[js] 第7天统计某一字符或字符串在另一个字符串中出现的次数 #21

haizhilin2013 · 2019-04-22T20:57:01Z

第7天统计某一字符或字符串在另一个字符串中出现的次数

linghucq1 · 2019-05-08T08:50:16Z

function strCount(str, target) {
  let count = 0
  if (!target) return count
  while(str.match(target)) {
    str = str.replace(target, '')
    count++
  }
  return count
}

console.log(strCount('abcdef abcdef a', 'abc'))

coderfe · 2019-05-09T05:56:52Z

function substrCount(str, target) {
  let count = 0;
  while (str.includes(target)) {
    const index = str.indexOf(target);
    count++;
    str = str.substring(index + target.length);
  }
  return count;
}

undefinedYu · 2019-05-10T02:11:42Z

这里找到一个好方法。希望能有用。
function substrCount(str, target) {
if(Object.prototype.toString.call(str).slice(8,-1) === 'String' && !str)
alert("请填写字符串")；
else
return (str.match(new RegExp(target, 'g')).length);
}

qqdnnd · 2019-05-16T09:49:14Z

function getStrTimes(str,rule){
    return rule !=='' && str.match(new RegExp(rule,'g'))? str.match(new RegExp(rule,'g')).length : 0;
}

Shoryukenbt · 2019-05-28T08:37:28Z

var childInNums = parent.split(child).length - 1, 这个没毛病吧

tiyunchen · 2019-05-31T09:20:20Z

var childInNums = parent.split(child).length - 1, 这个没毛病吧

感觉没毛病

myprelude · 2019-06-13T08:23:56Z

fucntion repeat(str,parentStr){
  return parentStr.split(str).length - 1
}

AricZhu · 2019-06-19T00:40:30Z

// 递归，一行代码实现
function strRepeatCount (subStr, str, count=0) {

return str.indexOf(subStr) !== -1? strRepeatCount(subStr, str.substring(str.indexOf(subStr)+subStr.length), ++count): count;

}

bWhirring · 2019-07-03T02:23:46Z

function count(str, param) {
  const reg = new RegExp(param, 'g');
  return str.match(reg).length;
}

Konata9 · 2019-07-04T08:43:56Z

const countAppears = (str, target) => {
  let count = 0;

  if (!str || !target) {
    return count;
  }

  const keyIndex = target.indexOf(str);
  if (keyIndex > -1) {
    count = 1 + countAppears(str, target.slice(keyIndex + 1));
  }

  return count;
};

const str = "abcaaadefg2333333333334abcddddea";

console.log(countAppears("2", str));
console.log(countAppears("b", str));
console.log(countAppears("d", str));
console.log(countAppears("a", str));
console.log(countAppears("f", str));

poporeki · 2019-07-04T15:18:30Z

var childInNums = parent.split(child).length - 1, 这个没毛病吧

这个好啊又简单

15190408121 · 2019-07-05T03:51:57Z

function strFind (str, target) {
var lengths = 0
if (!target) {
return lengths
}
while(str.match(target)) {
str = str.replace(target, '')
lengths++
}
return lengths
}
var str = "你好萨达所大所多所多所问问二位无你好萨达所大所多你好"
strFind(str, "你好")

taiyangxingchen · 2019-07-10T07:53:06Z

var childInNums = parent.split(child).length - 1, 这个没毛病吧

这个真心不错

shufangyi · 2019-07-18T06:23:40Z

const countString = (str, chars) => (
  (arr = str.match(new RegExp(chars, 'g'))), arr && chars ? arr.length : 0
)

pythonqi · 2019-07-24T06:18:05Z

fucntion repeat(str,parentStr){
  return parentStr.split(str).length - 1
}

function 关键字写错了

wyx2014 · 2019-07-24T09:28:00Z

function getCount1(str, target,count = 0 ) { var index = str.indexOf(target); if(index!== -1){ return getCount1(str.substring(index+target.length),target,++count) }else { return count; } }

Vi-jay · 2019-07-25T08:43:06Z

function getTargetCount(arr, target) {
  return arr
    .reduce((group, v) => group.set(v, (group.get(v) || 0) + 1), new Map())
    .get(target);
}
console.log(getTargetCount([1,1,2,5],1));// 2

wsb260 · 2019-07-31T07:14:11Z

// 方法1

var str = 'aabbcaacddaabbccdd'
var str2 = 'aa'

const contSvg = (s1,s2) => {
  var arr = [];
  arr = s1.split(s2);
  console.log(arr.length-1);
}

contSvg(str,str2) // 输出3

// 方法2

// eval() 函数可计算某个字符串，并执行其中的的 JavaScript 代码。
const strCount = (str, target) => {
  if (!target) return count
	 return  str.match(eval("/"+target+"/g")).length
}

console.log(strCount('abcd abcde a', 'abc')) // 输出2

nnnnnnnnnni · 2019-08-20T05:24:23Z

var childInNums = parent.split(child).length - 1, 这个没毛病吧

哈哈哈哈和你的想法一样

J1nvey · 2019-09-05T02:09:45Z

const letter = "a";
const str = "abcaabccabcaa"
const time = (str, letter) => str.split(letter).length - 1;
console.log(time(str, letter)) // 6

不知道会不会有没有其他特殊情况，这样写的话。。。

JJL-SH · 2019-09-05T09:14:50Z

function strCount(str, target) {
  let count = 0;
  while (str.includes(target)) {
    str = str.replace(target, "");
    count++;
  }
  return count;
}
function strCount(str, target) {
  return str.split(target).length - 1;
}
console.log(strCount("asdgqwidugasgdakisdggfjwdagasd", "gas"));

funlee · 2019-09-07T15:25:40Z

function count(str, string) {
  return Math.max(0, string.split(str).length - 1);
}

liaoyf · 2019-09-17T01:52:24Z

‘’aa‘’在“aaabcaaa”中出现几次？如果是两次上述答案没有问题，如果是四次
应该可以用零宽正向断言

function subCount(str,target){
  let matchs = str.match(new RegExp(`(?=${target})`, 'g'))
  return matchs ? matchs.length : 0
}

Daniel-Fang · 2019-09-19T09:06:30Z

fucntion repeat(str,parentStr){
  return parentStr.split(str).length - 1
}

function fn(str, substr) {
    if(substr === '') return 0;
    return str.split(substr).length - 1;
}

JYone223 · 2019-09-23T10:09:23Z

const subStrTimes = (str, sub) => {
  return Array.from(str.matchAll(sub), m => m[0]).length;
}

lijunren · 2019-09-27T02:23:29Z

function getTimes(str1, str2) {
if (str2.indexOf(str1) === -1) {
return 0;
}
const arr = str2.splite(str1);
return arr.length - 1;
}

username-xu · 2019-09-27T07:10:26Z

var str = 'aabbccabcaabbccabcabc15114816181abc64641414tregfagfczfavacabaaabc6486';
var strArr = str.split('abc')

console.log(strArr.length -1)

mopig · 2019-09-29T11:08:52Z

function counter(str, subStr, num = 0, index = 0) {
	let tempIndex = str.indexOf(subStr, index)
    return tempIndex > -1 ? counter(str, subStr, ++num, ++tempIndex) : num
}

Penndo · 2021-02-05T04:32:47Z

function statistics(str1,str2){
    let reg = new RegExp(`${str1}{1}`,'g');
    return str2.match(reg).length;
}

VaynePeng · 2021-02-14T13:17:18Z

function strCount(str, target) { let countArr = str.split(target) return countArr.length - 1 }

lxt-ing · 2021-04-14T08:52:56Z

function getSame(str,substr){ let reg = new RegExp(substr,'g'); return str.match(reg).length; }

378406712 · 2021-04-22T09:11:34Z

378406712 · 2021-04-22T09:14:23Z

@378406712

let index = 0;
const count = (str, anotherStr) => {
const len = str.length;
if (len === 0) return 0;
else if (len === 1) {
const counts = anotherStr.split("").filter((item) => item === str);
return counts.length;
} else {
const reg = eval(/${str}/);
const tag = anotherStr.search(reg);
if (tag > -1) {
index++;
const newStr = anotherStr.replace(reg, "");
count(str, newStr);
}
return index;
}
};
console.log(count("aaa", "aaabaaabaabaaa"));

...想复杂了

TheRest0 · 2021-04-23T05:18:12Z

function strCount(str, target) {
  let arr = str.split(target)
  return arr.length -1
}

cielaber · 2021-04-30T15:23:29Z

let counter = 0;//计数器
function count(str,string){
if(string.length<str.length) return;
if(string.indexOf(str)>-1){
let index = string.indexOf(str)
// console.log(index);
counter++;
count(str,string.slice(index+str.length))
}
}

4479zheng · 2021-05-27T14:13:37Z

function frequency(str1,str2){
let num = 0;
let postion = 0;
while(str1.indexOf(str2,postion)!=-1){
postion = str1.indexOf(str2,postion)+1;
str1.indexOf(str2,postion);
num++;
}
console.log(num);
}
frequency("jvkdasjvakj","jv")

shellyxiao48 · 2021-06-02T03:42:46Z

let num = 0;
function showNum(str, key) {
const len = key.length;
const index = str.indexOf(key)
if (index != -1) {
let newStr = str.substr(0, index) + str.substr(index + len)
num++;
return showNum(newStr, key)
}
return num
}

codeyourwayup · 2021-07-16T10:55:45Z

const countAppears = (char, str) => {

let count = 0;


function run(str) {

    if (!str.includes(char)) {
        return;
    }


    if (str.includes(char)) {
        count++;
        const newStr = str.replace(char, '');

        run(newStr);

    }



}

run(str);

return count;

};

const str = "ab2dx22d";

console.log(countAppears("d", str));
console.log(countAppears("2d", str));
console.log(countAppears("2", str));

Injured-Double · 2021-08-14T09:47:58Z

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


    function indexOfNum(obj, target, pos) {

        let count = 0;

        while (true) {
            const index = obj.indexOf(target, pos);

            if (index === -1) break;

            console.log(`Found as at ${index}`);
            pos = index + 1;
            count++;

        }


        return count;

    }


    console.log(indexOfNum(str, 'as', 0)); //3

FoxJokerEila · 2021-09-18T02:04:42Z

function count(son, mother) {
let c = 0
let arr = []
const first = son[0]
const len = son.length
for (let i = 0; i < mother.length; i++) {
if (mother[i] === first) {
arr.push(i)
}
}
for (item of arr) {
if (son == mother.substring(item, item + len)) {
c++
}
}
return c
}

console.log(count('abc', 'abcdef abcdef a'));

xuan-123 · 2021-10-11T07:48:27Z

function toCase(str,target){
return str.split(target).length-1
}

xuan-123 · 2021-10-11T07:49:09Z

@Shoryukenbt

var childInNums = parent.split(child).length - 1, 这个没毛病吧

哈哈我也是这么想的

amikly · 2021-10-25T12:10:16Z

split方法

// split 方法
function countStr(str1, str2) {
    return str2.split(str1).length - 1;
}
console.log(countStr("da", "dabshaudaoiddda"));

includes + indexOf + substring 方法

// includes 方法
function countStr(str1, str2) {
    let count = 0;
    while (str1 && str2 && str2.includes(str1)) {
      let index = str2.indexOf(str1);
      count++;
      str2 = str2.substring(index + str1.length);
    }
    return count;
 }
console.log(countStr("da", "dabshaudaoiddda"));

match + replace 方法

// match 方法
function countStr(str1, str2) {
    let count = 0;
    while (str1 && str2 && str2.match(str1)) {
    str2 = str2.replace(str1, "");
    count++;
}
	return count;
}
console.log(countStr(" ", " "));

递归

// 递归
function countStr(str1, str2, count = 0) {
    if (str1 && str2) {
    return str2.indexOf(str1) !== -1
    ? countStr(
    str1,
    str2.substring(str2.indexOf(str1) + str1.length),
    ++count
	)
	: count;
}
	return count;
}
console.log(countStr("da", "dabshaudaoiddda"));
console.log(countStr("", ""));

VaynePeng · 2021-11-02T09:37:27Z

function strCount(str, char) {
  return str.split(char).length - 1
}

strCount('abcabc', 'abc')

Rednaxela-2 · 2022-01-26T06:45:09Z

var childInNums = parent.split(child).length - 1, 这个没毛病吧

我试了几个方法，这个的速度也是最快的

github-cxtan · 2022-02-15T04:08:50Z

function GetCounts(srcData, DetectStr){
let reg = new RegExp(${DetectStr}, 'g');
return srcData.match(reg).length;
}

QiaoIsMySky · 2022-02-21T13:44:58Z

let str = 's'
let ostr = 'seowqjoewqskodkwqkdpsadpdkpwq'
let index = 0
ostr.split('').forEach(item => {
if (item === str){
index++
}
})
console.log(index);

sup194 · 2022-02-23T01:19:44Z

function strCount(str, target) {
let count = 0
if (!target) return count
while (str.match(target)) {
str.replact('')
count++
}
return count
}

yxllovewq · 2022-03-08T02:56:09Z

不使用正则：

const strCountOf = (str = '', fromStr = '') => {
  let count = 0;
  let start = false;
  let startId;
  for (let i = 0; i < fromStr.length; i++) {
    if (str[0] === fromStr[i]) {
      start = true;
      startId = i;
    }
    if (start) {
      if (str[i - startId] !== fromStr[i]) {
        start = false;
      }
      if (i === startId + str.length - 1) {
        count++;
        start = false;
      }
    }
  }
  return count;
}

使用正则：

const strCountOf = (str = '', fromStr = '') => [...fromStr.matchAll(new RegExp(str, 'g'))].length;

storm-cao · 2022-03-11T07:57:33Z

const countString = (childStr,parentStr) => parentStr.split(childStr).length - 1);

wenxd · 2022-05-20T06:43:40Z

function statCharInString(str, char) {
    let re = new RegExp(char, 'g')
    let num = str.match(re).length
    console.log(num);
  }
  statCharInString('asf sfas  sadf jjsfahf', 'as')

xiaoqiangz · 2022-05-22T02:22:29Z

// 统计某一字符或字符串在另一个字符串中出现的次数
function repatCount(str, target) {
// 第一种
// return str.split(target).length - 1
// 第二种
// let count = 0
// while (str.includes(target)) {
// count++
// str = str.replace(target, '')
// }
// return count
// 第三种
let reg = new RegExp(target, 'g')
return str.match(reg).length
}
console.log(repatCount('abcdeaveavffavvfa', 'av'))

xiaoxiaozhiya · 2022-05-28T23:37:22Z

方法一正则
function getTotalCount(str,target){ var reg=new RegExp(target,"g") var arr=str.match(reg) return arr.length }

方法二
利用字符串的位置关系（间接）
function countStr(str1, str2) { var arr=str2.split(str1) console.log(arr) return str2.split(str1).length - 1; }

方法三
一边判断一边丢

function countStr(str1, str2) { let count = 0; while (str1 && str2 && str2.includes(str1)) { let index = str2.indexOf(str1); count++; str2 = str2.substring(index + str1.length); console.log(str2) } return count; }

substring 方法不接受负的参数

www-wanglong · 2022-06-18T09:43:27Z

function count(str, target) {
let reg = new RegExp(str, 'g')
return target.match(reg).length
}

count('a', 'abcsadfwretaabcerwawabc')

wyy-g · 2022-09-08T00:18:06Z

function countStr(str, target){
var count = 0;
while(str.includes(target)){
count++;
str = str.replace(target, '');
}
}

mocc124 · 2022-12-04T14:34:15Z

function getCount(str, char) {
  let index = str.indexOf(char);
  if (index < 0) return 0;
  return 1 + getCount(str.slice(index + char.length), char);
}

tcxiao1 · 2023-08-10T06:48:02Z

function countStr(target, array, rightIndex, leftIndex, count) {

if (leftIndex > array.length) {
    return count;
}

let source = "";
for (let i = rightIndex; i < leftIndex; i++) {
    source = source + array[i];
}
if (source === target) {
    count++;
}

rightIndex++;
leftIndex++;
return countStr(target, array, rightIndex, leftIndex, count);

}

let target = "aa";
let src = "aaabbcaacddaabbccdd";
let countStr1 = countStr(target, Array.from(src), 0, target.length, 0);
console.log(countStr1)

双指针

lili-0923 · 2024-02-02T06:23:49Z

function countStr(str1, str2) {
return str2.split(str1).length - 1;
}

haizhilin2013 added the js JavaScript label Apr 22, 2019

zivenday mentioned this issue Aug 10, 2019

统计某一字符或字符串在另一个字符串中出现的次数？ zivenday/learning#21

Open

[js] 第7天 统计某一字符或字符串在另一个字符串中出现的次数 #21

[js] 第7天 统计某一字符或字符串在另一个字符串中出现的次数 #21

Comments

haizhilin2013 commented Apr 22, 2019

linghucq1 commented May 8, 2019

coderfe commented May 9, 2019 • edited

undefinedYu commented May 10, 2019

qqdnnd commented May 16, 2019

Shoryukenbt commented May 28, 2019

tiyunchen commented May 31, 2019

myprelude commented Jun 13, 2019

AricZhu commented Jun 19, 2019

bWhirring commented Jul 3, 2019

Konata9 commented Jul 4, 2019

poporeki commented Jul 4, 2019

15190408121 commented Jul 5, 2019

taiyangxingchen commented Jul 10, 2019

shufangyi commented Jul 18, 2019

pythonqi commented Jul 24, 2019

wyx2014 commented Jul 24, 2019

Vi-jay commented Jul 25, 2019

wsb260 commented Jul 31, 2019 • edited

nnnnnnnnnni commented Aug 20, 2019

J1nvey commented Sep 5, 2019

JJL-SH commented Sep 5, 2019

funlee commented Sep 7, 2019

liaoyf commented Sep 17, 2019

Daniel-Fang commented Sep 19, 2019

JYone223 commented Sep 23, 2019

lijunren commented Sep 27, 2019

username-xu commented Sep 27, 2019

mopig commented Sep 29, 2019

Penndo commented Feb 5, 2021

VaynePeng commented Feb 14, 2021

lxt-ing commented Apr 14, 2021

378406712 commented Apr 22, 2021

378406712 commented Apr 22, 2021

TheRest0 commented Apr 23, 2021

cielaber commented Apr 30, 2021

4479zheng commented May 27, 2021

shellyxiao48 commented Jun 2, 2021

codeyourwayup commented Jul 16, 2021

Injured-Double commented Aug 14, 2021

FoxJokerEila commented Sep 18, 2021

xuan-123 commented Oct 11, 2021

xuan-123 commented Oct 11, 2021

amikly commented Oct 25, 2021

split方法

includes + indexOf + substring 方法

match + replace 方法

递归

VaynePeng commented Nov 2, 2021

Rednaxela-2 commented Jan 26, 2022

github-cxtan commented Feb 15, 2022

QiaoIsMySky commented Feb 21, 2022

sup194 commented Feb 23, 2022 • edited

yxllovewq commented Mar 8, 2022

storm-cao commented Mar 11, 2022

wenxd commented May 20, 2022

xiaoqiangz commented May 22, 2022

xiaoxiaozhiya commented May 28, 2022

www-wanglong commented Jun 18, 2022

wyy-g commented Sep 8, 2022

mocc124 commented Dec 4, 2022

tcxiao1 commented Aug 10, 2023

lili-0923 commented Feb 2, 2024

[js] 第7天统计某一字符或字符串在另一个字符串中出现的次数 #21

[js] 第7天统计某一字符或字符串在另一个字符串中出现的次数 #21

coderfe commented May 9, 2019 •

edited

wsb260 commented Jul 31, 2019 •

edited

sup194 commented Feb 23, 2022 •

edited