How to count string occurrence in string?

S

41

848

How can I count the number of times a particular string occurs in another string. For example, this is what I am trying to do in Javascript:

var temp = "This is a string.";
alert(temp.count("is")); //should output '2'

Sinewy answered 24/10, 2010 at 18:39 Comment(4)

It depends on whether you accept overlapping instances, e.g. var t = "sss"; How many instances of the substring "ss" are in the string above? 1 or 2? Do you leapfrog over each instance, or move the pointer character-by-character, looking for the substring? – Mixup 24/10, 2010 at 19:30

An improved benchmark for this question's answers: jsperf.com/string-ocurrence-split-vs-match/2 (based of Kazzkiq's benchmark). – Effectual 27/5, 2015 at 15:8

Count Total Amount Of Specific Word In a String JavaScript https://mcmap.net/q/37463/-count-total-amount-of-specific-word-in-a-string-javascript – Biebel 29/11, 2020 at 18:57

this video seems vaguely related here - "Google Coding Interview With A Facebook Software Engineer" - youtube.com/watch?v=PIeiiceWe_w – Cymbal 13/12, 2020 at 9:7

S

1389

The g in the regular expression (short for global) says to search the whole string rather than just find the first occurrence. This matches is twice:

var temp = "This is a string.";
var count = (temp.match(/is/g) || []).length;
console.log(count);

And, if there are no matches, it returns 0:

var temp = "Hello World!";
var count = (temp.match(/is/g) || []).length;
console.log(count);

Suburbia answered 24/10, 2010 at 18:42 Comment(2)

Thanks for this.. I went with count = (str.match(/is/g) || []).length to handle if you don't have a match. – Apograph 29/10, 2012 at 12:19

Just to let you know: if you want to search using special characters, you have to escape them. For example: looking for how many '.' are, just use (text.match(/\./g) || []).length – Ced 9/1 at 13:51

O

279

/** 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://mcmap.net/q/37371/-how-to-count-string-occurrence-in-string
 */
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;
}

Usage

occurrences("foofoofoo", "bar"); //0

occurrences("foofoofoo", "foo"); //3

occurrences("foofoofoo", "foofoo"); //1

allowOverlapping

occurrences("foofoofoo", "foofoo", true); //2

Matches:

  foofoofoo
1 `----´
2    `----´

Unit Test

https://jsfiddle.net/Victornpb/5axuh96u/

Benchmark

I've made a benchmark test and my function is more then 10 times faster then the regexp match function posted by gumbo. In my test string is 25 chars length. with 2 occurences of the character 'o'. I executed 1 000 000 times in Safari.

Safari 5.1

Benchmark> Total time execution: 5617 ms (regexp)

Benchmark> Total time execution: 881 ms (my function 6.4x faster)

Firefox 4

Benchmark> Total time execution: 8547 ms (Rexexp)

Benchmark> Total time execution: 634 ms (my function 13.5x faster)

Edit: changes I've made

cached substring length

added type-casting to string.

added optional 'allowOverlapping' parameter

fixed correct output for "" empty substring case.

Gist

https://gist.github.com/victornpb/7736865

Oys answered 28/10, 2011 at 1:27 Comment(1)

I've tried different answers in this topic, and this answer turned out to behaving around 5-10% better performance in large amounts ( compared to other answers). – Paraphrastic 19/3, 2023 at 15:23

W

220

function countInstances(string, word) {
   return string.split(word).length - 1;
}
console.log(countInstances("This is a string", "is"))

Walden answered 24/10, 2010 at 18:43 Comment(3)

for me is working without the -1 – Gasify 5/10, 2022 at 13:41

Benchmarked it with jsbench.me against the solution "occurrences" from Vitim.us, almost same performance. A simple "split" is my preferred solution as it is the easiest one. The regexes are a bit slower. – Willtrude 7/1, 2023 at 11:24

@Gasify without -1 the count would be wrong – Consistent 15/4, 2023 at 20:6

I

116

You can try this:

var theString = "This is a string.";
console.log(theString.split("is").length - 1);

Impuissant answered 24/2, 2013 at 17:30 Comment(0)

W

43

My solution:

var temp = "This is a string.";

function countOccurrences(str, value) {
  var regExp = new RegExp(value, "gi");
  return (str.match(regExp) || []).length;
}

console.log(countOccurrences(temp, 'is'));

Washko answered 9/4, 2013 at 1:20 Comment(2)

maybe it would be better to return (str.match(regExp) || []).length; That way you don't evaluate the regular expression twice? – Midst 7/12, 2013 at 4:8

you also need to scape your string or countOcurrences('Hello...','.')==8 and not 3 – Oys 9/5, 2014 at 3:16

H

19

You can use match to define such function:

String.prototype.count = function(search) {
    var m = this.match(new RegExp(search.toString().replace(/(?=[.\\+*?[^\]$(){}\|])/g, "\\"), "g"));
    return m ? m.length:0;
}

Hermia answered 24/10, 2010 at 18:45 Comment(1)

If you wanted it to be uniform with JS's search semantics, the return line would be return m ? m.length:-1;. – Bautista 20/12, 2014 at 3:13

E

14

Just code-golfing Rebecca Chernoff's solution :-)

alert(("This is a string.".match(/is/g) || []).length);

Eba answered 13/8, 2011 at 18:16 Comment(0)

A

14

The non-regex version:

 var string = 'This is a string',
    searchFor = 'is',
    count = 0,
    pos = string.indexOf(searchFor);

while (pos > -1) {
    ++count;
    pos = string.indexOf(searchFor, ++pos);
}

console.log(count);   // 2

Antiperiodic answered 15/3, 2014 at 8:21 Comment(2)

1. It's only for single char search, too subtle 2. even OP asks for is occurences – Ahmedahmedabad 20/5, 2016 at 5:57

This is probably the fastest implementation here, but it would be even faster if you replaced "++pos" with "pos+=searchFor.length" – Sender 1/4, 2019 at 13:43

F

12

String.prototype.Count = function (find) {
    return this.split(find).length - 1;
}

console.log("This is a string.".Count("is"));

This will return 2.

Frank answered 10/9, 2016 at 3:53 Comment(0)

I

7

Here is the fastest function!

Why is it faster?

Doesn't check char by char (with 1 exception)
Uses a while and increments 1 var (the char count var) vs. a for loop checking the length and incrementing 2 vars (usually var i and a var with the char count)
Uses WAY less vars
Doesn't use regex!
Uses an (hopefully) highly optimized function

All operations are as combined as they can be, avoiding slowdowns due to multiple operations

String.prototype.timesCharExist=function(c){var t=0,l=0,c=(c+'')[0];while(l=this.indexOf(c,l)+1)++t;return t};

Here is a slower and more readable version:

    String.prototype.timesCharExist = function ( chr ) {
        var total = 0, last_location = 0, single_char = ( chr + '' )[0];
        while( last_location = this.indexOf( single_char, last_location ) + 1 )
        {
            total = total + 1;
        }
        return total;
    };

This one is slower because of the counter, long var names and misuse of 1 var.

To use it, you simply do this:

    'The char "a" only shows up twice'.timesCharExist('a');

Edit: (2013/12/16)

DON'T use with Opera 12.16 or older! it will take almost 2.5x more than the regex solution!

On chrome, this solution will take between 14ms and 20ms for 1,000,000 characters.

The regex solution takes 11-14ms for the same amount.

Using a function (outside String.prototype) will take about 10-13ms.

Here is the code used:

    String.prototype.timesCharExist=function(c){var t=0,l=0,c=(c+'')[0];while(l=this.indexOf(c,l)+1)++t;return t};

    var x=Array(100001).join('1234567890');

    console.time('proto');x.timesCharExist('1');console.timeEnd('proto');

    console.time('regex');x.match(/1/g).length;console.timeEnd('regex');

    var timesCharExist=function(x,c){var t=0,l=0,c=(c+'')[0];while(l=x.indexOf(c,l)+1)++t;return t;};

    console.time('func');timesCharExist(x,'1');console.timeEnd('func');

The result of all the solutions should be 100,000!

Note: if you want this function to count more than 1 char, change where is c=(c+'')[0] into c=c+''

Inspect answered 29/8, 2013 at 15:40 Comment(1)

the prototype was AN EXAMPLE! You can use the function as you please! You can even do this: var timesFunctionExist=function(x,c){var t=0,l=0,c=(c+'')[0];while(l=x.indexOf(c,l)+1)++t;return t}); alert(timesCharExist('The char "a" only shows up twice','a'));! (this will speed up a little more cause i wont be messing with prototypes). If you think I'm wrong, why don't you show it before throwing rocks at me? Prove to me that my function sucks and i will accept it. Show me a test case. And the length of vars does have influence on speed. You can test it. – Inspect 8/10, 2013 at 11:2

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7

var temp = "This is a string.";
console.log((temp.match(new RegExp("is", "g")) || []).length);

Serrated answered 18/11, 2015 at 12:56 Comment(0)

F

5

A simple way would be to split the string on the required word, the word for which we want to calculate the number of occurences, and subtract 1 from the number of parts:

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

Foreigner answered 1/6, 2021 at 20:15 Comment(0)

A

4

I think the purpose for regex is much different from indexOf. indexOf simply find the occurance of a certain string while in regex you can use wildcards like [A-Z] which means it will find any capital character in the word without stating the actual character.

Example:

 var index = "This is a string".indexOf("is");
 console.log(index);
 var length = "This is a string".match(/[a-z]/g).length;
 // where [a-z] is a regex wildcard expression thats why its slower
 console.log(length);

Apophasis answered 12/3, 2013 at 6:12 Comment(0)

T

3

Super duper old, but I needed to do something like this today and only thought to check SO afterwards. Works pretty fast for me.

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;
};

Turtleneck answered 5/4, 2013 at 4:16 Comment(0)

S

3

       var myString = "This is a string.";
        var foundAtPosition = 0;
        var Count = 0;
        while (foundAtPosition != -1)
        {
            foundAtPosition = myString.indexOf("is",foundAtPosition);
            if (foundAtPosition != -1)
            {
                Count++;
                foundAtPosition++;
            }
        }
        document.write("There are " + Count + " occurrences of the word IS");

Refer :- count a substring appears in the string for step by step explanation.

Shrub answered 19/11, 2014 at 17:13 Comment(0)

R

3

For anyone that finds this thread in the future, note that the accepted answer will not always return the correct value if you generalize it, since it will choke on regex operators like $ and .. Here's a better version, that can handle any needle:

function occurrences (haystack, needle) {
  var _needle = needle
    .replace(/\[/g, '\\[')
    .replace(/\]/g, '\\]')
  return (
    haystack.match(new RegExp('[' + _needle + ']', 'g')) || []
  ).length
}

Reliable answered 28/2, 2015 at 22:52 Comment(0)

S

3

Building upon @Vittim.us answer above. I like the control his method gives me, making it easy to extend, but I needed to add case insensitivity and limit matches to whole words with support for punctuation. (e.g. "bath" is in "take a bath." but not "bathing")

The punctuation regex came from: https://mcmap.net/q/37464/-how-can-i-strip-all-punctuation-from-a-string-in-javascript-using-regex (How can I strip all punctuation from a string in JavaScript using regex?)

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;
}

Please feel free to modify and refactor this answer if you spot bugs or improvements.

Sledgehammer answered 28/11, 2015 at 8:2 Comment(0)

F

3

We can use the js split function, and it's length minus 1 will be the number of occurrences.

var temp = "This is a string.";
alert(temp.split('is').length-1);

Fanchet answered 17/2, 2022 at 6:47 Comment(2)

Welcome. SO works differently than forums. SO is designed such that good answers should be upvoted, not duplicated. The answer you have suggested already exists, so you should upvote it instead. There are also answers using the same concept at its base, but also consider a more nuanced interpretation (eg, should ss in sss count as 1 or 2? ). So maybe upvote those as well, if you like. For onboarding, please read up on "how to answer" & "how to ask" topics in the help section, linked to at the top of every page. We appreciate & look forward to your future contributions. – Plenary 21/2, 2022 at 14:1

That said, great on posting in a concise, clear manner, with an attempt to provide an explanation, which many of the oldest answers failed to do. To be clear, their code-only answers are discouraged on SO (though it wasn't always well enforced back then). Looking forward to seeing more in the future. – Plenary 21/2, 2022 at 14:47

F

2

Try it

<?php 
$str = "33,33,56,89,56,56";
echo substr_count($str, '56');
?>

<script type="text/javascript">
var temp = "33,33,56,89,56,56";
var count = temp.match(/56/g);  
alert(count.length);
</script>

Fissionable answered 15/11, 2013 at 11:27 Comment(0)

C

2

Simple version without regex:

var temp = "This is a string.";

var count = (temp.split('is').length - 1);

alert(count);

Clypeate answered 8/6, 2016 at 13:21 Comment(0)

G

2

No one will ever see this, but it's good to bring back recursion and arrow functions once in a while (pun gloriously intended)

String.prototype.occurrencesOf = function(s, i) {
 return (n => (n === -1) ? 0 : 1 + this.occurrencesOf(s, n + 1))(this.indexOf(s, (i || 0)));
};

Grapery answered 11/2, 2019 at 3:25 Comment(0)

A

2

 function substrCount( str, x ) {
   let count = -1, pos = 0;
   do {
     pos = str.indexOf( x, pos ) + 1;
     count++;
   } while( pos > 0 );
   return count;
 }

Apartment answered 3/9, 2020 at 13:10 Comment(0)

F

2

ES2020 offers a new MatchAll which might be of use in this particular context.

Here we create a new RegExp, please ensure you pass 'g' into the function.

Convert the result using Array.from and count the length, which returns 2 as per the original requestor's desired output.

let strToCheck = RegExp('is', 'g')
let matchesReg = "This is a string.".matchAll(strToCheck)
console.log(Array.from(matchesReg).length) // 2