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Aggregate values from array of JSON objects based on timestamp

I am relatively new to programming and javascript and i want to achieve the following. Let's assume that we have the following data:

let data =
  [
    {
      "time": 1663163352001,
      "state": "10"
    },
    {
      "time": 1663163352002,
      "state": "20"
    },
    {
      "time": 1663163354002,
      "state": "10"
    },
    {
      "time": 1663163354002,
      "state": "10"
    },
    {
      "time": 1663163355033,
      "state": "30"
    },
    {
      "time": 1663163355035,
      "state": "10"
    },
    {
      "time": 1663163397035,
      "state": "30"
    },
    {
      "time": 1663163397044,
      "state": "50"
    }

  ]

i want to aggregate and calculate the averaged state based on timestamps on time interval (let's say 1 second) so the output should be

let data =
  [
    {
      "time": 1663163352000,
      "state": "15"
    },
    {
      "time": 1663163353000,
      "state": "0"
    },
    {
      "time": 1663163354000,
      "state": "10"
    },
    {
      "time": 1663163355000,
      "state": "20"
    },
    {
      "time": 1663163356000,
      "state": "0"
    },
    {
      "time": 1663163397000,
      "state": "40"
    }

  ]

I know my code is not right but at this is the logic that i have already implement.

//Inputs
var t_interval = 1000;


let aggregatedValues = [];
let count = 0;
let sum = 0;
//starting point converted in seconds
t0 = data[0].time - data[0].time % 1000;
//the timewindow
t_win = t0 + t_interval;

//Just for this case 
for (let i = 0; i < 7; i++) {

  if ( data[i].time < t_win) {
    count += 1;
    sum += parseFloat(data[i].state);
  } 
  else{
    if (sum === 0 || count === 0) {
      avg = NaN;
    }
    avg = sum / count;
    count = 1;
    sum = parseFloat(data[i].state);

    let temp = {
      "time": t_win - t_interval,
      "state": avg
    }
    aggregatedValues.push(temp);
    t_win = t_win + t_interval;
  }
  
}

Could you help me please? Thank you!

Via Active questions tagged javascript - Stack Overflow https://ift.tt/yLSj5RX
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