149

目前，我有一个这样的数组：

var uniqueCount = Array();

几个步骤后，我的数组如下所示：

uniqueCount = [a,b,c,d,d,e,a,b,c,f,g,h,h,h,e,a];

如何计算数组中有多少个 a、b、c？我想要这样的结果：

a = 3
b = 1
c = 2
d = 2

等等

4

34 回答 34

409 

const counts = {};
const sampleArray = ['a', 'a', 'b', 'c'];
sampleArray.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; });
console.log(counts)

于 2013-10-16T04:33:24.687 回答
90

像这样的东西：

uniqueCount = ["a","b","c","d","d","e","a","b","c","f","g","h","h","h","e","a"];
var count = {};
uniqueCount.forEach(function(i) { count[i] = (count[i]||0) + 1;});
console.log(count);

如果您不希望它在旧浏览器中中断，请使用简单的 for 循环而不是 forEach。

于 2013-10-16T04:33:23.673 回答
61

我偶然发现了这个（非常古老的）问题。有趣的是，缺少最明显和最优雅的解决方案（恕我直言）：Array.prototype.reduce(...)。自 2011 年左右（IE）或更早（所有其他）以来，所有主要浏览器都支持此功能：

var arr = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = arr.reduce(function(prev, cur) {
  prev[cur] = (prev[cur] || 0) + 1;
  return prev;
}, {});

// map is an associative array mapping the elements to their frequency:
console.log(map);
// prints {"a": 3, "b": 2, "c": 2, "d": 2, "e": 2, "f": 1, "g": 1, "h": 3}

编辑：

通过在箭头函数中使用逗号运算符，我们可以用一行代码编写它：

var arr = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = arr.reduce((cnt, cur) => (cnt[cur] = cnt[cur] + 1 || 1, cnt), {});

// map is an associative array mapping the elements to their frequency:
console.log(map);
// prints {"a": 3, "b": 2, "c": 2, "d": 2, "e": 2, "f": 1, "g": 1, "h": 3}

但是，由于这可能更难阅读/理解，因此应该坚持使用第一个版本。

于 2015-10-01T11:38:06.317 回答
34 

function count() {
    array_elements = ["a", "b", "c", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];

    array_elements.sort();

    var current = null;
    var cnt = 0;
    for (var i = 0; i < array_elements.length; i++) {
        if (array_elements[i] != current) {
            if (cnt > 0) {
                document.write(current + ' comes --> ' + cnt + ' times<br>');
            }
            current = array_elements[i];
            cnt = 1;
        } else {
            cnt++;
        }
    }
    if (cnt > 0) {
        document.write(current + ' comes --> ' + cnt + ' times');
    }

}

count();

演示小提琴

您也可以使用高阶函数来执行操作。 看到这个答案

于 2013-10-16T04:35:30.613 回答
23

简单更好，一个变量，一个函数:)

const arr = ["a", "b", "c", "d", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];

const counts = arr.reduce((acc, value) => ({
   ...acc,
   [value]: (acc[value] || 0) + 1
}), {});

console.log(counts);

于 2019-12-13T05:08:48.067 回答
11

基于reduce数组函数的单行

const uniqueCount =  ["a", "b", "c", "d", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
const distribution = uniqueCount.reduce((acum,cur) => Object.assign(acum,{[cur]: (acum[cur] || 0)+1}),{});
console.log(JSON.stringify(distribution,null,2));

于 2017-11-09T11:46:49.637 回答
8 

// Initial array
let array = ['a', 'b', 'c', 'd', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a'];

// Unique array without duplicates ['a', 'b', ... , 'h']
let unique = [...new Set(array)];

// This array counts duplicates [['a', 3], ['b', 2], ... , ['h', 3]] 
let duplicates = unique.map(value => [value, array.filter(str => str === value).length]);
于 2019-12-31T06:43:10.043 回答
8

似乎没有人对此使用Map()内置的响应，这往往是我的首选结合Array.prototype.reduce()：

const data = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
const result = data.reduce((a, c) => a.set(c, (a.get(c) || 0) + 1), new Map());
console.log(...result);

注意，如果想在旧版浏览器中使用它，则必须进行polyfill 。Map()

于 2020-02-21T12:59:22.157 回答
4

您可以拥有一个包含计数的对象。遍历列表并增加每个元素的计数：

var counts = {};

uniqueCount.forEach(function(element) {
  counts[element] = (counts[element] || 0) + 1;
});

for (var element in counts) {
  console.log(element + ' = ' + counts[element]);
}
于 2013-10-16T04:34:16.227 回答
4

我认为这是如何计算数组中具有相同值的出现的最简单方法。

var a = [true, false, false, false];
a.filter(function(value){
    return value === false;
}).length
于 2014-10-16T08:41:35.497 回答
4

您可以在不使用任何 for/while 循环或 forEach 的情况下解决它。

function myCounter(inputWords) {        
    return inputWords.reduce( (countWords, word) => {
        countWords[word] = ++countWords[word] || 1;
        return countWords;
    }, {});
}

希望对你有帮助！

于 2016-09-08T17:34:05.960 回答
4 

// new example.
var str= [20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5];

function findOdd(para) {
  var count = {};
  para.forEach(function(para) {
  count[para] = (count[para] || 0) + 1;
  });
  return count;
}

console.log(findOdd(str));

于 2019-03-21T08:23:05.763 回答
4 
const obj = {};
const uniqueCount = [ 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a' ];
for (let i of uniqueCount) obj[i] ? obj[i]++ : (obj[i] = 1);
console.log(obj);
于 2021-04-28T14:44:42.560 回答
3

你可以这样做：

uniqueCount = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = new Object();

for(var i = 0; i < uniqueCount.length; i++) {
 if(map[uniqueCount[i]] != null) {
    map[uniqueCount[i]] += 1;
} else {
    map[uniqueCount[i]] = 1;
    }
}

现在你有一张包含所有字符数的地图

于 2013-10-16T04:43:34.740 回答
3

在javascript中使用数组reduce方法很简单：

const arr = ['a','d','r','a','a','f','d'];
const result =  arr.reduce((json,val)=>({...json, [val]:(json[val] | 0) + 1}),{});
console.log(result)
//{ a:3,d:2,r:1,f:1 }

于 2019-11-15T12:59:17.280 回答
2

代码：

function getUniqueDataCount(objArr, propName) {
    var data = [];
    if (Array.isArray(propName)) {
        propName.forEach(prop => {
            objArr.forEach(function(d, index) {
                if (d[prop]) {
                    data.push(d[prop]);
                }
            });
        });
    } else {
        objArr.forEach(function(d, index) {
            if (d[propName]) {
                data.push(d[propName]);
            }
        });
    }

    var uniqueList = [...new Set(data)];

    var dataSet = {};
    for (var i = 0; i < uniqueList.length; i++) {
        dataSet[uniqueList[i]] = data.filter(x => x == uniqueList[i]).length;
    }

    return dataSet;
}

片段

var data= [
          {day:'Friday'   , name: 'John'      },
          {day:'Friday'   , name: 'John'      },
          {day:'Friday'   , name: 'Marium'    },
          {day:'Wednesday', name: 'Stephanie' },
          {day:'Monday'   , name: 'Chris'     },
          {day:'Monday'   , name: 'Marium'    },
          ];
          
console.log(getUniqueDataCount(data, ['day','name']));       
   
function getUniqueDataCount(objArr, propName) {
    var data = [];
    if (Array.isArray(propName)) {
        propName.forEach(prop => {
            objArr.forEach(function(d, index) {
                if (d[prop]) {
                    data.push(d[prop]);
                }
            });
        });
    } else {
        objArr.forEach(function(d, index) {
            if (d[propName]) {
                data.push(d[propName]);
            }
        });
    }

    var uniqueList = [...new Set(data)];

    var dataSet = {};
    for (var i = 0; i < uniqueList.length; i++) {
        dataSet[uniqueList[i]] = data.filter(x => x == uniqueList[i]).length;
    }

    return dataSet;
}

于 2020-08-10T07:46:43.923 回答
2 
uniqueCount = ["a","b","a","c","b","a","d","b","c","f","g","h","h","h","e","a"];
var count = {};
uniqueCount.forEach((i) => { count[i] = ++count[i]|| 1});
console.log(count);
于 2020-04-27T15:07:48.463 回答
2

包含字母的数组中的重复项：

var arr = ["a", "b", "a", "z", "e", "a", "b", "f", "d", "f"],
  sortedArr = [],
  count = 1;

sortedArr = arr.sort();

for (var i = 0; i < sortedArr.length; i = i + count) {
  count = 1;
  for (var j = i + 1; j < sortedArr.length; j++) {
    if (sortedArr[i] === sortedArr[j])
      count++;
  }
  document.write(sortedArr[i] + " = " + count + "<br>");
}

包含数字的数组中的重复项：

var arr = [2, 1, 3, 2, 8, 9, 1, 3, 1, 1, 1, 2, 24, 25, 67, 10, 54, 2, 1, 9, 8, 1],
  sortedArr = [],
  count = 1;
sortedArr = arr.sort(function(a, b) {
  return a - b
});
for (var i = 0; i < sortedArr.length; i = i + count) {
  count = 1;
  for (var j = i + 1; j < sortedArr.length; j++) {
    if (sortedArr[i] === sortedArr[j])
      count++;
  }
  document.write(sortedArr[i] + " = " + count + "<br>");
}

于 2017-07-21T13:04:15.817 回答
1

简化的 sheet.js 答案

var counts = {};
var aarr=['a','b','a'];
aarr.forEach(x=>counts[x]=(counts[x] || 0)+1 );
console.log(counts)

于 2020-05-07T20:21:43.470 回答
1 
var uniqueCount = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
// here we will collect only unique items from the array
var uniqueChars = [];

// iterate through each item of uniqueCount
for (i of uniqueCount) {
// if this is an item that was not earlier in uniqueCount, 
// put it into the uniqueChars array
  if (uniqueChars.indexOf(i) == -1) {
    uniqueChars.push(i);
  } 
}
// after iterating through all uniqueCount take each item in uniqueChars
// and compare it with each item in uniqueCount. If this uniqueChars item 
// corresponds to an item in uniqueCount, increase letterAccumulator by one.
for (x of uniqueChars) {
  let letterAccumulator = 0;
  for (i of uniqueCount) {
    if (i == x) {letterAccumulator++;}
  }
  console.log(`${x} = ${letterAccumulator}`);
}
于 2017-03-24T19:30:07.287 回答
1

var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h ','h','h','e','a'];

var newArr = [];
testArray.forEach((item) => {
    newArr[item] = testArray.filter((el) => {
            return el === item;
    }).length;
})
console.log(newArr);
于 2019-10-18T14:14:19.900 回答
0

好答案的组合： 

var count = {};
var arr = ['a', 'b', 'c', 'd', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a'];
var iterator = function (element) {
    count[element] = (count[element] || 0) + 1;
}

if (arr.forEach) {
    arr.forEach(function (element) {
        iterator(element);
    });
} else {
    for (var i = 0; i < arr.length; i++) {
        iterator(arr[i]);
    }
}

希望它有帮助。

于 2013-10-16T08:09:30.913 回答
0

例如创建一个文件demo.js并在控制台中使用节点运行它，demo.js您将以矩阵的形式出现元素。

var multipleDuplicateArr = Array(10).fill(0).map(()=>{return Math.floor(Math.random() * Math.floor(9))});
console.log(multipleDuplicateArr);

var resultArr = Array(Array('KEYS','OCCURRENCE'));

for (var i = 0; i < multipleDuplicateArr.length; i++) {
  var flag = true;
  for (var j = 0; j < resultArr.length; j++) {
     if(resultArr[j][0] == multipleDuplicateArr[i]){
       resultArr[j][1] = resultArr[j][1] + 1;
       flag = false;
      }
  }
  if(flag){
    resultArr.push(Array(multipleDuplicateArr[i],1));
  }
}

console.log(resultArr);

您将在控制台中获得如下结果：

[ 1, 4, 5, 2, 6, 8, 7, 5, 0, 5 ] . // multipleDuplicateArr
[ [ 'KEYS', 'OCCURENCE' ],        // resultArr
  [ 1, 1 ],
  [ 4, 1 ],
  [ 5, 3 ],
  [ 2, 1 ],
  [ 6, 1 ],
  [ 8, 1 ],
  [ 7, 1 ],
  [ 0, 1 ] ]
于 2018-09-01T14:29:21.193 回答
0

最快的方法：

计算复杂度为 O(n)。

function howMuchIsRepeated_es5(arr) {
	const count = {};
	for (let i = 0; i < arr.length; i++) {
		const val = arr[i];
		if (val in count) {
			count[val] = count[val] + 1;
		} else {
			count[val] = 1;
		}
	}

	for (let key in count) {
		console.log("Value " + key + " is repeated " + count[key] + " times");
	}
}

howMuchIsRepeated_es5(['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']);

最短的代码：

使用 ES6。

function howMuchIsRepeated_es6(arr) {
	// count is [ [valX, count], [valY, count], [valZ, count]... ];
	const count = [...new Set(arr)].map(val => [val, arr.join("").split(val).length - 1]);

	for (let i = 0; i < count.length; i++) {
		console.log(`Value ${count[i][0]} is repeated ${count[i][1]} times`);
	}
}

howMuchIsRepeated_es6(['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']);

于 2019-07-01T17:57:43.900 回答
0

声明一个对象arr来保存唯一集作为键。使用 map 循环遍历数组来填充arr。如果之前没有找到该键，则添加该键并分配零值。在每次迭代中增加键的值。

给定测试数组：

var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];

解决方案：

var arr = {};
testArray.map(x=>{ if(typeof(arr[x])=="undefined") arr[x]=0; arr[x]++;});

JSON.stringify(arr)将输出

{"a":3,"b":2,"c":2,"d":2,"e":2,"f":1,"g":1,"h":3}

Object.keys(arr)将返回["a","b","c","d","e","f","g","h"]

要查找任何项目的出现，例如 barr['b']将输出2

于 2019-10-18T14:22:49.253 回答
0

通过使用 array.map 我们可以减少循环，请参阅jsfiddle

function Check(){
    var arr = Array.prototype.slice.call(arguments);
    var result = [];
    for(i=0; i< arr.length; i++){
        var duplicate = 0;
        var val = arr[i];
        arr.map(function(x){
            if(val === x) duplicate++;
        })
        result.push(duplicate>= 2);
    }
    return result;
}

去测试：

var test = new Check(1,2,1,4,1);
console.log(test);
于 2017-08-12T22:53:36.223 回答
0 

var string = ['a','a','b','c','c','c','c','c','a','a','a'];

function stringCompress(string){

var obj = {},str = "";
string.forEach(function(i) { 
  obj[i] = (obj[i]||0) + 1;
});

for(var key in obj){
  str += (key+obj[key]);
}
  console.log(obj);
  console.log(str);
}stringCompress(string)

/*
Always open to improvement ,please share 
*/

于 2018-07-15T03:17:45.307 回答
0

示例如何返回字符串中最常用的字符。

function maxChar(str) {
    const charMap = {};

    let maxCharacter = '';
    let maxNumber = 0;

    for (let item of str) {
        charMap[item] = charMap[item] + 1 || 1;
    }

    for (let char in charMap) {
        if (charMap[char] > maxNumber) {
            maxNumber = charMap[char];
            maxCharacter = char;
        }
    }

    return maxCharacter;
}


console.log(maxChar('abcccccccd'))

于 2021-03-07T06:01:02.623 回答
0 
let arr=[1,2,3,3,4,5,5,6,7,7]
let obj={}
for(var i=0;i<arr.length;i++){
    obj[arr[i]]=obj[arr[i]]!=null ?obj[arr[i]]+1:1 //stores duplicate in an obj

}
console.log(obj)
//returns object {1:1,:1,3:2,.....}
于 2021-01-13T17:14:52.360 回答
0 
var arr = ['a','d','r','a','a','f','d'];  

//call function and pass your array, function will return an object with array values as keys and their count as the key values.
duplicatesArr(arr);

function duplicatesArr(arr){
    var obj = {}
    for(var i = 0; i < arr.length; i++){
        obj[arr[i]] = [];
        for(var x = 0; x < arr.length; x++){
            (arr[i] == arr[x]) ? obj[arr[i]].push(x) : '';
        }
        obj[arr[i]] = obj[arr[i]].length;
    }

    console.log(obj);
    return obj;
}
于 2019-08-21T10:01:52.413 回答
0 

Str= ['a','b','c','d','d','e','a','h','e','a'];
var obj= new Object();

for(var i = 0; i < Str.length; i++) {
 if(obj[Str[i]] != null) {
    obj[Str[i]] += 1;
} else {
    obj[Str[i]] = 1;
    }
}
console.log(obj);

现在您可以获得重复项目的地图

于 2021-10-13T07:05:45.890 回答
0

计算字符串中提供的字母

function countTheElements(){
    
    var str = "ssdefrcgfrdesxfdrgs";
    var arr = [];
    var count = 0;
    
    for(var i=0;i<str.length;i++){
        arr.push(str[i]);
        
    }
        arr.sort();
    for(var i=0;i<arr.length;i++){     
        if(arr[i] == arr[i-1]){
            count++;
        }else{        
            count = 1;        
        }
            if(arr[i] != arr[i+1]){
                console.log(arr[i] +": "+(count));
            }    
            
       }
    }
        countTheElements()
于 2021-02-08T12:47:38.527 回答
0

您可以使用 array.reduce() 方法简单地做到这一点

const votes = ['Yes', 'Yes', 'Yes', 'No', 'No', 'Absent'];

const result = votes.reduce((prevValue, vote) => {

    if (prevValue[vote]) {

        prevValue[vote]++;

    } else {

        prevValue[vote] = 1;

    }

    return prevValue;

}, {});

console.log(result);

输出：{是：3，否：2，缺席：1}

于 2022-03-01T06:57:33.803 回答
-1 
public class CalculateCount {
public static void main(String[] args) {
    int a[] = {1,2,1,1,5,4,3,2,2,1,4,4,5,3,4,5,4};
    Arrays.sort(a);
    int count=1;
    int i;
    for(i=0;i<a.length-1;i++){
        if(a[i]!=a[i+1]){
            System.out.println("The Number "+a[i]+" appears "+count+" times");
            count=1;                
        }
        else{
            count++;
        }
    }
    System.out.println("The Number "+a[i]+" appears "+count+" times");

}

}

于 2017-07-19T14:07:08.213 回答