如何计算特定字符串在另一个字符串中出现的次数。例如,这就是我试图在Javascript中做的事情:
var temp = "This is a string.";
alert(temp.count("is")); //should output '2'
如何计算特定字符串在另一个字符串中出现的次数。例如,这就是我试图在Javascript中做的事情:
var temp = "This is a string.";
alert(temp.count("is")); //should output '2'
当前回答
函数countInstance(字符串,单词){返回字符串.split(word).length-1;}console.log(countInstance(“This is a string”,“is”))
其他回答
/** Function that count occurrences of a substring in a string;
* @param {String} string The string
* @param {String} subString The sub string to search for
* @param {Boolean} [allowOverlapping] Optional. (Default:false)
*
* @author Vitim.us https://gist.github.com/victornpb/7736865
* @see Unit Test https://jsfiddle.net/Victornpb/5axuh96u/
* @see https://stackoverflow.com/a/7924240/938822
*/
function occurrences(string, subString, allowOverlapping) {
string += "";
subString += "";
if (subString.length <= 0) return (string.length + 1);
var n = 0,
pos = 0,
step = allowOverlapping ? 1 : subString.length;
while (true) {
pos = string.indexOf(subString, pos);
if (pos >= 0) {
++n;
pos += step;
} else break;
}
return n;
}
用法
occurrences("foofoofoo", "bar"); //0
occurrences("foofoofoo", "foo"); //3
occurrences("foofoofoo", "foofoo"); //1
允许重叠
occurrences("foofoofoo", "foofoo", true); //2
比赛:
foofoofoo
1 `----´
2 `----´
单元测试
https://jsfiddle.net/Victornpb/5axuh96u/
基准
我做了一个基准测试,我的功能超过了10倍比gumbo发布的regexp匹配函数更快。在我的测试中字符串长度为25个字符。字符“o”出现2次。我在Safari中执行了1000 000次。Safari 5.1基准>执行总时间:5617ms(正则表达式)基准测试>执行总时间:881毫秒(我的功能快6.4倍)Firefox 4基准>执行总时间:8547毫秒(Rexep)基准测试>总执行时间:634毫秒(我的功能更快13.5倍)编辑:我所做的更改缓存的子字符串长度为字符串添加了类型转换。添加了可选的“allowOverlapping”参数修复了“”空子字符串大小写的正确输出。
Gist
https://gist.github.com/victornpb/7736865
试试看:
function countString(str, search){
var count=0;
var index=str.indexOf(search);
while(index!=-1){
count++;
index=str.indexOf(search,index+1);
}
return count;
}
这是我的解决方案。我希望这会对某人有所帮助
const countOccurence = (string, char) => {
const chars = string.match(new RegExp(char, 'g')).length
return chars;
}
我认为regex的目的与indexOf有很大不同。indexOf只需查找某个字符串的出现,而在正则表达式中,您可以使用[a-Z]之类的通配符,这意味着它将查找单词中的任何大写字符,而无需说明实际字符。
例子:
var index=“This is a string”.indexOf(“is”);console.log(索引);var length=“This is a string”.match(/[a-z]/g).length;//其中[a-z]是正则表达式通配符,这就是为什么其速度较慢的原因console.log(长度);
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)