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
太老了,但我今天需要做这样的事情,只想事后检查一下。对我来说工作很快。
String.prototype.count = function(substr,start,overlap) {
overlap = overlap || false;
start = start || 0;
var count = 0,
offset = overlap ? 1 : substr.length;
while((start = this.indexOf(substr, start) + offset) !== (offset - 1))
++count;
return count;
};
其他回答
参数:ustring:超集字符串countChar:子字符串
一个计算JavaScript中子字符串出现次数的函数:
函数subStringCount(ustring,countChar){var correspCount=0;var corresp=false;变量量=0;var prevChar=空;对于(var i=0;i!=ustring.length;i++){如果(ustring.charAt(i)==countChar.charAt(0)&&corresp==false){corresp=真;correspCount+=1;如果(correspCount==countChar.length){数量+=1;corresp=false;correspCount=0;}prevChar=1;}否则如果(ustring.charAt(i)==countChar.charAt(prevChar)&&corresp==true){correspCount+=1;如果(correspCount==countChar.length){数量+=1;corresp=false;correspCount=0;prevChar=空;}其他{prevChar+=1;}}其他{corresp=false;correspCount=0;}} 回报金额;}console.log(subStringCount(“Hello World,Hello World”,“ll”));
对于将来找到此线程的任何人,请注意,如果您对其进行概括,则接受的答案不会总是返回正确的值,因为它会阻塞正则表达式运算符,如$和。。这里有一个更好的版本,可以处理任何针头:
function occurrences (haystack, needle) {
var _needle = needle
.replace(/\[/g, '\\[')
.replace(/\]/g, '\\]')
return (
haystack.match(new RegExp('[' + _needle + ']', 'g')) || []
).length
}
一种简单的方法是将字符串拆分为所需单词,即我们要计算出现次数的单词,然后从部分数中减去1:
function checkOccurences(string, word) {
return string.split(word).length - 1;
}
const text="Let us see. see above, see below, see forward, see backward, see left, see right until we will be right";
const count=countOccurences(text,"see "); // 2
基于@Vittim.us的上述回答。我喜欢他的方法给我的控制,使其易于扩展,但我需要添加不区分大小写的功能,并将匹配限制在支持标点符号的整个单词中。(例如,“洗澡”是指“洗澡”,而不是“洗澡”)
标点正则表达式来自:https://stackoverflow.com/a/25575009/497745(如何使用正则表达式从JavaScript字符串中删除所有标点符号?)
function keywordOccurrences(string, subString, allowOverlapping, caseInsensitive, wholeWord)
{
string += "";
subString += "";
if (subString.length <= 0) return (string.length + 1); //deal with empty strings
if(caseInsensitive)
{
string = string.toLowerCase();
subString = subString.toLowerCase();
}
var n = 0,
pos = 0,
step = allowOverlapping ? 1 : subString.length,
stringLength = string.length,
subStringLength = subString.length;
while (true)
{
pos = string.indexOf(subString, pos);
if (pos >= 0)
{
var matchPos = pos;
pos += step; //slide forward the position pointer no matter what
if(wholeWord) //only whole word matches are desired
{
if(matchPos > 0) //if the string is not at the very beginning we need to check if the previous character is whitespace
{
if(!/[\s\u2000-\u206F\u2E00-\u2E7F\\'!"#$%&\(\)*+,\-.\/:;<=>?@\[\]^_`{|}~]/.test(string[matchPos - 1])) //ignore punctuation
{
continue; //then this is not a match
}
}
var matchEnd = matchPos + subStringLength;
if(matchEnd < stringLength - 1)
{
if (!/[\s\u2000-\u206F\u2E00-\u2E7F\\'!"#$%&\(\)*+,\-.\/:;<=>?@\[\]^_`{|}~]/.test(string[matchEnd])) //ignore punctuation
{
continue; //then this is not a match
}
}
}
++n;
} else break;
}
return n;
}
如果发现错误或改进,请随时修改和重构此答案。
//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;
}