CODEWITHSUNDEEP
javascriptarraysd3.jsnestedsum
Vinicius Folha
Vinicius Folha

Reputation: 109

Multiple sums of object efficiently d3.js

I have a object like: 

{
  [

      {  id: "AL01",
         list: [
                  { speed : 5
                    value : 10} , 
                  { speed : 7
                    value : 15}
               ]
     }, 
     {  id: "AB01",
        list: [
                { speed : 20
                  value : 1} , 
                { speed : 8
                  value : 10}
              ]
     }

 ]

}

and i would like to have this result like:

{[ {  id: "AL01", speed: 12, value: 25}, {  id: "AB01", speed: 28, value: 11}]}

How can i get this efficiently? If is it possible to run only one time the functions map or forEach? I'm wondering something which go by only one time per each object.

Upvotes: 3

Views: 123

Answers (2)

Redu
Redu

Reputation: 26191

Your data structure is not correct. You have to have a property for the outermost array in the data object. Accordingly one can do as follows;

var data = {groups: [ {id: "AL01", list: [{ speed : 5,  value : 10}, { speed : 7, value : 15}]}, {id: "AB01", list: [{ speed : 20, value : 1 }, { speed : 8, value : 10}]}]},
  result = {groups: data.groups.map(g => Object.assign({id: g.id},g.list.reduce((r,c) => ({speed : r.speed + c.speed, value : r.value + c.value}))))};
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 2

Nina Scholz
Nina Scholz

Reputation: 386680

You need some more loops to get the result.

Basically the outer array, then iterate list and use another array for the keys. Then collect all data and return a new object with a new structure.

var array = [{ id: "AL01", list: [{ speed: 5, value: 10 }, { speed: 7, value: 15 }] }, { id: "AB01", list: [{ speed: 20, value: 1 }, { speed: 8, value: 10 }] }],
    result = array.map(function (a) {
        var temp = { id: a.id, speed: 0, value: 0 };
        a.list.forEach(function (b) {
            ['speed', 'value'].forEach(function (k) {
                temp[k] += b[k];
            });
        });
        return temp;
    });

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

Upvotes: 1

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