Question

我需要从另一个数组hexCode

中获取特定的对象

这是我正在过滤的数据

{
  dateCreated: "2019-10-30 08:29:54 PM" 
   hexCode: "#ed4c67" 
  id: 1 
  leaveTag: "Vacation Leave" 
  leaveTagColor: "Bara Red"
}
{
  dateCreated: "2019-10-30 08:30:05 PM" 
  hexCode: "#ea2027"
     id: 2 
  leaveTag: "Sick Leave"
     leaveTagColor: "Red Pigment"
}

这是我要插入hexCode的数据。然后映射leaveType以将其与上面的数据进行比较。可以吗？

{
  leaveType: "Emergency Leave"
  email: "sfe.bruce.logan@gmail.com"
  hexCode: ???
}
{
  leaveType: "Sick Leave"
  email: "sfe.bruce.logan@gmail.com"
  hexCode: ???
}

Answer 1

您可以像下面这样简单地进行查找，但是在这种情况下，时间复杂度更高（尽管可读性更高，并且没有空间复杂度）

或者您可以从第一个数据数组创建类型为leaveType : hexCode的Map，然后使用Map查找相应的值并填充第二个数组。请检查@CodeManiac的答案是否适用。

const data = [{
  dateCreated: "2019-10-30 08:29:54 PM",
  hexCode: "#ed4c67",
  id: 1,
  leaveTag: "Vacation Leave",
  leaveTagColor: "Bara Red"
},
{
  dateCreated: "2019-10-30 08:30:05 PM",
  hexCode: "#ea2027",
  id: 2,
  leaveTag: "Sick Leave",
  leaveTagColor: "Red Pigment"
}]

const arr = [{
  leaveType: "Emergency Leave",
  email: "sfe.bruce.logan@gmail.com",
},
{
  leaveType: "Sick Leave",
  email: "sfe.bruce.logan@gmail.com"
}]

arr.forEach(leave => {
 const matchedLeave = data.find(lv => lv.leaveTag === leave.leaveType);
 if(matchedLeave) {
   leave.hexCode = matchedLeave.hexCode;
 }
});
 
console.log(arr);

Answer 2

您可以使用Map和map

将leaveTag映射到各自的hexCode
遍历数据，检查leaveType中是否存在Map（如果存在），请使用Map中的值，否则将其保留为原始值

const data = [{dateCreated: "2019-10-30 08:29:54 PM",hexCode: "#ed4c67",id: 1,leaveTag: "Vacation Leave",leaveTagColor: "Bara Red"}, {dateCreated: "2019-10-30 08:30:05 PM",hexCode: "#ea2027",id: 2,leaveTag: "Sick Leave",leaveTagColor: "Red Pigment",}]
const desiredFormat = [{leaveType: "Emergency Leave",email: "sfe.bruce.logan@gmail.com",hexCode: ''}, {leaveType: "Sick Leave",email: "sfe.bruce.logan@gmail.com",hexCode: ''}]

const mapper = new Map(data.map(({leaveTag,hexCode}) => [leaveTag, hexCode]))

const final = desiredFormat.map(({ hexCode, ...rest }) => {
  return {
    ...rest,
    hexCode: mapper.get(rest.leaveType) || hexCode
  }
})
console.log(final)

Answer 3

我只使用数组映射的方法

const obj1 = [{
  dateCreated: "2019-10-30 08:29:54 PM",
  hexCode: "#ed4c67",
  id: 1,
  leaveTag: "Vacation Leave",
  leaveTagColor: "Bara Red"
},
{
  dateCreated: "2019-10-30 08:30:05 PM",
  hexCode: "#ea2027",
  id: 2,
  leaveTag: "Sick Leave",
  leaveTagColor: "Red Pigment"
}]


const obj2 = [{
  leaveType: "Emergency Leave",
  email: "sfe.bruce.logan@gmail.com",
  hexCode: null
},
{
  leaveType: "Sick Leave",
  email: "sfe.bruce.logan@gmail.com",
  hexCode: null
}]
const result = obj2.map(item => {
  const findedHex = obj1.find(o2item => o2item.leaveTag == item.leaveType)
  item.hexCode = findedHex && findedHex.hexCode || null;
  return item;

})
console.log(result)

筛选对象并将其从数组分配给另一个对象

3 个答案: