Sum javascript object propertyA values with same object propertyB in array of objects

10

How would one take a javascript array of objects such as:

my objArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:42},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23},

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:54}]

and merge duplicate keys by summing the values.
In order to get something like this:

my reducedObjArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:96},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23}]

I have tried iterating and adding to a new array, but this didn't work:

var reducedObjArr = ;

var item = null, key = null;

for(var i=0; i<objArr.length; i++) {

   item=objArr[i];

   key = Object.keys(item)[0];

   item=item[key];



   if(!result[key]){

       result[key] = item;

   }else{

       result[key] += item;}

   }a

javascript arrays json object reduce

share|improve this question

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why are you doing key = Object.keys(item)[0]; item=item[key]; ? You already know the name is key, so just do item.key or objArr[i].key. Also, using the [0] index won't necessarily always give you the same property.
  
  – user2736012
  Oct 7 '13 at 19:50
  
  
  

  
  
  
  

add a comment | 

10

How would one take a javascript array of objects such as:

my objArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:42},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23},

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:54}]

and merge duplicate keys by summing the values.
In order to get something like this:

my reducedObjArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:96},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23}]

I have tried iterating and adding to a new array, but this didn't work:

var reducedObjArr = ;

var item = null, key = null;

for(var i=0; i<objArr.length; i++) {

   item=objArr[i];

   key = Object.keys(item)[0];

   item=item[key];



   if(!result[key]){

       result[key] = item;

   }else{

       result[key] += item;}

   }a

javascript arrays json object reduce

share|improve this question

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why are you doing key = Object.keys(item)[0]; item=item[key]; ? You already know the name is key, so just do item.key or objArr[i].key. Also, using the [0] index won't necessarily always give you the same property.
  
  – user2736012
  Oct 7 '13 at 19:50
  
  
  

  
  
  
  

add a comment | 

10

10

10

5

How would one take a javascript array of objects such as:

my objArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:42},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23},

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:54}]

and merge duplicate keys by summing the values.
In order to get something like this:

my reducedObjArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:96},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23}]

I have tried iterating and adding to a new array, but this didn't work:

var reducedObjArr = ;

var item = null, key = null;

for(var i=0; i<objArr.length; i++) {

   item=objArr[i];

   key = Object.keys(item)[0];

   item=item[key];



   if(!result[key]){

       result[key] = item;

   }else{

       result[key] += item;}

   }a

javascript arrays json object reduce

share|improve this question

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

How would one take a javascript array of objects such as:

my objArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:42},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23},

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:54}]

and merge duplicate keys by summing the values.
In order to get something like this:

my reducedObjArr = [

{key:Mon Sep 23 2013 00:00:00 GMT-0400, val:96},

{key:Mon Sep 24 2013 00:00:00 GMT-0400, val:78},

{key:Mon Sep 25 2013 00:00:00 GMT-0400, val:23}]

I have tried iterating and adding to a new array, but this didn't work:

var reducedObjArr = ;

var item = null, key = null;

for(var i=0; i<objArr.length; i++) {

   item=objArr[i];

   key = Object.keys(item)[0];

   item=item[key];



   if(!result[key]){

       result[key] = item;

   }else{

       result[key] += item;}

   }a

javascript arrays json object reduce

javascript arrays json object reduce

share|improve this question

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

share|improve this question

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

share|improve this question

share|improve this question

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

asked Oct 7 '13 at 19:45

AlecPerkeyAlecPerkey

189719

189719

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why are you doing key = Object.keys(item)[0]; item=item[key]; ? You already know the name is key, so just do item.key or objArr[i].key. Also, using the [0] index won't necessarily always give you the same property.
  
  – user2736012
  Oct 7 '13 at 19:50
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why are you doing key = Object.keys(item)[0]; item=item[key]; ? You already know the name is key, so just do item.key or objArr[i].key. Also, using the [0] index won't necessarily always give you the same property.
  
  – user2736012
  Oct 7 '13 at 19:50
  
  
  

  
  
  
  

Why are you doing key = Object.keys(item)[0]; item=item[key]; ? You already know the name is key, so just do item.key or objArr[i].key. Also, using the [0] index won't necessarily always give you the same property.

– user2736012
Oct 7 '13 at 19:50

Why are you doing key = Object.keys(item)[0]; item=item[key]; ? You already know the name is key, so just do item.key or objArr[i].key. Also, using the [0] index won't necessarily always give you the same property.

– user2736012
Oct 7 '13 at 19:50

add a comment | 

6 Answers
6

active

oldest

votes

10

You should be assigning each object not found to the result with its .key property.

If it is found, then you need to add its .val.

var temp = {};

var obj = null;

for(var i=0; i < objArr.length; i++) {

   obj=objArr[i];



   if(!temp[obj.key]) {

       temp[obj.key] = obj;

   } else {

       temp[obj.key].val += obj.val;

   }

}

var result = ;

for (var prop in temp)

    result.push(temp[prop]);

---

Also, part of the problem was that you were reusing the item variable to reference the value of .key, so you lost reference to the object.

share|improve this answer

edited Oct 7 '13 at 20:13

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

add a comment | 

5

Rather than using a for loop and pushing values, you can directly use map and reduce:

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

share|improve this answer

edited Jun 3 '16 at 21:20

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

add a comment | 

4

You could use a hash table for the grouping by key.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Another approach is to collect all key/value pairs in a Map and format the final array with Array.from and a callback for the objects.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

share|improve this answer

edited Feb 5 at 7:31

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to add multiple keys?
  
  – klent
  Aug 20 '18 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @klent, replace val in this[o.key].val += o.val; with your property to add.
  
  – Nina Scholz
  Aug 20 '18 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.
  
  – klent
  Aug 20 '18 at 8:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.
  
  – Nina Scholz
  Aug 20 '18 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz
  
  – Zum Dummi
  Jan 15 at 9:01
  

  
  
  
  

 | 
show 5 more comments

3

var targetObj = {};

for (var i = 0; i < objArr.length; i++) {

    if (!targetObj.hasOwnProperty(objArr[i].key)) {

        targetObj[objArr[i].key] = 0;

    }

    targetObj[objArr[i].key] += objArr[i].val;

}

http://jsfiddle.net/HUMxp/

share|improve this answer

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OP wants the same key/value pairs, but with the val consolidated.
  
  – user2736012
  Oct 7 '13 at 20:11
  

  
  
  
  

add a comment | 

0

Here is an alternative for you, but similar to that of Explosion Pills, reuses the original array rather than creating a new one or a different object. The sort may not be necessary and will slow things down a little, but it could be removed.

Javascript

function reduceMyObjArr(arr) {

    var temp = {},

        index;



    for (index = arr.length - 1; index >= 0; index -= 1) {

        key = arr[index].key;

        if (temp.hasOwnProperty(key)) {

            arr[temp[key]].val += arr[index].val;

            arr.splice(index, 1);

        } else {

            temp[key] = index;

        }

    }



    arr.sort(function (a, b) {

        if (a.key === b.key) {

            return 0;

        }



        if (a.key < b.key) {

            return -1;

        }



        return 1;

    });



    return arr;

}



var myObjArr = [{

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 42

}, {

    key: "Mon Sep 24 2013 00: 00: 00 GMT - 0400",

    val: 78

}, {

    key: "Mon Sep 25 2013 00: 00: 00 GMT - 0400",

    val: 23

}, {

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 54

}];



reduceMyObjArr(myObjArr);



console.log(myObjArr);

jsFiddle

And a jsperf that compares this (with and without the sort) against the accepted answer. You can improve the performance test by extending the data set.

share|improve this answer

edited May 23 '17 at 11:33

Community♦

11

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

add a comment | 

0

you can also try using javascript linq framework which is exactly same as sql statement which is given desired output with less written code and effective and found at linq.js

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

which is pretty easy as compare to looping.
i hope this may help.

share|improve this answer

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

add a comment | 

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

10

You should be assigning each object not found to the result with its .key property.

If it is found, then you need to add its .val.

var temp = {};

var obj = null;

for(var i=0; i < objArr.length; i++) {

   obj=objArr[i];



   if(!temp[obj.key]) {

       temp[obj.key] = obj;

   } else {

       temp[obj.key].val += obj.val;

   }

}

var result = ;

for (var prop in temp)

    result.push(temp[prop]);

---

Also, part of the problem was that you were reusing the item variable to reference the value of .key, so you lost reference to the object.

share|improve this answer

edited Oct 7 '13 at 20:13

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

add a comment | 

10

You should be assigning each object not found to the result with its .key property.

If it is found, then you need to add its .val.

var temp = {};

var obj = null;

for(var i=0; i < objArr.length; i++) {

   obj=objArr[i];



   if(!temp[obj.key]) {

       temp[obj.key] = obj;

   } else {

       temp[obj.key].val += obj.val;

   }

}

var result = ;

for (var prop in temp)

    result.push(temp[prop]);

---

Also, part of the problem was that you were reusing the item variable to reference the value of .key, so you lost reference to the object.

share|improve this answer

edited Oct 7 '13 at 20:13

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

add a comment | 

10

10

10

You should be assigning each object not found to the result with its .key property.

If it is found, then you need to add its .val.

var temp = {};

var obj = null;

for(var i=0; i < objArr.length; i++) {

   obj=objArr[i];



   if(!temp[obj.key]) {

       temp[obj.key] = obj;

   } else {

       temp[obj.key].val += obj.val;

   }

}

var result = ;

for (var prop in temp)

    result.push(temp[prop]);

---

Also, part of the problem was that you were reusing the item variable to reference the value of .key, so you lost reference to the object.

share|improve this answer

edited Oct 7 '13 at 20:13

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

You should be assigning each object not found to the result with its .key property.

If it is found, then you need to add its .val.

var temp = {};

var obj = null;

for(var i=0; i < objArr.length; i++) {

   obj=objArr[i];



   if(!temp[obj.key]) {

       temp[obj.key] = obj;

   } else {

       temp[obj.key].val += obj.val;

   }

}

var result = ;

for (var prop in temp)

    result.push(temp[prop]);

---

Also, part of the problem was that you were reusing the item variable to reference the value of .key, so you lost reference to the object.

share|improve this answer

edited Oct 7 '13 at 20:13

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

share|improve this answer

share|improve this answer

edited Oct 7 '13 at 20:13

edited Oct 7 '13 at 20:13

edited Oct 7 '13 at 20:13

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

answered Oct 7 '13 at 19:53

user2736012user2736012

3,380811

3,380811

add a comment | 

add a comment | 

5

Rather than using a for loop and pushing values, you can directly use map and reduce:

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

share|improve this answer

edited Jun 3 '16 at 21:20

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

add a comment | 

5

Rather than using a for loop and pushing values, you can directly use map and reduce:

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

share|improve this answer

edited Jun 3 '16 at 21:20

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

add a comment | 

5

5

5

Rather than using a for loop and pushing values, you can directly use map and reduce:

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

share|improve this answer

edited Jun 3 '16 at 21:20

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

Rather than using a for loop and pushing values, you can directly use map and reduce:

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

let objArr = [

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42},

  {key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78},

  {key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23},

  {key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}

];



// first, convert data into a Map with reduce

let counts = objArr.reduce((prev, curr) => {

  let count = prev.get(curr.key) || 0;

  prev.set(curr.key, curr.val + count);

  return prev;

}, new Map());



// then, map your counts object back to an array

let reducedObjArr = [...counts].map(([key, value]) => {

  return {key, value}

})



console.log(reducedObjArr);

share|improve this answer

edited Jun 3 '16 at 21:20

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

share|improve this answer

share|improve this answer

edited Jun 3 '16 at 21:20

edited Jun 3 '16 at 21:20

edited Jun 3 '16 at 21:20

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

answered Jun 3 '16 at 21:15

HammsHamms

3,5511221

3,5511221

add a comment | 

add a comment | 

4

You could use a hash table for the grouping by key.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Another approach is to collect all key/value pairs in a Map and format the final array with Array.from and a callback for the objects.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

share|improve this answer

edited Feb 5 at 7:31

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to add multiple keys?
  
  – klent
  Aug 20 '18 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @klent, replace val in this[o.key].val += o.val; with your property to add.
  
  – Nina Scholz
  Aug 20 '18 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.
  
  – klent
  Aug 20 '18 at 8:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.
  
  – Nina Scholz
  Aug 20 '18 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz
  
  – Zum Dummi
  Jan 15 at 9:01
  

  
  
  
  

 | 
show 5 more comments

4

You could use a hash table for the grouping by key.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Another approach is to collect all key/value pairs in a Map and format the final array with Array.from and a callback for the objects.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

share|improve this answer

edited Feb 5 at 7:31

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to add multiple keys?
  
  – klent
  Aug 20 '18 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @klent, replace val in this[o.key].val += o.val; with your property to add.
  
  – Nina Scholz
  Aug 20 '18 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.
  
  – klent
  Aug 20 '18 at 8:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.
  
  – Nina Scholz
  Aug 20 '18 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz
  
  – Zum Dummi
  Jan 15 at 9:01
  

  
  
  
  

 | 
show 5 more comments

4

4

4

You could use a hash table for the grouping by key.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Another approach is to collect all key/value pairs in a Map and format the final array with Array.from and a callback for the objects.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

share|improve this answer

edited Feb 5 at 7:31

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

You could use a hash table for the grouping by key.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Another approach is to collect all key/value pairs in a Map and format the final array with Array.from and a callback for the objects.

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54}],

    grouped = ;



array.forEach(function (o) {

    if (!this[o.key]) {

        this[o.key] = { key: o.key, val: 0 };

        grouped.push(this[o.key]);

    }

    this[o.key].val += o.val;

}, Object.create(null));



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var array = [{ key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 42 }, { key: 'Mon Sep 24 2013 00:00:00 GMT-0400', val: 78 }, { key: 'Mon Sep 25 2013 00:00:00 GMT-0400', val: 23 }, { key: 'Mon Sep 23 2013 00:00:00 GMT-0400', val: 54 }],

    grouped = Array.from(

        array.reduce((m, { key, val }) => m.set(key, (m.get(key) || 0) + val), new Map),

        ([key, val]) => ({ key, val })

    );



console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

share|improve this answer

edited Feb 5 at 7:31

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

share|improve this answer

share|improve this answer

edited Feb 5 at 7:31

edited Feb 5 at 7:31

edited Feb 5 at 7:31

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

answered Jan 13 '17 at 19:39

Nina ScholzNina Scholz

193k15104177

193k15104177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to add multiple keys?
  
  – klent
  Aug 20 '18 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @klent, replace val in this[o.key].val += o.val; with your property to add.
  
  – Nina Scholz
  Aug 20 '18 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.
  
  – klent
  Aug 20 '18 at 8:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.
  
  – Nina Scholz
  Aug 20 '18 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz
  
  – Zum Dummi
  Jan 15 at 9:01
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to add multiple keys?
  
  – klent
  Aug 20 '18 at 2:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @klent, replace val in this[o.key].val += o.val; with your property to add.
  
  – Nina Scholz
  Aug 20 '18 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.
  
  – klent
  Aug 20 '18 at 8:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.
  
  – Nina Scholz
  Aug 20 '18 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz
  
  – Zum Dummi
  Jan 15 at 9:01
  

  
  
  
  

how to add multiple keys?

– klent
Aug 20 '18 at 2:55

how to add multiple keys?

– klent
Aug 20 '18 at 2:55

@klent, replace val in this[o.key].val += o.val; with your property to add.

– Nina Scholz
Aug 20 '18 at 8:30

@klent, replace val in this[o.key].val += o.val; with your property to add.

– Nina Scholz
Aug 20 '18 at 8:30

Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.

– klent
Aug 20 '18 at 8:40

Sorry, I've meant how to add another condition before it adds for example if their's another property called 'category'. I want to check key and category.

– klent
Aug 20 '18 at 8:40

1

1

take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.

– Nina Scholz
Aug 20 '18 at 8:46

take an array for the keys to group, like groups = ['foo', 'bar'], inside the forEach use var key = groups.map(k => o[k]).join('|'); and then use this[key] instead of this[o.key]. maybe omit this at all and use a global variable for the object for grouping.

– Nina Scholz
Aug 20 '18 at 8:46

how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz

– Zum Dummi
Jan 15 at 9:01

how to sum same values in object if my obj something like this obj = [ { menu: "apple", amount: 3}, { menu: "apple", amount: 1}, {menu: "melon", amount: 10}, {menu: "stawberry", amount: 7} ]; i want the ouput is [ {menu: apple, total: 4}, {menu: melon, total: 10}, and.......... ] @NinaScholz

– Zum Dummi
Jan 15 at 9:01

 | 
show 5 more comments

3

var targetObj = {};

for (var i = 0; i < objArr.length; i++) {

    if (!targetObj.hasOwnProperty(objArr[i].key)) {

        targetObj[objArr[i].key] = 0;

    }

    targetObj[objArr[i].key] += objArr[i].val;

}

http://jsfiddle.net/HUMxp/

share|improve this answer

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OP wants the same key/value pairs, but with the val consolidated.
  
  – user2736012
  Oct 7 '13 at 20:11
  

  
  
  
  

add a comment | 

3

var targetObj = {};

for (var i = 0; i < objArr.length; i++) {

    if (!targetObj.hasOwnProperty(objArr[i].key)) {

        targetObj[objArr[i].key] = 0;

    }

    targetObj[objArr[i].key] += objArr[i].val;

}

http://jsfiddle.net/HUMxp/

share|improve this answer

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OP wants the same key/value pairs, but with the val consolidated.
  
  – user2736012
  Oct 7 '13 at 20:11
  

  
  
  
  

add a comment | 

3

3

3

var targetObj = {};

for (var i = 0; i < objArr.length; i++) {

    if (!targetObj.hasOwnProperty(objArr[i].key)) {

        targetObj[objArr[i].key] = 0;

    }

    targetObj[objArr[i].key] += objArr[i].val;

}

http://jsfiddle.net/HUMxp/

share|improve this answer

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

var targetObj = {};

for (var i = 0; i < objArr.length; i++) {

    if (!targetObj.hasOwnProperty(objArr[i].key)) {

        targetObj[objArr[i].key] = 0;

    }

    targetObj[objArr[i].key] += objArr[i].val;

}

http://jsfiddle.net/HUMxp/

share|improve this answer

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

share|improve this answer

share|improve this answer

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

answered Oct 7 '13 at 19:53

Explosion PillsExplosion Pills

151k38228317

151k38228317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OP wants the same key/value pairs, but with the val consolidated.
  
  – user2736012
  Oct 7 '13 at 20:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OP wants the same key/value pairs, but with the val consolidated.
  
  – user2736012
  Oct 7 '13 at 20:11
  

  
  
  
  

OP wants the same key/value pairs, but with the val consolidated.

– user2736012
Oct 7 '13 at 20:11

OP wants the same key/value pairs, but with the val consolidated.

– user2736012
Oct 7 '13 at 20:11

add a comment | 

0

Here is an alternative for you, but similar to that of Explosion Pills, reuses the original array rather than creating a new one or a different object. The sort may not be necessary and will slow things down a little, but it could be removed.

Javascript

function reduceMyObjArr(arr) {

    var temp = {},

        index;



    for (index = arr.length - 1; index >= 0; index -= 1) {

        key = arr[index].key;

        if (temp.hasOwnProperty(key)) {

            arr[temp[key]].val += arr[index].val;

            arr.splice(index, 1);

        } else {

            temp[key] = index;

        }

    }



    arr.sort(function (a, b) {

        if (a.key === b.key) {

            return 0;

        }



        if (a.key < b.key) {

            return -1;

        }



        return 1;

    });



    return arr;

}



var myObjArr = [{

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 42

}, {

    key: "Mon Sep 24 2013 00: 00: 00 GMT - 0400",

    val: 78

}, {

    key: "Mon Sep 25 2013 00: 00: 00 GMT - 0400",

    val: 23

}, {

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 54

}];



reduceMyObjArr(myObjArr);



console.log(myObjArr);

jsFiddle

And a jsperf that compares this (with and without the sort) against the accepted answer. You can improve the performance test by extending the data set.

share|improve this answer

edited May 23 '17 at 11:33

Community♦

11

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

add a comment | 

0

Here is an alternative for you, but similar to that of Explosion Pills, reuses the original array rather than creating a new one or a different object. The sort may not be necessary and will slow things down a little, but it could be removed.

Javascript

function reduceMyObjArr(arr) {

    var temp = {},

        index;



    for (index = arr.length - 1; index >= 0; index -= 1) {

        key = arr[index].key;

        if (temp.hasOwnProperty(key)) {

            arr[temp[key]].val += arr[index].val;

            arr.splice(index, 1);

        } else {

            temp[key] = index;

        }

    }



    arr.sort(function (a, b) {

        if (a.key === b.key) {

            return 0;

        }



        if (a.key < b.key) {

            return -1;

        }



        return 1;

    });



    return arr;

}



var myObjArr = [{

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 42

}, {

    key: "Mon Sep 24 2013 00: 00: 00 GMT - 0400",

    val: 78

}, {

    key: "Mon Sep 25 2013 00: 00: 00 GMT - 0400",

    val: 23

}, {

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 54

}];



reduceMyObjArr(myObjArr);



console.log(myObjArr);

jsFiddle

And a jsperf that compares this (with and without the sort) against the accepted answer. You can improve the performance test by extending the data set.

share|improve this answer

edited May 23 '17 at 11:33

Community♦

11

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

add a comment | 

0

0

0

Here is an alternative for you, but similar to that of Explosion Pills, reuses the original array rather than creating a new one or a different object. The sort may not be necessary and will slow things down a little, but it could be removed.

Javascript

function reduceMyObjArr(arr) {

    var temp = {},

        index;



    for (index = arr.length - 1; index >= 0; index -= 1) {

        key = arr[index].key;

        if (temp.hasOwnProperty(key)) {

            arr[temp[key]].val += arr[index].val;

            arr.splice(index, 1);

        } else {

            temp[key] = index;

        }

    }



    arr.sort(function (a, b) {

        if (a.key === b.key) {

            return 0;

        }



        if (a.key < b.key) {

            return -1;

        }



        return 1;

    });



    return arr;

}



var myObjArr = [{

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 42

}, {

    key: "Mon Sep 24 2013 00: 00: 00 GMT - 0400",

    val: 78

}, {

    key: "Mon Sep 25 2013 00: 00: 00 GMT - 0400",

    val: 23

}, {

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 54

}];



reduceMyObjArr(myObjArr);



console.log(myObjArr);

jsFiddle

And a jsperf that compares this (with and without the sort) against the accepted answer. You can improve the performance test by extending the data set.

share|improve this answer

edited May 23 '17 at 11:33

Community♦

11

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

Here is an alternative for you, but similar to that of Explosion Pills, reuses the original array rather than creating a new one or a different object. The sort may not be necessary and will slow things down a little, but it could be removed.

Javascript

function reduceMyObjArr(arr) {

    var temp = {},

        index;



    for (index = arr.length - 1; index >= 0; index -= 1) {

        key = arr[index].key;

        if (temp.hasOwnProperty(key)) {

            arr[temp[key]].val += arr[index].val;

            arr.splice(index, 1);

        } else {

            temp[key] = index;

        }

    }



    arr.sort(function (a, b) {

        if (a.key === b.key) {

            return 0;

        }



        if (a.key < b.key) {

            return -1;

        }



        return 1;

    });



    return arr;

}



var myObjArr = [{

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 42

}, {

    key: "Mon Sep 24 2013 00: 00: 00 GMT - 0400",

    val: 78

}, {

    key: "Mon Sep 25 2013 00: 00: 00 GMT - 0400",

    val: 23

}, {

    key: "Mon Sep 23 2013 00: 00: 00 GMT - 0400",

    val: 54

}];



reduceMyObjArr(myObjArr);



console.log(myObjArr);

jsFiddle

And a jsperf that compares this (with and without the sort) against the accepted answer. You can improve the performance test by extending the data set.

share|improve this answer

edited May 23 '17 at 11:33

Community♦

11

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

share|improve this answer

share|improve this answer

edited May 23 '17 at 11:33

Community♦

11

edited May 23 '17 at 11:33

Community♦

11

edited May 23 '17 at 11:33

Community♦

11

11

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

answered Oct 7 '13 at 21:00

Xotic750Xotic750

16.4k74165

16.4k74165

add a comment | 

add a comment | 

0

you can also try using javascript linq framework which is exactly same as sql statement which is given desired output with less written code and effective and found at linq.js

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

which is pretty easy as compare to looping.
i hope this may help.

share|improve this answer

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

add a comment | 

0

you can also try using javascript linq framework which is exactly same as sql statement which is given desired output with less written code and effective and found at linq.js

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

which is pretty easy as compare to looping.
i hope this may help.

share|improve this answer

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

add a comment | 

0

0

0

you can also try using javascript linq framework which is exactly same as sql statement which is given desired output with less written code and effective and found at linq.js

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

which is pretty easy as compare to looping.
i hope this may help.

share|improve this answer

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

you can also try using javascript linq framework which is exactly same as sql statement which is given desired output with less written code and effective and found at linq.js

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

which is pretty easy as compare to looping.
i hope this may help.

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

var objArr = 

[

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:42},

{key:'Mon Sep 24 2013 00:00:00 GMT-0400', val:78},

{key:'Mon Sep 25 2013 00:00:00 GMT-0400', val:23},

{key:'Mon Sep 23 2013 00:00:00 GMT-0400', val:54}

];





var aggregatedObject = Enumerable.From(objArr)

        .GroupBy("$.key", null,

                 function (key, g) {

                     return {

                       key: key,

                       contributions: g.Sum("$.val")

                     }

        })

        .ToArray();



console.log(aggregatedObject);

<script src="http://cdnjs.cloudflare.com/ajax/libs/linq.js/2.2.0.2/linq.min.js"></script>

share|improve this answer

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

share|improve this answer

share|improve this answer

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

answered Jul 10 '16 at 18:32

Sanjay RadadiyaSanjay Radadiya

932820

932820

add a comment | 

add a comment | 

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