Re indexing array-object in jquery

Question

I have an array with object like this

var arr = [
  {'key1':'A','key2':1,'key3':0.04,'key4':10},
  {'key1':'B','key2':1,'key3':0.03,'key4':20},
  {'key1':'B','key2':2,'key3':0.05,'key4':30},
  {'key1':'A','key2':1,'key3':0.03,'key4':40},
  ...
];

Inside array there can be unlimited objects. I need to create new array where every key1 and key2 must save unique and if same key1 and key2,then key3 must be added. My result for above array look like this

//Can be array or object
var a = [
  ['A'][1] => {'key3':0.07,'key4':50},
  ['B'][1] => {'key3':0.07,'key4':20},
  ['B'][2] => {'key3':0.05,'key4':30}
];

My result can be array or object. What I have tried is

var a = {};
arr.forEach(function(obj){
  if(a[obj.key1]){
    if(a[obj.key1][obj.key2]){
      a[obj.key1][obj.key2] += obj.key3;
    }else{
      a[obj.key1][obj.key2] = obj.key3;
    }
  }else{
    a[obj.key1] = obj.key1;
  }
});
console.log(a);

And I am getting only depth of result. Please suggest me how can I get my result output from given array object. Thanks in advance.

sorry, my mistake.Please check the edit.

Dev
– Dev

2017-03-29 09:18:26 +00:00
Commented Mar 29, 2017 at 9:18 — Dev
– Dev, Commented Mar 29, 2017 at 9:18

Nina Scholz · Accepted Answer · 2017-03-29 14:33:27Z

2

For more nested objects, you could use an array for grouping the keys and one for the keys with the value to add. Then build a new object, if necessary and assign the values.

var array = [{ key1: 'A', key2: 1, key3: 0.04, key4: 10 }, { key1: 'B', key2: 1, key3: 0.03, key4: 20 }, { key1: 'B', key2: 2, key3: 0.05, key4: 30 }, { key1: 'A', key2: 1, key3: 0.03, key4: 40 }],
    groupKeys = ['key1', 'key2'],
    valueKeys = ['key3', 'key4'],
    result = Object.create(null);

array.forEach(function (a) {
    var object = groupKeys.reduce(function (r, k) {
        return r[a[k]] = r[a[k]] || {};
    }, result);
    valueKeys.forEach(function (k) {
        object[k] = (object[k] || 0) + a[k];
    });
});

console.log(result);

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

edited Mar 29, 2017 at 14:33

answered Mar 29, 2017 at 9:21

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Dev Over a year ago

if i need to add more value inside like ['A'][1][0.07]=> 4,How can i achieve this.Is works but I don't get how its work.Thank You.

Dev Over a year ago

It works and Thanks. But I prefer before one solution without define key.

Nina Scholz Over a year ago

where do you know, which keys is a grouping key and which is for the value?

Kinjal Akhani · Accepted Answer · 2017-03-29 11:04:14Z

0

var arr = [
    {'key1':'A','key2':1,'key3':0.04},
    {'key1':'B','key2':1,'key3':0.03},
    {'key1':'B','key2':2,'key3':0.05},
    {'key1':'A','key2':1,'key3':0.03}
];
var tempArr = $.extend(true, [], arr);
var newArray = new Array();
var len = arr.length;
for(var i=0;i<len;i++){
    var totalk3 = arr[i].key3;

    for(var j=i+1;j<tempArr.length;j++){

        if(arr[i].key1 == tempArr[j].key1 && arr[i].key2 == tempArr[j].key2){
            totalk3 = totalk3 + tempArr[j].key3;
            arr.splice(j,1);
        }
    }

    var json1 = {};
    var json2 = {};
    json2[arr[i].key2] = totalk3;
    json1[arr[i].key1]=json2;
    newArray[i] = json1;

    len = arr.length;
}
console.log(newArray);

answered Mar 29, 2017 at 11:04

Kinjal Akhani

16812 bronze badges

2 Comments

ry8806 Over a year ago

please explain why this answers the question

Kinjal Akhani Over a year ago

Iterated with two loops the outer one is for all individual JSON object for main Array [arr] and inner one is for addition of key3's from json's where key1 and key2 is same in any case and after addition will remove that [arr] json from outer loop and its length is also updated.

Collectives™ on Stack Overflow

Re indexing array-object in jquery

2 Answers 2

3 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related