How to create a distinct list from array and sum numeric values

Question

I have the following array:

var array = []
array.push({payslipId: 1759});
array.push({payDate: "2019-04-19T00:00:00+00:00"});
array.push({periodsPerYear: 52});
array.push({taxPeriod: 2});
array.push({payslipId: 1760});
array.push({payDate: "2019-04-19T00:00:00+00:00"});
array.push({periodsPerYear: 52});
array.push({taxPeriod: 2});

What I am hoping to be able to do is create an array of distinct names, and where the values are numeric, the total off all in the above array would be shown.

The only way I can think to do this is to create a new array, and add each name to it, then loop around the above array, and add to the value to the name, if the value is numeric.

I am not overly familiar with javascript, so am hoping there is a neater way, a bit like linq in C#?

Desired output:

[{ payslipId: 3519 }, { periodsPerYear: 104} , { taxPeriod: 4}]

The array has four different key-value pairs: payslipId, payDate, 'periodsPerYear` and taxPeriod. When you say you want to add up numeric values do you mean all fields except payDate? Why do wan to add payslipId with period related fields? — fiveobjects
– fiveobjects, Commented Aug 29, 2019 at 18:13
Can you show how your desired output looks like ? IMO you're looking for Object — Code Maniac
– Code Maniac, Commented Aug 29, 2019 at 18:14
@PM77-1 hmm, I ideally don't want to have to tell a function what the fields are, I'd like to group it, as the array can be different, and I want a generic function. — Keith
– Keith, Commented Aug 29, 2019 at 18:27

Nina Scholz · Accepted Answer · 2019-08-29 19:13:33Z

1

You could take a Map and reduce the array by taking the only key/value pair of the object.

function group(array) {
    return Array.from(
        array.reduce((m, o) => {
            var [k, v] = Object.entries(o)[0];
            return typeof v === 'number'
                ? m.set(k, (m.get(k) || 0) + v)
                : m;
        }, new Map),
        ([k, v]) => ({ [k]: v })
    );
}

var array = [{ payslipId: 1759 }, { payDate: "2019-04-19T00:00:00+00:00" }, { periodsPerYear: 52 }, { taxPeriod: 2 }, { payslipId: 1760 }, { payDate: "2019-04-19T00:00:00+00:00" }, { periodsPerYear: 52 }, { taxPeriod: 2 }],
    result = group(array);

console.log(result);

answered Aug 29, 2019 at 19:13

Nina Scholz

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Comments

fiveobjects · Accepted Answer · 2019-08-29 19:04:48Z

As I have mentioned in the comment to the original post, you can use reduce() function for aggregation. You may apply any condition you need to for aggregation. As you have mentioned, I have considered number fields only.

Here you go:

var array = []
array.push({ payslipId: 1759 });
array.push({ payDate: "2019-04-19T00:00:00+00:00" });
array.push({ periodsPerYear: 52 });
array.push({ taxPeriod: 2 });
array.push({ payslipId: 1760 });
array.push({ payDate: "2019-04-19T00:00:00+00:00" });
array.push({ periodsPerYear: 52 });
array.push({ taxPeriod: 2 });

//console.log(array);
result = array.reduce((aggr, item) => {
    itemKey = Object.keys(item)[0];
    if (Number.isInteger(item[itemKey])) {
        if (!aggr[itemKey]) {
            aggr[itemKey] = 0;
        }
        aggr[itemKey] = parseInt(aggr[itemKey]) + parseInt(item[itemKey]);
    }
    return aggr;
}, {});
console.log(result);

Addis · Accepted Answer · 2019-08-29 23:34:47Z

0

Another option could be:

let array2 = [array[0]];
array.shift();

array.forEach(obj => {
  if(typeof(obj[Object.keys(obj)[0]]) !== 'number')  return;
  //test if the object with the same property name already exists to array2 or not. If it doesn't exist, it will be added to array2. If it does only the value will be added to the existing element
  let test = array2.reduce((accum,obj2) => {
    return accum && (Object.keys(obj2)[0] !== Object.keys(obj)[0]);
  }, true);
  if(test)     array2.push(obj);
  else  array2.find(obj2 => obj2.hasOwnProperty(Object.keys(obj)[0]))[Object.keys(obj)[0]] += obj[Object.keys(obj)[0]];
});

console.log(array2);

edited Aug 29, 2019 at 23:34

answered Aug 29, 2019 at 23:23

Addis

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