21

我想使用 JavaScript 计算给定字符串中每个字符的出现次数。

例如：

var str = "I want to count the number of occurances of each char in this string";

输出应该是：

h = 4;
e = 4; // and so on

我尝试搜索谷歌，但没有找到任何答案。我想实现这样的目标；顺序无关紧要。

4

19 回答 19

19

这在 JavaScript（或任何其他支持地图的语言）中非常非常简单：

// The string
var str = "I want to count the number of occurances of each char in this string";

// A map (in JavaScript, an object) for the character=>count mappings
var counts = {};

// Misc vars
var ch, index, len, count;

// Loop through the string...
for (index = 0, len = str.length; index < len; ++index) {
    // Get this character
    ch = str.charAt(index); // Not all engines support [] on strings

    // Get the count for it, if we have one; we'll get `undefined` if we
    // don't know this character yet
    count = counts[ch];

    // If we have one, store that count plus one; if not, store one
    // We can rely on `count` being falsey if we haven't seen it before,
    // because we never store falsey numbers in the `counts` object.
    counts[ch] = count ? count + 1 : 1;
}

现在counts拥有每个角色的属性；每个属性的值是计数。你可以像这样输出：

for (ch in counts) {
    console.log(ch + " count: " + counts[ch]);
}
于 2013-10-20T18:10:23.197 回答
19

更短的答案，减少：

let s = 'hello';
var result = [...s].reduce((a, e) => { a[e] = a[e] ? a[e] + 1 : 1; return a }, {}); 
console.log(result); // {h: 1, e: 1, l: 2, o: 1}
于 2019-04-05T17:06:19.910 回答
6 
let str = "atul kumar srivastava";
let obj ={};
for(let s of str)if(!obj[s])obj[s] = 1;else obj[s] = obj[s]  + 1;
console.log(obj)
于 2020-03-12T06:15:40.367 回答
4

我遍历每个字符并将其与计数一起放入嵌套对象中。如果字符已经存在于对象中，我只需增加计数。这是 myObj 的样子：

myObj = {
char1 = { count : <some num> },
char2 = { count : <some num> },
....
}

这是代码：

function countChar(str) {
    let myObj= {};
    for (let s of str) {
        if ( myObj[s] ? myObj[s].count ++ : myObj[s] = { count : 1 } );
    }
    return myObj;
}

var charCount = countChar('abcceddd');
于 2021-04-11T11:11:07.903 回答
3

您可以在 Javascript 中使用 ES6 中的地图。在我看来，提供了一个更简洁的代码。这是我将如何去做

function countChrOccurence ('hello') {
 let charMap = new Map();
 const count = 0;
  for (const key of str) {
   charMap.set(key,count); // initialize every character with 0. this would make charMap to be 'h'=> 0, 'e' => 0, 'l' => 0, 
  }

  for (const key of str) {
    let count = charMap.get(key);
    charMap.set(key, count + 1);
  }
// 'h' => 1, 'e' => 1, 'l' => 2, 'o' => 1

  for (const [key,value] of charMap) {
    console.log(key,value);
  }
// ['h',1],['e',1],['l',2],['o',1]
}
于 2020-07-07T18:08:56.860 回答
3

这对我来说效果很好：

function Char_Count(str1) {
    var chars = {};
    str1.replace(/\S/g, function(l){chars[l] = (isNaN(chars[l]) ? 1 : chars[l] + 1);});
    return chars;
}

var myString = "This is my String";
console.log(Char_Count(myString));
于 2017-09-19T10:21:59.890 回答
2 
str = "aaabbbccccdefg";

words = str.split("");

var obj = [];

var counter = 1, jump = 0;

for (let i = 0; i < words.length; i++) {
    if (words[i] === words[i + 1]) {
        counter++;
        jump++;
    }
    else {
        if (jump > 0) {
            obj[words[i]] = counter;
            jump = 0;
            counter=1
        }
        else
            obj[words[i]] = 1;
    }

}
console.log(obj);
于 2021-03-22T13:51:15.423 回答
2

我给你非常非常简单的代码。

 // Converts String To Array
        var SampleString= Array.from("saleem");

        // return Distinct count as a object
        var allcount = _.countBy(SampleString, function (num) {
            return num;
        });

        // Iterating over object and printing key and value
        _.map(allcount, function(cnt,key){
            console.log(key +":"+cnt);
        });

        // Printing Object
        console.log(allcount);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

    <p>Set the variable to different value and then try...</p>

于 2017-10-18T12:59:15.777 回答
2

1-liner ES6 方式：

const some_string = 'abbcccdddd';
const charCountIndex = [ ...some_string ].reduce( ( a, c ) => ! a[ c ] ? { ...a, [ c ]: 1 } : { ...a, [ c ]: a[ c ] + 1 }, {} );
console.log( charCountIndex )
于 2021-04-13T12:13:44.507 回答
2

您可以将对象用于任务。

第 1 步 - 创建一个对象

第 2 步 - 遍历字符串

第 3 步 - 在对象中添加字符作为键和字符计数作为值

var obj={}

function countWord(arr)
{
for(let i=0;i<arr.length;i++)
{
if(obj[arr[i]]) //check if character is present in the obj as key
{
    obj[arr[i]]=obj[arr[i]]+1; //if yes then update its value
}
else
{
    obj[arr[i]]=1; //initialise it with a value 1

}
}
}
于 2021-07-14T07:27:57.073 回答
1 
let newStr= "asafasdhfasjkhfweoiuriujasfaksldjhalsjkhfjlkqaofadsfasasdfas";
       
function checkStringOccurnace(newStr){
    let finalStr = {};
    let checkArr = [];
    let counterArr = [];
    for(let i = 0; i < newStr.length; i++){
        if(checkArr.indexOf(newStr[i]) == -1){
            checkArr.push(newStr[i])
            let counter = 0;
            counterArr.push(counter + 1)
            finalStr[newStr[i]] = 1;
        }else if(checkArr.indexOf(newStr[i]) > -1){
            let index = checkArr.indexOf(newStr[i])
            counterArr[index] = counterArr[index] + 1;
            finalStr[checkArr[index]] = counterArr[index];
        }
    }
    return finalStr;
}

let demo = checkStringOccurnace(newStr);
console.log(" finalStr >> ", demo);
于 2021-03-23T16:32:00.533 回答
1 
    function cauta() {

        var str = document.form.stringul.value;
        str = str.toLowerCase();
        var tablou = [];

        k = 0;
        //cautarea caracterelor unice
        for (var i = 0, n = 0; i < str.length; i++) {
            for (var j = 0; j < tablou.length; j++) {
                if (tablou[j] == str[i]) k = 1;
            }
            if (k != 1) {
                if (str[i] != ' ')
                    tablou[n] = str[i]; n++;
            }
            k = 0;
        }
        //numararea aparitilor
        count = 0;
        for (var i = 0; i < tablou.length; i++) {
            if(tablou[i]!=null){
            char = tablou[i];
            pos = str.indexOf(char);
            while (pos > -1) {
                ++count;
                pos = str.indexOf(char, ++pos);

            }

            document.getElementById("rezultat").innerHTML += tablou[i] + ":" + count + '\n';
            count = 0;
        }
        }

    }

此函数将每个唯一的字符放入数组中，然后在 str 中查找每个字符的外观。在我的案例中，我获取数据并将其放入

于 2017-02-06T19:21:41.497 回答
1 
    package com.company;

import java.util.HashMap;


 public class Main {

    public static void main(String[] args) {
    // write your code here
    HashMap<Character, Integer> sHashMap = new HashMap();  // using hashMap<key , value > here key = character and  value = count

    String arr = "HelloWorld";

    for (int i = 0; i < arr.length(); i++) {
        boolean flag = sHashMap.containsKey(arr.charAt(i));  // check if char is already  present 

    if (flag == true)
        {
            int Count = sHashMap.get(arr.charAt(i)); // get the char count
            sHashMap.put(arr.charAt(i), ++Count); //   increment the count and update in hashMap
        } 
        else 
        {
            sHashMap.put(arr.charAt(i), 1); //if  char not present then insert into hashMap
        }
    }

     System.out.println(sHashMap);
    //OutPut would be like ths {r=1, d=1, e=1, W=1, H=1, l=3, o=2}

}

}
于 2019-02-20T04:26:32.630 回答
1

我用检查“空白空间”和“特殊字符”进行了尝试：

function charCount(str){
    const requiredString = str.toLowerCase();

    const leng = str.length;

    let output = {};

    for(let i=0; i<leng; i++){
        const activeCharacter = requiredString[i];
        if(/[a-z0-9]/.test(activeCharacter)){
            output.hasOwnProperty(activeCharacter) ? output[activeCharacter]++ : output[activeCharacter] = 1;
        }
    }
    return output;
}
于 2021-06-11T17:10:05.227 回答
1

我使用了 Map 对象，该 map 对象不允许您设置任何重复键，这使我们的工作变得容易。我正在检查 key 是否已经存在于 map 中，如果没有，我正在插入并将计数设置为 1，如果它已经存在，我正在获取值然后递增 

const str = "Hello H"
    const strTrim = str.replace(/\s/g,'') // HelloH
    const strArr=strTrim.split('')

    let myMap = new Map(); // Map object 

    strArr.map(ele=>{
    let count =0
    if(!myMap.get(ele)){
    myMap.set(ele,++count)
    }else {
    let cnt=myMap.get(ele)
    myMap.set(ele,++cnt)
    }
    console.log("map",myMap)
    })
于 2020-03-04T11:53:02.883 回答
1

希望这可以帮助某人

function getNoOfOccurences(str){
    var temp = {};
    for(var oindex=0;oindex<str.length;oindex++){
        if(typeof temp[str.charAt(oindex)] == 'undefined'){
            temp[str.charAt(oindex)] = 1;
        }else{
            temp[str.charAt(oindex)] = temp[str.charAt(oindex)]+1;
        }
    }
    return temp;
}
于 2018-07-27T12:47:23.767 回答
1 

 // Converts String To Array
        var SampleString= Array.from("saleem");

        // return Distinct count as a object
        var allcount = _.countBy(SampleString, function (num) {
            return num;
        });

        // Iterating over object and printing key and value
        _.map(allcount, function(cnt,key){
            console.log(key +":"+cnt);
        });

        // Printing Object
        console.log(allcount);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

    <p>Set the variable to different value and then try...</p>

于 2017-10-18T13:42:14.053 回答
0

为了检查和避免空格并将大写字母转换为小写和计数，我们可以像下面那样做。

function CountNumberOfCharacters(str) {
  let countObject = {};
  const lowerCaseString = str.toLowerCase();
  for (let char of lowerCaseString) {
    if (!countObject[char] && char !== ' ') {
      countObject[char] = 1;
    } else if (char !== ' ') {
      countObject[char] = countObject[char] + 1;
    }
  }
  return countObject;
}
于 2021-10-09T17:15:04.957 回答
0

使用扩展...运算符而不是拆分字符串.split('')：

''.split('')
//=> ["\ud83c", "\udf2f", "\ud83c", "\udf2f", "\ud83c", "\udf63", "\ud83c", "\udf7b"]

对比

[...'']
//=> ["", "", "", ""]

对比

''.charAt(0)
//=> "\ud83c"

然后减少：

[...''].reduce((m, c) => (m[c] = (m[c] || 0) + 1, m), {})
//=> {'': 2, '': 1, '': 1}
于 2021-07-16T00:21:13.993 回答