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JavaScript : within a string, count the number of occurances of a character / character counting and string-position
//www.lsauer.com 2012
//Answer to:
//http://stackoverflow.com/questions/881085/count-the-number-of-occurances-of-a-character-in-a-string-in-javascript/10671743#10671743
//There are at least four ways. The best option, which should also be the fastest -owing to the native RegEx engine -, is placed at the top. //jsperf.com is currently down, otherwise I would provide you with performance statistics.
#1.
("this is foo bar".match(/o/g)||[]).length
//>2
#2.
"this is foo bar".split("o").length-1
//>2
//split is not recommended. Resource hungry. Allocates new instances of 'Array' for each match. Don't try that for a >100MB file via FileReader.
//You can actually easily observe the EXACT resource usage using **Chrome's profiler** option.
#3.
var stringsearch = "o"
,str = "this is foo bar";
for(var count=-1,index=-2; index != -1; count++,index=str.indexOf(stringsearch,index+1) );
//>count:2
#4.
//searching for a single character
var stringsearch = "o"
,str = "this is foo bar";
for(var i=count=0; i<str.length; count+=+(stringsearch===str[i++]));
//>count:2
#5.
//element mapping and filtering; not recommended due to its overall resource pre-allocation vs. Pythonian 'generators'
//provides the position within the string
var str = "this is foo bar"
str.split('').map( function(e,i){ if(e === 'o') return i;} )
.filter(Boolean)
//>[9, 10]
[9, 10].length
//>2
#6
//'deleting' the character out of the string and measuring the distance in length
var str = "this is foo bar";
str.length - str.replace(/o/g,'').length
//>2
#7
//based on typed arrays; str2buffer is taken from 'is-lib'; See: https://gist.github.com/lsauer
//Converts an ASCII string to an typed-Array buffer
str2buffer = function(s){ var bu = new ArrayBuffer(s.length), aUint8 = new Uint8Array(bu );
for(var i=0; i<bu.byteLength; aUint8[i]=s.charCodeAt(i),i++);return aUint8;
};
var bstr = str2buffer ("this is foo bar")
,schar = 'o'.charCodeAt()
,cnt=0;
for(var i=0;i<bstr.byteLength;schar!==bstr[i++]||cnt++);
//>cnt
2
#8
//based on untyped Arrays. Is expected to be slower. Analogous to #7
var ubstr = "this is foo bar".split('').map( function(e,i){ return e.charCodeAt();} )
//>[116, 104, 105, 115, 32, 105, 115, 32, 102, 111, 111, 32, 98, 97, 114]
,schar = 'o'.charCodeAt()
,cnt=0;
for(var i=0;i<ubstr.length;schar!==ubstr[i++]||cnt++);
//>cnt
2
#9
//using reduce. Note: Element map functions are powerful but slow as they involve their own heap/stack allocation
//see: http://stackoverflow.com/questions/10293378/what-is-the-most-efficient-way-of-merging-1-2-and-7-8-into-1-7-2-8/17910641#17910641
var str = "this is foo bar",
schar = 'o';
str.split('').reduce(
function(p,c,i,a){ if(c === schar || p === schar){return isNaN(parseInt(p))? 1:+p+1;} return p;}
)
//Note: faster: c === schar || p === schar; slower: (c+p).indexOf(schar)>-1
#10. dictionary character histogram
var str = "this is foo bar",
schar = 'o',
hist={};
for(si in str){
hist[str[si]] = hist[str[si]] ? 1+hist[str[si]]:1;
}
//>hist[schar]
2
//Changelog > 11/2013:
//
// 24/11/2013 #3 bug fixed in initial index position; pointed out by Augustus@Stackoverflow
@FenrisWolfe

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@FenrisWolfe FenrisWolfe commented Jun 28, 2018

Good code.
How would you deal with finding the maximum occurrence of substrings in a larger string without looking for a specific string?
EX: "can can ran tan van"
The output being "an".

@yosvelquintero

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@yosvelquintero yosvelquintero commented Aug 23, 2018

Just another way:

const str = 'this is foo bar';
const count = [...str].filter(l => l === 'o').length;

console.log(count);
@ykarkar

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@ykarkar ykarkar commented Jan 31, 2019

Can you explain the number 4 for loop.
count+=+(stringsearch===str[i++])

@ByFaizan

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@ByFaizan ByFaizan commented Oct 24, 2019

Here's Another Way;

var text = "The quick brown fox jumps over the lazy dog";
text = text.toLowerCase();
var textLen = text.length;
var searchFor = "the";
var indexOfSearch = text.indexOf(searchFor);
var counter = 0;

for (var i = 0; i < textLen; i++) {
if (text.indexOf(text.charAt(i)) === indexOfSearch) {
counter++
}
}

console.log(counter);

@MNSY

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@MNSY MNSY commented Nov 28, 2019

which one is faster?

@ghost

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@ghost ghost commented Jan 4, 2020

var text = "The quick brown fox jumps over the lazy dog";
document.write("Text: "+text+"
"+"There are "+text.match(/the/gi).length+" of word the");

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