计算reduce的每次交互

Question

我有一个对象列表，例如

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
    }
}

我将其减少为按货币求和

 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,
        }
      },
      {},
    )

它返回一个像

这样的对象

{
EUR: 522.3
GBP: 2,098.17
}

我还想返回每种货币的总设备数，例如：

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

尝试添加另一个循环，但它没有按预期工作

Answer 1

分两部分完成此操作要容易得多。

首先

reduce

将其存储为具有分组值的数组。

然后循环（也可以减少）对象，并获取数组的和，并将

${array.length} devices

添加到字符串中：

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)

Answer 2

解决方案如下：

const result = Object.values(usageCosts).reduce((acc, stat) => {
    if (stat.currency && !acc[stat.currency]) {
        acc[stat.currency] = {
            total: 0,
            devices: 0,
        }
    }

    return {
        ...acc,
        [stat.currency]: {
            total: acc[stat.currency].total + stat.yearlyUsage,
            devices: acc[stat.currency].devices + 1,
        }
    };
}, {});

结果，你将得到以下结构：

{
  GBP: { total: 2098.17, devices: 2 },
  EUR: { total: 522.3, devices: 1 }
}