextString=0;
arr.forEach(function(x) {
if (isNumber(x)) {
sorted.push(String(numbers[nextNumber++]));
} else {
sorted.push(strings[nextString++]);
}
});
return sorted;
}
// XXX: lots of pitfalls but good enough for this exercise.
function isNumber(x) {
return Number(x).toString() === x;
}
The Big-O performance is bound by the underlying sort algorithm, so likely O(n log n).
Same idea and performance as the other answer. But written in something like coffeescript so it's easier to read
Here's a clumsy variation on quicksort that will sort either the numbers or the words in situ. (I modified a regular JavaScript quicksort posted by Paul Lewis; not sure if all the kinks are completely ironed out...).
function isNumber(x,y) {
return y ? Number(x).toString() !== x : Number(x).toString() === x;
}
function less(a,b,y){
return y ? a < b : Number(a) < Number(b)
}
function swap(a, i, j) { var t = a[i]; a[i] = a[j]; a[j] = t; }
function partition(array, pivot, left, right, what) {
var store = left,
pivotValue = array[pivot];
swap(array, pivot, right);
for (var v = left; v < right; v++) {
if (less(array[v],pivotValue,what) && isNumber(array[v],what)) {
swap(array, v, store);
store++;
}
}
while(!isNumber(array[store],what))
store++;
swap(array, right, store);
return store;
}
function doubleQSort(array, left, right, what) {
while(!isNumber(array[right],what) && right > left)
right--;
while(!isNumber(array[left],what) && left < right)
left++;
var pivot = null;
if (left < right) {
pivot = (right + left) >> 1;
while(!isNumber(array[pivot],what))
pivot--;
newPivot = partition(array, pivot, left, right, what);
doubleQSort(array, left, newPivot - 1,what);
doubleQSort(array, newPivot + 1, right,what);
}
}
Output:
var things = ['dog', 'asdf','31','11','6','fat','cat', '4'];
doubleQSort(things,0,things.length - 1,'words');
doubleQSort(things,0,things.length - 1);
console.log(things);
[ "asdf", "cat", "4", "6", "11", "dog", "fat", "31" ]
I'd like to sum the values of an object.
I'm used to python where it would just be:
sample = { 'a': 1 , 'b': 2 , 'c':3 };
summed = sum(sample.itervalues())
The following code works, but it's a lot of code:
function obj_values(object) {
var results = [];
for (var property in object)
results.push(object[property]);
return results;
}
function list_sum( list ){
return list.reduce(function(previousValue, currentValue, index, array){
return previousValue + currentValue;
});
}
function object_values_sum( obj ){
return list_sum(obj_values(obj));
}
var sample = { a: 1 , b: 2 , c:3 };
var summed = list_sum(obj_values(a));
var summed = object_values_sum(a)
Am i missing anything obvious, or is this just the way it is?
It can be as simple as that:
const sumValues = obj => Object.values(obj).reduce((a, b) => a + b, 0);
Quoting MDN:
The Object.values() method returns an array of a given object's own enumerable property values, in the same order as that provided by a for...in loop (the difference being that a for-in loop enumerates properties in the prototype chain as well).
from Object.values() on MDN
The reduce() method applies a function against an accumulator and each value of the array (from left-to-right) to reduce it to a single value.
from Array.prototype.reduce() on MDN
You can use this function like that:
sumValues({a: 4, b: 6, c: -5, d: 0}); // gives 5
Note that this code uses some ECMAScript features which are not supported by some older browsers (like IE). You might need to use Babel to compile your code.
You could put it all in one function:
function sum( obj ) {
var sum = 0;
for( var el in obj ) {
if( obj.hasOwnProperty( el ) ) {
sum += parseFloat( obj[el] );
}
}
return sum;
}
var sample = { a: 1 , b: 2 , c:3 };
var summed = sum( sample );
console.log( "sum: "+summed );
For fun's sake here is another implementation using Object.keys() and Array.reduce() (browser support should not be a big issue anymore):
function sum(obj) {
return Object.keys(obj).reduce((sum,key)=>sum+parseFloat(obj[key]||0),0);
}
let sample = { a: 1 , b: 2 , c:3 };
console.log(`sum:${sum(sample)}`);
But this seems to be way slower: jsperf.com
If you're using lodash you can do something like
_.sum(_.values({ 'a': 1 , 'b': 2 , 'c':3 }))
Now you can make use of reduce function and get the sum.
const object1 = { 'a': 1 , 'b': 2 , 'c':3 }
console.log(Object.values(object1).reduce((a, b) => a + b, 0));
A regular for loop is pretty concise:
var total = 0;
for (var property in object) {
total += object[property];
}
You might have to add in object.hasOwnProperty if you modified the prototype.
Honestly, given our "modern times" I'd go with a functional programming approach whenever possible, like so:
const sumValues = (obj) => Object.keys(obj).reduce((acc, value) => acc + obj[value], 0);
Our accumulator acc, starting with a value of 0, is accumulating all looped values of our object. This has the added benefit of not depending on any internal or external variables; it's a constant function so it won't be accidentally overwritten... win for ES2015!
Any reason you're not just using a simple for...in loop?
var sample = { a: 1 , b: 2 , c:3 };
var summed = 0;
for (var key in sample) {
summed += sample[key];
};
http://jsfiddle.net/vZhXs/
let prices = {
"apple": 100,
"banana": 300,
"orange": 250
};
let sum = 0;
for (let price of Object.values(prices)) {
sum += price;
}
alert(sum)
I am a bit tardy to the party, however, if you require a more robust and flexible solution then here is my contribution. If you want to sum only a specific property in a nested object/array combo, as well as perform other aggregate methods, then here is a little function I have been using on a React project:
var aggregateProperty = function(obj, property, aggregate, shallow, depth) {
//return aggregated value of a specific property within an object (or array of objects..)
if ((typeof obj !== 'object' && typeof obj !== 'array') || !property) {
return;
}
obj = JSON.parse(JSON.stringify(obj)); //an ugly way of copying the data object instead of pointing to its reference (so the original data remains unaffected)
const validAggregates = [ 'sum', 'min', 'max', 'count' ];
aggregate = (validAggregates.indexOf(aggregate.toLowerCase()) !== -1 ? aggregate.toLowerCase() : 'sum'); //default to sum
//default to false (if true, only searches (n) levels deep ignoring deeply nested data)
if (shallow === true) {
shallow = 2;
} else if (isNaN(shallow) || shallow < 2) {
shallow = false;
}
if (isNaN(depth)) {
depth = 1; //how far down the rabbit hole have we travelled?
}
var value = ((aggregate == 'min' || aggregate == 'max') ? null : 0);
for (var prop in obj) {
if (!obj.hasOwnProperty(prop)) {
continue;
}
var propValue = obj[prop];
var nested = (typeof propValue === 'object' || typeof propValue === 'array');
if (nested) {
//the property is an object or an array
if (prop == property && aggregate == 'count') {
value++;
}
if (shallow === false || depth < shallow) {
propValue = aggregateProperty(propValue, property, aggregate, shallow, depth+1); //recursively aggregate nested objects and arrays
} else {
continue; //skip this property
}
}
//aggregate the properties value based on the selected aggregation method
if ((prop == property || nested) && propValue) {
switch(aggregate) {
case 'sum':
if (!isNaN(propValue)) {
value += propValue;
}
break;
case 'min':
if ((propValue < value) || !value) {
value = propValue;
}
break;
case 'max':
if ((propValue > value) || !value) {
value = propValue;
}
break;
case 'count':
if (propValue) {
if (nested) {
value += propValue;
} else {
value++;
}
}
break;
}
}
}
return value;
}
It is recursive, non ES6, and it should work in most semi-modern browsers. You use it like this:
const onlineCount = aggregateProperty(this.props.contacts, 'online', 'count');
Parameter breakdown:
obj = either an object or an array
property = the property within the nested objects/arrays you wish to perform the aggregate method on
aggregate = the aggregate method (sum, min, max, or count)
shallow = can either be set to true/false or a numeric value
depth = should be left null or undefined (it is used to track the subsequent recursive callbacks)
Shallow can be used to enhance performance if you know that you will not need to search deeply nested data. For instance if you had the following array:
[
{
id: 1,
otherData: { ... },
valueToBeTotaled: ?
},
{
id: 2,
otherData: { ... },
valueToBeTotaled: ?
},
{
id: 3,
otherData: { ... },
valueToBeTotaled: ?
},
...
]
If you wanted to avoid looping through the otherData property since the value you are going to be aggregating is not nested that deeply, you could set shallow to true.
Use Lodash
import _ from 'Lodash';
var object_array = [{a: 1, b: 2, c: 3}, {a: 4, b: 5, c: 6}];
return _.sumBy(object_array, 'c')
// return => 9
I came across this solution from #jbabey while trying to solve a similar problem. With a little modification, I got it right. In my case, the object keys are numbers (489) and strings ("489"). Hence to solve this, each key is parse. The following code works:
var array = {"nR": 22, "nH": 7, "totB": "2761", "nSR": 16, "htRb": "91981"}
var parskey = 0;
for (var key in array) {
parskey = parseInt(array[key]);
sum += parskey;
};
return(sum);
A ramda one liner:
import {
compose,
sum,
values,
} from 'ramda'
export const sumValues = compose(sum, values);
Use:
const summed = sumValues({ 'a': 1 , 'b': 2 , 'c':3 });
We can iterate object using in keyword and can perform any arithmetic operation.
// input
const sample = {
'a': 1,
'b': 2,
'c': 3
};
// var
let sum = 0;
// object iteration
for (key in sample) {
//sum
sum += (+sample[key]);
}
// result
console.log("sum:=>", sum);
A simple solution would be to use the for..in loop to find the sum.
function findSum(obj){
let sum = 0;
for(property in obj){
sum += obj[property];
}
return sum;
}
var sample = { a: 1 , b: 2 , c:3 };
console.log(findSum(sample));
function myFunction(a) { return Object.values(a).reduce((sum, cur) => sum + cur, 0); }
Sum the object key value by parse Integer. Converting string format to integer and summing the values
var obj = {
pay: 22
};
obj.pay;
console.log(obj.pay);
var x = parseInt(obj.pay);
console.log(x + 20);
function totalAmountAdjectives(obj) {
let sum = 0;
for(let el in obj) {
sum += el.length;
}
return sum;
}
console.log(totalAmountAdjectives({ a: "apple" }))
A simple and clean solution for typescrip:
const sample = { a: 1, b: 2, c: 3 };
const totalSample = Object.values(sample).reduce(
(total: number, currentElement: number) => total + currentElement
);
console.log(totalSample);
Good luck!