Count the number of interactions for each reduce operation

Question

I have a list of objects as follows:

const usageCosts = {
    224910186407: {
        deviceId: "224910186407",
        currency: "GBP",
        yearlyUsage: 1480.81
    },
    224910464538: {
        deviceId: "224910464538",
        currency: "GBP",
        yearlyUsage: 617.36
    },
    224910464577: {
        deviceId: "224910464577",
        currency: "EUR",
        yearlyUsage: 522.3
    }
}

I'm doing a sum by currency, like this:

const totalYearlyCost = Object.values(usageCosts).reduce(
      (acc: { [key: string]: any }, stat: any) => {
        if (stat.currency && !acc[stat.currency]) {
          acc[stat.currency] = 0
        }
        return {
          ...acc,
          [stat.currency!]: acc[stat.currency!] stat.yearlyUsage,
        }
      },
      {},
    )

It returns an object as follows:

{
EUR: 522.3
GBP: 2,098.17
}

I also want to return the total number of devices per currency, something like:

{
EUR: 522.3 (1 device)
GBP: 2,098.17 (2 devices)
}

Tried adding another loop, but the results were not as expected.

Answer 1

This mission will be easier if divided into two parts.

First, reduce it into an array containing the grouped values.

Then loop through (you can also use reduce) the object and get the sum of the array and add ${array.length} devices to the string:

const usageCosts = {
    224910186407: {
        deviceId: "224910186407",
        currency: "GBP",
        yearlyUsage: 1480.81
    },
    224910464538: {
        deviceId: "224910464538",
        currency: "GBP",
        yearlyUsage: 617.36
    },
    224910464577: {
        deviceId: "224910464577",
        currency: "EUR",
        yearlyUsage: 522.3
    }
}

let grouped = Object.values(usageCosts).reduce((p, c) => {
    if (!p[c.currency]) p[c.currency] = [];
    p[c.currency].push(c.yearlyUsage);
    return p;
}, {});

for (var key in grouped) {
    grouped[key] = `${grouped[key].reduce((a,b)=>a+b)} (${grouped[key].length}) devices`;
}

console.log(grouped)