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

此函数在三种模式下工作:查找字符串中单个字符的频率,查找字符串中相邻子字符串的频率,然后如果它与一个匹配,则会直接向前移动到它后面的下一个,第三个与前一个相似,但它也会计算给定字符串中的交叉子字符串

函数substringFrequency(字符串、子字符串、连接){let索引允许发生频率=0for(设i=0;i<string.length;i++){index=string.indexOf(substring,i)如果(索引!=-1){if((子字符串长度==1)||连接==true){i=索引}其他{i=索引+1}发生频率++}其他{打破} }return(发生频率)}console.log(substringFrequency('vvv','v'))console.log(substringFrequency('vvv','vv'))console.log(substringFrequency('vvv','vv'))

其他回答

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。

看到这篇帖子。

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
}

相同的算法可以被布置得更短:

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 );
}

你可以试试这个:

var theString=“这是一个字符串。”;console.log(String.split(“is”).length-1);

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
}