Question

假设我们有对象的初始数组：

{vendor:"vendor1", item:"item1", price:1100, rank:0},
{vendor:"vendor1", item:"item2",price:3200, rank:0},
{vendor:"vendor1", item:"item3", price:1100, rank:0},

{vendor:"vendor2", item:"item1", price:2000, rank:0},
{vendor:"vendor2", item:"item2",price:2000, rank:0},
{vendor:"vendor2", item:"item3", price:3200, rank:0},

{vendor:"vendor3", item:"item1", price:3200, rank:0},
{vendor:"vendor3", item:"item2",price:1100, rank:0},
{vendor:"vendor3", item:"item3", price:2000, rank:0},

与所有具有相同项目的供应商相比，如何根据其项目类别填充基于价格属性的排名。

所需结果：

{vendor:"vendor1", item:"item1", price:1100, rank:1},
{vendor:"vendor1", item:"item2",price:3200, rank:3},
{vendor:"vendor1", item:"item3", price:1100, rank:1},

{vendor:"vendor2", item:"item1", price:2000, rank:2},
{vendor:"vendor2", item:"item2",price:2000, rank:2},
{vendor:"vendor2", item:"item3", price:3200, rank:3},

{vendor:"vendor3", item:"item1", price:3200, rank:3},
{vendor:"vendor3", item:"item2",price:1100, rank:1},
{vendor:"vendor3", item:"item3", price:2000, rank:2},

Answer 1

STEPS：

获取独特的物品。
按唯一商品将整个数组过滤成小的对象数组，然后按价格排序。
更新每个单独数组中每个对象的排名。
将它们重新串联在一起。
按供应商排序将其排序为原始订单。

let arr = [{vendor:"vendor1", item:"item1", price:1000, rank:0},
{vendor:"vendor1", item:"item2",price:3000, rank:0},
{vendor:"vendor1", item:"item3", price:1000, rank:0},
{vendor:"vendor2", item:"item1", price:2000, rank:0},
{vendor:"vendor2", item:"item2",price:2000, rank:0},
{vendor:"vendor2", item:"item3", price:3000, rank:0},
{vendor:"vendor3", item:"item1", price:3000, rank:0},
{vendor:"vendor3", item:"item2",price:1000, rank:0},
{vendor:"vendor3", item:"item3", price:2000, rank:0},]
let items = [...new Set(arr.map(o => o.item))]
let resultArr = []
items.forEach(item => {
  let filteredArr = arr.filter(o => o.item === item)
  filteredArr.sort((a,b) => (a.price > b.price) ? 1 : ((b.price > a.price) ? -1 : 0))
  for(let i = 0; i < filteredArr.length; i++){
    filteredArr[i].rank = i + 1
  }
  resultArr = resultArr.concat(filteredArr)
})
resultArr.sort((a,b) => (a.vendor > b.vendor) ? 1 : ((b.vendor > a.vendor) ? -1 : 0))
console.log(resultArr)

Answer 2

尝试此.rank是根据价格值创建的

var arr = [{vendor:"vendor1", item:"item1", price:1000, rank:0}, {vendor:"vendor1", item:"item2",price:1000, rank:0}, {vendor:"vendor1", item:"item3", price:1000, rank:0}, {vendor:"vendor2", item:"item1", price:2000, rank:0}, {vendor:"vendor2", item:"item2",price:2000, rank:0}, {vendor:"vendor2", item:"item3", price:2000, rank:0}, {vendor:"vendor3", item:"item1", price:3000, rank:0}, {vendor:"vendor3", item:"item2",price:3000, rank:0}, {vendor:"vendor3", item:"item3", price:3000, rank:0}];
arr.forEach(a => a.rank = (a.price/1000))
console.log(arr)

Answer 3

您可以收集分组的price，对其进行排序，然后将更正后的索引应用为rank。

var data = [{ vendor: "vendor1", item: "item1", price: 1100, rank: 0 }, { vendor: "vendor1", item: "item2", price: 3200, rank: 0 }, { vendor: "vendor1", item: "item3", price: 1100, rank: 0 }, { vendor: "vendor2", item: "item1", price: 2000, rank: 0 }, { vendor: "vendor2", item: "item2", price: 2000, rank: 0 }, { vendor: "vendor2", item: "item3", price: 3200, rank: 0 }, { vendor: "vendor3", item: "item1", price: 3200, rank: 0 }, { vendor: "vendor3", item: "item2", price: 1100, rank: 0 }, { vendor: "vendor3", item: "item3", price: 2000, rank: 0 }],
    collection = data.reduce((r, o) => ((r[o.item] = r[o.item] || []).push(o.price), r), {});

Object.values(collection).forEach(a => a.sort((a, b) => a - b));

data.forEach(o => o.rank = collection[o.item].indexOf(o.price) + 1);

console.log(data);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

您可以基于item和price sort数组。然后，循环遍历已排序的数组，并根据先前的rank与当前的item是否相同来分配item：

const arr = [{vendor:"vendor1",item:"item1",price:1100,rank:0},{vendor:"vendor1",item:"item2",price:3200,rank:0},{vendor:"vendor1",item:"item3",price:1100,rank:0},{vendor:"vendor2",item:"item1",price:2000,rank:0},{vendor:"vendor2",item:"item2",price:2000,rank:0},{vendor:"vendor2",item:"item3",price:3200,rank:0},{vendor:"vendor3",item:"item1",price:3200,rank:0},{vendor:"vendor3",item:"item2",price:1100,rank:0},{vendor:"vendor3",item:"item3",price:2000,rank:0}]

const output = arr.sort((a, b) => a.item.localeCompare(b.item) || a.price - b.price)
  .map((o, i) => {
    const prev = arr[arr.length - 1]
    const rank = prev && prev.item === o.item
                   ? prev.rank + 1
                   : 1;
                   
    return { ...o, rank }
  })

console.log(output)

Answer 5

您可以使用 map 函数

var arr = [{vendor:"vendor1", item:"item1", price:1000, rank:0},
{vendor:"vendor1", item:"item2",price:3000, rank:0},
{vendor:"vendor1", item:"item3", price:1000, rank:0},

{vendor:"vendor2", item:"item1", price:2000, rank:0},
{vendor:"vendor2", item:"item2",price:2000, rank:0},
{vendor:"vendor2", item:"item3", price:3000, rank:0},

{vendor:"vendor3", item:"item1", price:3000, rank:0},
{vendor:"vendor3", item:"item2",price:1000, rank:0},
{vendor:"vendor3", item:"item3", price:2000, rank:0}];
arr.map(a => a.rank = (a.price/1000))
console.log(arr)

或使用 forEach 函数

var arr = [{vendor:"vendor1", item:"item1", price:1000, rank:0},
{vendor:"vendor1", item:"item2",price:3000, rank:0},
{vendor:"vendor1", item:"item3", price:1000, rank:0},

{vendor:"vendor2", item:"item1", price:2000, rank:0},
{vendor:"vendor2", item:"item2",price:2000, rank:0},
{vendor:"vendor2", item:"item3", price:3000, rank:0},

{vendor:"vendor3", item:"item1", price:3000, rank:0},
{vendor:"vendor3", item:"item2",price:1000, rank:0},
{vendor:"vendor3", item:"item3", price:2000, rank:0}];
arr.forEach(a => a.rank = (a.price/1000))
console.log(arr)

打字稿按分区/组排名

5 个答案: