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如何统计字符串在另一个字符串中出现的次数？

javascriptregexstring
839

我该如何在一个字符串中统计另一个特定字符串出现的次数？例如，我想在Javascript中实现以下操作：

var temp = "This is a string.";
alert(temp.count("is")); //should output '2'
23
这取决于你是否接受重叠实例，例如：var t = "sss"；上述字符串中子串"ss"出现了几次？是1次还是2次？你是跳过每个实例，还是逐个字符地移动指针来寻找子串呢？ - Tim
4
这个问题的改进基准：http://jsperf.com/string-ocurrence-split-vs-match/2 （基于Kazzkiq的基准测试）。 - idmean
在 JavaScript 中计算字符串中特定单词的总数 https://stackoverflow.com/a/65036248/4752258 - Farbod Aprin
这个视频似乎与此相关 - “Google编程面试与Facebook软件工程师” - https://www.youtube.com/watch?v=PIeiiceWe_w - Deryck
41个回答
0

var countInstances = function(body, target) {
  var globalcounter = 0;
  var concatstring  = '';
  for(var i=0,j=target.length;i<body.length;i++){
    concatstring = body.substring(i-1,j);
    
    if(concatstring === target){
       globalcounter += 1;
       concatstring = '';
    }
  }
  
  
  return globalcounter;
 
};

console.log(   countInstances('abcabc', 'abc')   ); // ==> 2
console.log(   countInstances('ababa', 'aba')   ); // ==> 2
console.log(   countInstances('aaabbb', 'ab')   ); // ==> 1

0

这个函数将告诉你子字符串是否在字符串中以及出现了多少次。

const wordInText = (wordToFind, wholeText) => {
    const wordToFindRegex = new RegExp(wordToFind, 'gi');
    const occurences = wholeText.match(wordToFindRegex) ?? [];
    return {isWordInText: occurences.length > 0, occurences: occurences.length};
 }
 
 console.log(wordInText("is", "This cow jumped over this moon"))

0
参数： ustring：超集字符串 countChar：子字符串
JavaScript 中计算子字符串出现次数的函数： 

function subStringCount(ustring, countChar){
  var correspCount = 0;
  var corresp = false;
  var amount = 0;
  var prevChar = null;
  
 for(var i=0; i!=ustring.length; i++){

     if(ustring.charAt(i) == countChar.charAt(0) && corresp == false){
       corresp = true;
       correspCount += 1;
       if(correspCount == countChar.length){
         amount+=1;
         corresp = false;
         correspCount = 0;
       }
       prevChar = 1;
     }
     else if(ustring.charAt(i) == countChar.charAt(prevChar) && corresp == true){
       correspCount += 1;
       if(correspCount == countChar.length){
         amount+=1;
         corresp = false;
         correspCount = 0;
         prevChar = null;
       }else{
         prevChar += 1 ;
       }
     }else{
       corresp = false;
       correspCount = 0;
     }
 } 
 return amount;
}

console.log(subStringCount('Hello World, Hello World', 'll'));

0

var str = 'stackoverflow';
var arr = Array.from(str);
console.log(arr);

for (let a = 0; a <= arr.length; a++) {
  var temp = arr[a];
  var c = 0;
  for (let b = 0; b <= arr.length; b++) {
    if (temp === arr[b]) {
      c++;
    }

  }
  console.log(`the ${arr[a]} is counted for ${c}`)
}

1
请不要仅仅发布代码作为答案，还要提供代码的解释以及它是如何解决问题的。带有解释的答案通常更有帮助和更高质量，并且更有可能吸引赞同。 - Ran Marciano
0

这个函数有三种模式：在字符串中查找单个字符的频率，查找字符串中的连续子字符串，如果匹配，则继续向右移动到其后面的下一个子字符串，第三种模式与前一种类似，但它还将计算给定字符串中相交的子字符串。

function substringFrequency(string , substring , conjunction){
    let index 
    let occurenceFrequency  = 0
    for (let i=0 ; i < string.length  ; i++){
        index = string.indexOf(substring , i)
        if (index != -1){
            if ((substring.length == 1 ) || conjunction == true) {
                i = index 
            }else {
                i = index + 1
            }
            occurenceFrequency++
        }else{
            break
        } 
    }
    return (occurenceFrequency)
}

console.log(substringFrequency('vvvv' , 'v' ))
console.log(substringFrequency('vvvv' , 'vv'))
console.log(substringFrequency('vvvv' , 'vv' , true))

@Calculuswhiz将其与for循环代码块更加集成化了。 - polendina
@Calculuswhiz，一开始我编辑是因为你的评论要求仅在for循环体内集成变量i，但在这个过程中我注意到交叉子字符串被省略了，只有单独的子字符串被计算。所以我默认计算单独的子字符串，因为这更有意义，但由于原始问题没有明确指定任何行为，我不想强加任何行为以适应不同的用例，所以我添加了更多语句。 - polendina
0
const getLetterMatchCount = (guessedWord, secretWord) => {
  const secretLetters = secretWord.split('');
  const guessedLetterSet = new Set(guessedWord);
  return secretLetters.filter(letter => guessedLetterSet.has(letter)).length;
};
const str = "rahul";
const str1 = "rajendra";

getLetterMatchCount(str, str1)
1
有这么多其他的答案，如果您能解释一下您的代码是如何工作的以及它相对于其他提出的解决方案有什么优势，那就太好了。 - Adrian Mole
0

substr_count从php翻译成Javascript

function substr_count (haystack, needle, offset, length) { 
  // eslint-disable-line camelcase
  //  discuss at: https://locutus.io/php/substr_count/
  // original by: Kevin van Zonneveld (https://kvz.io)
  // bugfixed by: Onno Marsman (https://twitter.com/onnomarsman)
  // improved by: Brett Zamir (https://brett-zamir.me)
  // improved by: Thomas
  //   example 1: substr_count('Kevin van Zonneveld', 'e')
  //   returns 1: 3
  //   example 2: substr_count('Kevin van Zonneveld', 'K', 1)
  //   returns 2: 0
  //   example 3: substr_count('Kevin van Zonneveld', 'Z', 0, 10)
  //   returns 3: false

  var cnt = 0

  haystack += ''
  needle += ''
  if (isNaN(offset)) {
    offset = 0
  }
  if (isNaN(length)) {
    length = 0
  }
  if (needle.length === 0) {
    return false
  }
  offset--

  while ((offset = haystack.indexOf(needle, offset + 1)) !== -1) {
    if (length > 0 && (offset + needle.length) > length) {
      return false
    }
    cnt++
  }

  return cnt
}

查看 Locutus 对 Php 的 substr_count 函数的翻译

0
//Try this code

const countSubStr = (str, search) => {
    let arrStr = str.split('');
    let i = 0, count = 0;

    while(i < arrStr.length){
        let subStr = i + search.length + 1 <= arrStr.length ?
                  arrStr.slice(i, i+search.length).join('') :
                  arrStr.slice(i).join('');
        if(subStr === search){
            count++;
            arrStr.splice(i, search.length);
        }else{
            i++;
        }
    }
    return count;
  }
-1

发现了这篇文章。

let str = 'As sly as a fox, as strong as an ox';

let target = 'as'; // let's look for it

let pos = 0;
while (true) {
  let foundPos = str.indexOf(target, pos);
  if (foundPos == -1) break;

  alert( `Found at ${foundPos}` );
  pos = foundPos + 1; // continue the search from the next position
}

同样的算法可以更简洁地呈现：

let str = "As sly as a fox, as strong as an ox";
let target = "as";

let pos = -1;
while ((pos = str.indexOf(target, pos + 1)) != -1) {
  alert( pos );
}
-1
回答Leandro Batista的问题： 只是正则表达式有一个问题。 

 "use strict";
 var dataFromDB = "testal";
 
  $('input[name="tbInput"]').on("change",function(){
 var charToTest = $(this).val();
 var howManyChars = charToTest.length;
 var nrMatches = 0;
 if(howManyChars !== 0){
  charToTest = charToTest.charAt(0);
  var regexp = new RegExp(charToTest,'gi');
  var arrMatches = dataFromDB.match(regexp);
  nrMatches = arrMatches ? arrMatches.length : 0;
 }
  $('#result').html(nrMatches.toString());

  });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="main">
What do you wanna count <input type="text" name="tbInput" value=""><br />
Number of occurences = <span id="result">0</span>
</div>

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