javascript - 过滤和映射后如何获取用户ID

Question

我正在尝试为每个服务创建一个包含名称和用户 ID 的对象，我希望该对象看起来像这样:

const object = {
    name: Netflix, user: (user id who pays max price for this provider)
    name: Comcast, user: (same)
    name: Verizon, user: (same)
}

我已经尝试更改映射返回中的对象属性，但它不起作用，而且我已经完成了一半的对象我的意思是提供者的名称现在我需要另一个键，值对

const services = [
     { userid: 1, providerId: 1, amount: 250000 },
     { userid: 4, providerId: 3, amount: 280900 },
     { userid: 6, providerId: 3, amount: 31000 },
     { userid: 2, providerId: 2, amount: 58600 },
     { userid: 3, providerId: 1, amount: 13000 },
     { userid: 5, providerId: 2, amount: 5000 },
     { userid: 3, providerId: 3, amount: 59900 },
     { userid: 6, providerId: 3, amount: 9500 }
]

const providers = [
    { id: 1, name: Netflix },
    { id: 2, name: Comcast },
    { id: 3, name: Verizon }
]

这是我的功能

function getUserId(providerId) {
    return services.filter(function(obj) {
        if (obj.providerId == providerId)
            return obj.providerId;
    }).map(function(obj) { return obj.amount });
}

function getMaxUserId(providerId) {
    return Math.max(...getUserId(providerId));
}

providers.forEach(prov => {
    object[prov.name] = getUserId(prov.id);
})

如您所见，我首先过滤整个数组以寻找具有特定 providerId 的提供者，然后我创建一个新数组，其中包含 map 函数，其中填充了该特定提供者的所有“金额”，最后我搜索了该提供者的最大金额数组，这一切都很好，它为每个提供者返回正确的最大金额，但我还想获得为每个提供者支付最多的用户 ID

实际上我得到的对象是这样的:

[
   {name: Netflix, user: 250000},
   {name: Comcast, user: 58600},
   {name: Verizon, user: 280900}
]

我需要这个:

[
   {name: Netflix, user: 1},
   {name: Comcast, user: 2},
   {name: Verizon, user: 4}
]

Answer 1

您还可以根据数量排序您的服务。然后您可以通过 providerId 对它们进行分组，最后映射提供程序以获得您想要的输出:

const services = [ { id: 1, providerId: 1, amount: 250000 }, { id: 4, providerId: 3, amount: 280900 }, { id: 6, providerId: 3, amount: 31000 }, { id: 2, providerId: 2, amount: 58600 }, { id: 3, providerId: 1, amount: 13000 }, { id: 5, providerId: 2, amount: 5000 }, { id: 3, providerId: 3, amount: 59900 }, { id: 6, providerId: 3, amount: 9500 } ]
const providers = [ { id: 1, name: 'Netflix' }, { id: 2, name: 'Comcast' }, { id: 3, name: 'Verizon' } ] 

let grouped = services
  .sort((a,b) => b.amount - a.amount)  // sort `desc`
  .reduce((r,c) => ((r[c.providerId] = r[c.providerId] || []).push(c), r), {})
	
let result = providers.map(p => ({name: p.name, user: grouped[p.id][0].id}))

console.log(result)

您也可以跳过上面的sort 并检查amount 并基于push 或unshift :

const services = [ { id: 1, providerId: 1, amount: 250000 }, { id: 4, providerId: 3, amount: 280900 }, { id: 6, providerId: 3, amount: 31000 }, { id: 2, providerId: 2, amount: 58600 }, { id: 3, providerId: 1, amount: 13000 }, { id: 5, providerId: 2, amount: 5000 }, { id: 3, providerId: 3, amount: 59900 }, { id: 6, providerId: 3, amount: 9500 } ]
const providers = [ { id: 1, name: 'Netflix' }, { id: 2, name: 'Comcast' }, { id: 3, name: 'Verizon' } ] 

let groupedById = services.reduce((acc, cur) => { 
  let k = cur.providerId
  acc[k] = acc[k] || []
  if(acc[k][0] && acc[k][0].amount > cur.amount) acc[k].push(cur)
  else acc[k].unshift(cur)
  return acc
}, {})

let result = providers.map(({name, id}) => ({name, user: groupedById[id][0].id}))

console.log(result)

这样你只有一个 Array.reduce 和一个 Array.map 这将比你的原始版本 性能更高 forEach、filter 和 map。