'reduce an Object based on DATE and TIME + calculate a VALUE [duplicate]


I have an object and I want to reduce the date and time together and calculate a number.
At the moment I'm trying the following code (but the 'time' value is missing)...

const myObject = [
    {
        "date": "2023-01-01",
        "time": "18:00",
        "somevalue": 2
    },
    {
        "date": "2023-01-01",
        "time": "18:00",
        "somevalue": 5
    },
    {
        "date": "2023-01-02",
        "time": "19:00",
        "somevalue": 10
    },
    {
        "date": "2023-01-03",
        "time": "06:00",
        "somevalue": 13
    }
];

const all = myObject.reduce( (prev, value) => (
  { ...prev, [value.date]: (prev[value.date] || 0) + value.somevalue } ), {});

// to array
console.log(
  Object.entries(all).map( ([key, value]) => ( {date: key, somevalue: value} ) )
);

My question is: How can I add the "time" value. Like this:

[
    {
        "date": "2023-01-01",
        "time": "18:00",
        "somevalue": 7
    },
    {
        "date": "2023-01-02",
        "time": "19:00",
        "somevalue": 10
    },
    {
        "date": "2023-01-03",
        "time": "06:00",
        "somevalue": 13
    }
]


Solution 1:[1]

Maybe something like this?

const myObject = [ { date: "2023-01-01", time: "18:00", somevalue: 2 }, { date: "2023-01-01", time: "18:00", somevalue: 5 }, { date: "2023-01-02", time: "19:00", somevalue: 10 }, { date: "2023-01-03", time: "06:00", somevalue: 13 }];

const output = myObject.reduce((arr, acc) => {
  const index = arr.findIndex(({ date, time }) => date === acc.date && time === acc.time); //find duplicated elements by date
  if (index > -1) { // if you have duplicated element
    arr[index] = {
      ...arr[index], // merge objects
      somevalue: arr[index].somevalue + acc.somevalue // add value
    };
  } else {
    arr = [...arr, acc]; //if no duplicates values, just add to array. 
  }
  return arr;
}, []);

console.log(output)

I use reduce to merge duplicated elements by "date" and add the someValue.

Sources

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Source: Stack Overflow

Solution Source
Solution 1