合并 2 个数组并创建一个新数组 [关闭]

Question

我有 2 个数组这个

let array1 = [
    {
        designation: "SSE",
        emailId: "abc@gmail.com",
        employeeId: 1997,
        firstName: "user2",
    },
    {
        designation: "DEVELOPER",
        emailId: "ab@gmail.com",
        employeeId: 19,
        firstName: "user1",
    },
    {
        designation: "DEVELOPER",
        emailId: "ab@gmail.com",
        employeeId: 191,
        firstName: "user1",
    },
];

let array2 = [
    {
        designation: "SSE",
        emailId: "abc@gmail.com",
        employeeId: 199,
        firstName: "user2",
    },
    {
        designation: "DEVELOPER",
        emailId: "ab@gmail.com",
        employeeId: 19,
        firstName: "user1",
    },
    {
        designation: "DEVELOPER",
        emailId: "ab@gmail.com",
        employeeId: 191,
        firstName: "user1",
    },
    {
        designation: "TESTER",
        emailId: "ab@gmail.com",
        employeeId: 1221,
        firstName: "user1",
    },
];

看，我在两个数组中都有一些共同的记录，而 array2 中缺少一个。如果employeeId 不匹配，我想将它们合并为一个。像这样。

let array3 = [
   {employeeId: 1997, isActive: false}, 
   {employeeId: 199, isActive: true}, 
   {employeeId: 19, isActive: true}, 
   {employeeId: 191, isActive: true},
   {employeeId: 1221, isActive: false}
]

array3[3].isActive 为 false，因为它在两个数组中都不匹配。

Answer 1

1. 从array1获取一个employeeId数组
2. 从array2 获取一组employeeId
3. 创建一个包含所有 (unique) emplyeeId 的数组
4. Create an array with the ID's that are not matching
5. map() 将所有 id (array_0_ids) 转换为：
   1. 创建一个对象
   2. 将isActive 设置为!non_matching.includes(emplyeeId) 的结果

let array1 = [{designation: "SSE", emailId: "abc@gmail.com", employeeId: 199, firstName: "user2", }, {designation: "DEVELOPER", emailId: "ab@gmail.com", employeeId: 19, firstName: "user1", }, {designation: "DEVELOPER", emailId: "ab@gmail.com", employeeId: 191, firstName: "user1", }, ];
let array2 = [{designation: "SSE", emailId: "abc@gmail.com", employeeId: 199, firstName: "user2", }, {designation: "DEVELOPER", emailId: "ab@gmail.com", employeeId: 19, firstName: "user1", }, {designation: "DEVELOPER", emailId: "ab@gmail.com", employeeId: 191, firstName: "user1", }, {designation: "TESTER", emailId: "ab@gmail.com", employeeId: 1221, firstName: "user1", }, ];

let array_1_ids = array1.map(i => i.employeeId);
let array_2_ids = array2.map(i => i.employeeId);
let array_0_ids = [ ...array_1_ids, ...array_2_ids ].filter((v, i, s) => s.indexOf(v) === i);

let non_matching = array_2_ids.filter(id => !array_1_ids.includes(id));

let result = array_0_ids.map(emplyeeId => {
  return {
    emplyeeId,
    isActive: !non_matching.includes(emplyeeId)
  };
});

console.log(result);

[
  {"emplyeeId": 199, "isActive": true },
  {"emplyeeId": 19, "isActive": true}, 
  {"emplyeeId": 191, "isActive": true},
  {"emplyeeId": 1221, "isActive": false} 
]

Answer 2

let result = [];

const employeeIdArray2 = array2.map(e => e.employeeId);
const employeeIdArray1 = array1.map(e => e.employeeId);

const missingEmployeeIdArray = employeeIdArray2.length > employeeIdArray1.length ? employeeIdArray1 : employeeIdArray2;

result = array2.map(e => {
  if (missingEmployeeIdArray.includes(e.employeeId)) {
    return {
      ...e,
      isActive: true
    }
  } else {
    return {
      ...e,
      isActive: false
    }
  }
})

Answer 3

您可以使用来自 lodash 的 uniqBy method。

const mergedArray = uniqBy([...array1, ...array2], "employeeId");

查看stackblitz 上的实时示例。

Answer 4

mergeArr = ():any[] => {
return this.array1.map(
  e => {
    if(this.array2.map(x=>x.employeeId).includes(e.employeeId)){
      return {employeeId:e.employeeId, isActive:true};
    }else{
      return {employeeId:e.employeeId, isActive:false};
    }
  }
)

}

这可能行得通..

更新了一个..

mergeArr = ():any[] => {
let addedEmployees = new Set<number>();
let arr3 =  this.array1.map(
  e => {
    addedEmployees.add(e.employeeId);
    if(this.array2.map(x=>x.employeeId).includes(e.employeeId)){
      return {employeeId:e.employeeId, isActive:true};
    }else{
      return {employeeId:e.employeeId, isActive:false};
    }
  }
)

this.array2.forEach(e => {
  if(!addedEmployees.has(e.employeeId)){
    arr3.push({employeeId:e.employeeId, isActive:false});
  }
})

return arr3;

}

Answer 5

1. 创建一个对象，其中键是employeeId，值是出现次数
2. 通过迭代对象条目创建最终数组，当出现次数为 2 时将 isActive 设置为 true。

const array1 = [
  {
    designation: "SSE",
    emailId: "abc@gmail.com",
    employeeId: 199,
    firstName: "user2",
  },
  {
    designation: "DEVELOPER",
    emailId: "ab@gmail.com",
    employeeId: 19,
    firstName: "user1",
  },
  {
    designation: "DEVELOPER",
    emailId: "ab@gmail.com",
    employeeId: 191,
    firstName: "user1",
  },
];

const array2 = [
  {
    designation: "SSE",
    emailId: "abc@gmail.com",
    employeeId: 199,
    firstName: "user2",
  },
  {
    designation: "DEVELOPER",
    emailId: "ab@gmail.com",
    employeeId: 19,
    firstName: "user1",
  },
  {
    designation: "DEVELOPER",
    emailId: "ab@gmail.com",
    employeeId: 191,
    firstName: "user1",
  },
  {
    designation: "TESTER",
    emailId: "ab@gmail.com",
    employeeId: 1221,
    firstName: "user1",
  },
];

const reducer = (acc, e) => ({
  ...acc,
  [e.employeeId]: (acc[e.employeeId] + 1) || 1,
}); 

const arr3 = Object.entries([...array1, ...array2].reduce(reducer, {}))
  .map(([employeeId, count]) => ({ employeeId, isActive: count == 2 }));
  
console.log(arr3);

Answer 6

将您的问题解释为“如果两个数组中都存在 employeeId，则 true，否则 false”。

这可以通过构建地图来解决。您甚至可以在之后继续使用地图，而不是在最后转换回 Array

// Set up the initial Map using `array1`'s values, assume everything is different
const employeeStatus = new Map(
  array1.map(({ employeeId }) => [employeeId, false]),
);

// if `array2` has the same employeeId then set `true`, otherwise add a false entry
array2.forEach(({ employeeId }) => {
  // for something more complex, you could also do some other logic here
  const seen = employeeStatus.has(employeeId);
  employeeStatus.set(employeeId, seen);
});


// convert the Map to the output data structure
const result = [...employeeStatus].map(([employeeId, isActive]) => ({
  employeeId,
  isActive,
}));

这给了

[
  { employeeId: 1997, isActive: false },
  { employeeId: 19, isActive: true },
  { employeeId: 191, isActive: true },
  { employeeId: 199, isActive: false },
  { employeeId: 1221, isActive: false },
]