Javascript 如何计算数组中每个项目的出现次数？

How to count the number of occurrences of each item in an array?

提问by Aadi

I have an array as follows,

我有一个数组如下，

var arr = ['ab','pq','mn','ab','mn','ab']

Expected result

预期结果

arr['ab'] = 3
arr['pq'] = 1
arr['mn'] = 2

Tried as follows,

尝试如下，

$.each(arr, function (index, value) {
    if (value) 
        arr[value] = (resultSummary[value]) ? arr[value] + 1 : 1;
});

console.log(arr.join(','));

回答by Fabrizio Calderan

no need to use jQuery for this task — this example will build an object with the amount of occurencies of every different element in the array in O(n)

不需要使用 jQuery 来完成这个任务——这个例子将构建一个对象，其中包含数组中每个不同元素的出现次数 O(n)

var occurrences = { };
for (var i = 0, j = arr.length; i < j; i++) {
   occurrences[arr[i]] = (occurrences[arr[i]] || 0) + 1;
}

console.log(occurrences);        // {ab: 3, pq: 1, mn: 2}
console.log(occurrences['mn']);  // 2

Example fiddle

示例小提琴

---

---

You could also use Array.reduceto obtain the same result and avoid a for-loop

您也可以使用Array.reduce来获得相同的结果并避免for-loop

var occurrences = arr.reduce(function(obj, item) {
  obj[item] = (obj[item] || 0) + 1;
  return obj;
}, {});

console.log(occurrences);        // {ab: 3, pq: 1, mn: 2}
console.log(occurrences['mn']);  // 2

Example fiddle

示例小提琴

回答by Dmytro Kozlovskyi

I think this is the simplest way how to count occurrences with same value in array.

我认为这是如何计算数组中具有相同值的出现次数的最简单方法。

var a = [true, false, false, false];
a.filter(function(value){
    return value === false;
}).length                                      

回答by code_monk

if you don't want a solution that requires a library, and don't have to support legacy javascript:

如果您不想要需要库的解决方案，并且不必支持旧版 javascript：

var report = {};

arr.forEach(function(el){
  report[el] = report[el] + 1 || 1;
});

Or if you want to do it using jQuery:

或者，如果您想使用 jQuery 执行此操作：

var report = {};

$.each(arr,function(i,el){
  report[el] = report[el] + 1 || 1;
});

This uses short-circuit logicto test for conditions and set values. I think it's a pretty concise and readable way to do javascript.

这使用短路逻辑来测试条件和设定值。我认为这是一种非常简洁易读的 javascript 方法。

console.log( report );

回答by rsp

If you are using the Sugarlibrary it's just:

如果您使用的是Sugar库，它只是：

arr.count('ab');

DEMO

演示

See: Array count API.

请参阅：数组计数 API。

回答by meadlai

var result = {};
function count(input){
  var tmp = 0;
  if(result.hasOwnProperty(input)){
     tmp = result[input];
     result[input]=tmp+1;
  }else{
    result[input]=1;
  }
}

above function will help you to count the num of the same string in an Array.

上面的函数将帮助您计算数组中相同字符串的数量。

回答by Hymanwin tung

Try $.grep

试试$.grep

Example:

例子：

var arr = ['ab','pq','mn','ab','mn','ab']

jQuery.grep(arr, function(a){
   return a == 'ab'
}).length // 3

if supporting ie8 is not necessary.

如果不需要支持 ie8。

['ab','pq','mn','ab','mn','ab'].filter(function(value){
    return value == 'ab'
}).length // 3

回答by codebox

Could this be what you are trying to do?

这可能是你想要做的吗？

$.each(arr, function(index, value) {
    if (!resultSummary[value]){
        resultSummary[value] = 0;
    }
    resultSummary[value] += 1;
});

this code will count the occurrences of each string in the array and store the results in resultsArray

此代码将计算数组中每个字符串的出现次数并将结果存储在 resultsArray

回答by spaceman12

var arr = ['ab','pq','mn','ab','mn','ab']



function getCount(arr,val)
{
    var ob={};
    var len=arr.length;
    for(var k=0;k<len;k++)
    {
        if(ob.hasOwnProperty(arr[k]))
        {
            ob[arr[k]]++;
            continue;
        }
        ob[arr[k]]=1;
    }
    return ob[val];
}



//run test
alert(getCount(arr,'ab'));//3

var arr = ['ab','pq','mn','ab','mn','ab']



function getCount(arr,val)
{
    var ob={};
    var len=arr.length;
    for(var k=0;k<len;k++)
    {
        if(ob.hasOwnProperty(arr[k]))
        {
            ob[arr[k]]++;
            continue;
        }
        ob[arr[k]]=1;
    }
    return ob[val];
}



//run test
alert(getCount(arr,'ab'));//3

Demo :http://jsfiddle.net/dmx4y/2/

演示：http: //jsfiddle.net/dmx4y/2/

回答by Prashant_M

var arr = ['ab','pq','mn','ab','mn','ab'];
var result = { };
   for(i=0;i<arr.length;++i) 
   {
       if(!result[arr[i]])
           result[arr[i]]=0;
       ++result[arr[i]];
   }

for (var i in result){
console.log(i+":"+result[i]);
}