如何计算 JavaScript 中对象数组中重复名称的出现次数？

给定一个对象数组，任务是根据其值查找给定键的出现。

例子：

Input : arr = [
   {
      employeeName: "Ram",
      employeeId: 23
   },
   {
      employeeName: "Shyam",
      employeeId: 24
   },
   {
      employeeName: "Ram",
      employeeId: 21
   },
   {
      employeeName: "Ram",
      employeeId: 25
   },
   {
      employeeName: "Kisan",
      employeeId: 22
   },
   {
      employeeName: "Shyam",
      employeeId: 20
   }
]

key = "employeeName"

Output: [
    {employeeName: "Ram", occurrences: 3},
    {employeeName: "Shyam", occurrences: 2},
    {employeeName: "Kisan", occurrences:  1}
] 

为了实现这一点，我们有以下方法：

方法一：

在这种方法中，我们遵循以下步骤。

• 创建一个空的输出数组。
• 使用 forEach 迭代输入数组。
• 检查输出数组是否包含任何包含提供的键值的对象
• 如果不是，则创建一个新对象并初始化该对象，其中键（提供的键名）设置为值（当前迭代的对象的键值）并将发生设置为值 1
• 如果是，则迭代输出数组并搜索当前迭代的键是否等于输入数组迭代的键，然后将出现次数增加到 1。

Javascript

function findOcc(arr, key){
  let arr2 = [];
    
  arr.forEach((x)=>{
       
    // Checking if there is any object in arr2
    // which contains the key value
     if(arr2.some((val)=>{ return val[key] == x[key] })){
         
       // If yes! then increase the occurrence by 1
       arr2.forEach((k)=>{
         if(k[key] === x[key]){ 
           k["occurrence"]++
         }
      })
         
     }else{
       // If not! Then create a new object initialize 
       // it with the present iteration key's value and 
       // set the occurrence to 1
       let a = {}
       a[key] = x[key]
       a["occurrence"] = 1
       arr2.push(a);
     }
  })
    
  return arr2
}
  
  
let arr = [
   {
      employeeName: "Ram",
      employeeId: 23
   },
   {
      employeeName: "Shyam",
      employeeId: 24
   },
   {
      employeeName: "Ram",
      employeeId: 21
   },
   {
      employeeName: "Ram",
      employeeId: 25
   },
   {
      employeeName: "Kisan",
      employeeId: 22
   },
   {
      employeeName: "Shyam",
      employeeId: 20
   }
]
  
let key = "employeeName"
console.log(findOcc(arr, key))

输出：

[
   {
      employeeName: "Ram",
      occurrence: 3
   }, 
   {
      employeeName: "Shyam",
     occurrence: 2
   }, 
   {
    employeeName: "Kisan",
    occurrence: 1
   }
]