下面是ES2017+获取O(N)中所有数组项计数的方法:

const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const counts = {};

arr.forEach((el) => {
  counts[el] = counts[el] ? (counts[el] + 1) : 1;
});

你也可以选择对输出进行排序:

const countsSorted = Object.entries(counts).sort(([_, a], [__, b]) => a - b);

console.log(countsSorted)用于示例数组:

[
  [ '2', 3 ],
  [ '1', 1 ],
  [ '3', 1 ],
  [ '5', 1 ],
  [ '8', 1 ],
  [ '9', 1 ]
]

一行程序功能

const countBy = (a,f)=>a.reduce((p,v,i,x)=>p+!!f(v,i,x), 0)
countBy([1,2,3,4,5], v=>v%2===0) // 2

非常简单:

var count = 0;
for(var i = 0; i < array.length; ++i){
    if(array[i] == 2)
        count++;
}

下面是ES2017+获取O(N)中所有数组项计数的方法:

const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const counts = {};

arr.forEach((el) => {
  counts[el] = counts[el] ? (counts[el] + 1) : 1;
});

你也可以选择对输出进行排序:

const countsSorted = Object.entries(counts).sort(([_, a], [__, b]) => a - b);

console.log(countsSorted)用于示例数组:

[
  [ '2', 3 ],
  [ '1', 1 ],
  [ '3', 1 ],
  [ '5', 1 ],
  [ '8', 1 ],
  [ '9', 1 ]
]

最好把它包装成函数:

let countNumber = (array,specificNumber) => {
    return array.filter(n => n == specificNumber).length
}

countNumber([1,2,3,4,5],3) // returns 1

var arrayCount = [1,2,3,2,5,6,2,8]; Var co = 0; 函数findElement () { arrayCount。Find(函数(值，索引){ If (value == 2) 公司+ +; })； Console.log ('found' + ' ' + co + '元素值为2'); ｝

我会这样做:

var arraycount = [1，2，3，4，5，6，7，8]; 函数 countarr（）{ 变量 dd = 0; arrayCount.forEach（ function（s）{ DD++; }); 控制台.log（dd）; }