How to group by nested by key in array and calculate sum , avg in javascript?

Question

I have an array of data in the below format

let people=[
  {
    aggregationType: "1",
    coefficient: 0.03,
    id: "p1",
    name: "petter",
    type: "Number",
    age: 14
  },
  {
    aggregationType: "0",
    coefficient: 1,
    id: "p2",
    name: "mary",
    type: "Number",
    value: 24
  },
  {
    aggregationType: "1",
    coefficient: 0.03,
    id: "p1",
    name: "Amee",
    type: "Number",
    value: 32
  },
  {
    aggregationType: "0",
    coefficient: 1,
    id: "p2",
    name: "Mtp",
    type: "Number",
    value: 33
  },
  {
    aggregationType: "1",
    coefficient: 0.03,
    id: "p1",
    name: "Louis",
    type: "Number",
    value: 44
  },
]

I want to grouped the dataset by the 'id' and check the condition: if key aggregationType = 0 then calculate sum age, if key aggregationType = 1 then calculate Avg age. And multiply by key 'coefficient' so that the desired result should be like this.

output=[
  [
    "p1", 0.9  // equivalent to ((14 + 32 + 44)/3)*0.03 aggregationType = 1 ==> calculate avg age,coefficient: 0.03
  ],
  [
    "p2", 57  // equivalent to (24+33)*1 = 57  aggregationType = 0 ==> calculate sum age, coefficient: 1
  ]
]

I tried to write this function but I don't know how to continue.

let result = {};

for ( let { id, aggregationType, value } of task ) 
{ 
  result[id] = result[id] || [];
  result[id].push({aggregationType, value }); 
}
console.log(result);
/*
// group by id
result = {
  p1: [
    {
      aggregationType: "1",
      value: 14
    },
    {
      aggregationType: "1",
      value: 32
    },
    {
      aggregationType: "1",
      value: 44
    }
  ],
  p2: [
    {
      aggregationType: "0",
      value: 24
    },
    {
      aggregationType: "0",
      value: 33
    },
    
  ]
}

*/
let newResult=[];
for (let [key, value] of Object.entries(result)) {
    newResult = [key, value.reduce((a, e) => {
    if (e.aggregationType === '0') {
        a += e.value ;
    } else {
       (a += e.value) / value.length;
    }
     return a;
}, 0)]
}

Please help me, thank you!

@trincot Your answer results as I desired. I have not used the set, get in js ever so I will learn more — oneeyes
– oneeyes, Commented Aug 19, 2020 at 1:42
Using a Map (with get, set) is a way to avoid a time complexity of O(n²). When input is very large, this can make a huge difference in performance. — trincot
– trincot, Commented Aug 19, 2020 at 6:12

trincot · Accepted Answer · 2020-08-18 18:11:15Z

1

As you could have more than two distinct id values, I would suggest creating a Map keyed by id where in the corresponding values you store the corresponding aggregation type, coeefficient, and keep track of sum and count.

Then after the accumulation you can extract the values from this Map into the desired output format:

function aggregate(people) {
    let map = new Map;
    for (let {aggregationType, coefficient, id, age} of people) {
        let entry = map.get(id);
        if (!entry) map.set(id, entry = { aggregationType, coefficient, sum: 0, count: 0 });
        entry.sum += age;
        entry.count++;
    }
    return Array.from(map, ([id, { aggregationType, coefficient, sum, count}]) => 
        [id, (+aggregationType ? sum / count : sum) * coefficient]
    );
}

let people=[{aggregationType: "1",coefficient: 0.03,id: "p1",name: "petter",type: "Number",age: 14},{aggregationType: "0",coefficient: 1,id: "p2",name: "mary",type: "Number",age: 24},{aggregationType: "1",coefficient: 0.03,id: "p1",name: "Amee",type: "Number",age: 32},{aggregationType: "0",coefficient: 1,id: "p2",name: "Mtp",type: "Number",age: 33},{aggregationType: "1",coefficient: 0.03,id: "p1",name: "Louis",type: "Number",age: 44},]

console.log(aggregate(people));

answered Aug 18, 2020 at 18:11

trincot

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2 Comments

minhminh Over a year ago

hello sir. What do you mean by code +aggregationType . ([id, (+aggregationType ? sum / count : sum) * coefficient])

trincot Over a year ago

The + is the unary plus: it converts its argument to number, so it will be 0 or 1. 0 is falsy, 1 is truthy for the ternary operator, so that means when aggregationType is "1", this will evaluate to sum / count, and when it is "0", just sum.

Giovanni Esposito · Accepted Answer · 2020-08-18 19:19:43Z

0

Ciao, you could try to use filter function to filter your people based on aggregation and then use a forEach to calculate result:

let people=[
          {
            aggregationType: "1",
            coefficient: 0.03,
            id: "p1",
            name: "petter",
            type: "Number",
            value: 14
          },
          {
            aggregationType: "0",
            coefficient: 1,
            id: "p2",
            name: "mary",
            type: "Number",
            value: 24
          },
          {
            aggregationType: "1",
            coefficient: 0.03,
            id: "p1",
            name: "Amee",
            type: "Number",
            value: 32
          },
          {
            aggregationType: "0",
            coefficient: 1,
            id: "p2",
            name: "Mtp",
            type: "Number",
            value: 33
          },
          {
            aggregationType: "1",
            coefficient: 0.03,
            id: "p1",
            name: "Louis",
            type: "Number",
            value: 44
          },
        ]
        let result = [];
        let aggregationType = ["0", "1"];  //put here all the aggregation types you need
        aggregationType.forEach(aggregation => {
           let peopleFiltered = people.filter(el => el.aggregationType === aggregation)
           let sum = 0;
           let id = "";
           peopleFiltered.forEach(el => {
              sum = sum + (el.value * el.coefficient);
              id = el.id;
           })
           let partialResult = [];
           if (aggregation === "0") {
              partialResult = [ id, sum ]
           }
           else if(aggregation === "1") {
              partialResult = [ id, sum/peopleFiltered.length ]
           }
// extend this 'if...else if' to add new calculation for different aggregation
           
           result.push(partialResult)
        })

        
        console.log(result)

edited Aug 18, 2020 at 19:19

answered Aug 18, 2020 at 17:57

Giovanni Esposito

11.3k1 gold badge19 silver badges33 bronze badges

7 Comments

oneeyes Over a year ago

id p1 and p2 not fixed, i can have p3 p4 .... please help me.. thank u

Giovanni Esposito Over a year ago

@oneeyes I updated my answer with a dynamic solution (you have only to add all the aggregation you want to aggregationType array and extend if...esleif with all the aggregation calculations you need)

trincot Over a year ago

This solution has a time complexity of O(n²), while it can be done with O(n) time complexity.

oneeyes Over a year ago

@Giovanni Esposito may i ask you more. In case the people array has both aggregationType =1 and = 0, the program is desired . if aggregationType can = 0 or 1 then your program will not run as expected. For example : jsfiddle.net/681gktox/14

oneeyes Over a year ago

@Giovanni Esposito I want in case aggregationType = 0 or =1, which is grouped by p1,p2 and calculations. If remove "0" or "1" in aggregationType array , In case the array people has aggregationtype = 1 and 0, the result is not desired

|

caramba · Accepted Answer · 2020-08-18 18:25:19Z

let people=[{aggregationType: "1",coefficient: 0.03,id: "p1",name: "petter",type: "Number",value: 14},{aggregationType: "0",coefficient: 1,id: "p2",name: "mary",type: "Number",value: 24},{aggregationType: "1",coefficient: 0.03,id: "p1",name: "Amee",type: "Number",value: 32},{aggregationType: "0",coefficient: 1,id: "p2",name: "Mtp",type: "Number",value: 33},{aggregationType: "1",coefficient: 0.03,id: "p1",name: "Louis",type: "Number",value: 44}];

var data = {};
people.forEach(function(item, index){
    data[item.id] = data[item.id] || {};
    data[item.id]['result'] = data[item.id]['result'] || 0;
    data[item.id]['values'] = data[item.id]['values'] || [];
    data[item.id]['values'].push(item.value)
    if('1' === item.aggregationType) {
        data[item.id]['result'] =  data[item.id]['values'].reduce(function(a,b){ return a + b},0) / data[item.id]['values'].length * item.coefficient;
    } else {
         data[item.id]['result'] = data[item.id]['result'] + item.value
    }

});

console.log(data.p1.result)
console.log(data.p2.result)

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