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'
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);
/** 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;
}
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
function countInstances(string, word) {
return string.split(word).length - 1;
}
console.log(countInstances("This is a string", "is"))
You can try this:
var theString = "This is a string.";
console.log(theString.split("is").length - 1);
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'));
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;
}
Just code-golfing Rebecca Chernoff's solution :-)
alert(("This is a string.".match(/is/g) || []).length);
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
String.prototype.Count = function (find) {
return this.split(find).length - 1;
}
console.log("This is a string.".Count("is"));
This will return 2.
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+''
var temp = "This is a string.";
console.log((temp.match(new RegExp("is", "g")) || []).length);
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
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);
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;
};
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.
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://stackoverflow.com/a/25575009/497745 (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.
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
}
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>
Simple version without regex:
var temp = "This is a string.";
var count = (temp.split('is').length - 1);
alert(count);
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)));
};
function substrCount( str, x ) {
let count = -1, pos = 0;
do {
pos = str.indexOf( x, pos ) + 1;
count++;
} while( pos > 0 );
return count;
}
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
Now this is a very old thread i've come across but as many have pushed their answer's, here is mine in a hope to help someone with this simple code.
var search_value = "This is a dummy sentence!";
var letter = 'a'; /*Can take any letter, have put in a var if anyone wants to use this variable dynamically*/
letter = letter && "string" === typeof letter ? letter : "";
var count;
for (var i = count = 0; i < search_value.length; count += (search_value[i++] == letter));
console.log(count);
I'm not sure if it is the fastest solution but i preferred it for simplicity and for not using regex (i just don't like using them!)
You could try this
let count = s.length - s.replace(/is/g, "").length;
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);
Here is my solution. I hope it would help someone
const countOccurence = (string, char) => {
const chars = string.match(new RegExp(char, 'g')).length
return chars;
}
var countInstances = function(body, target) {
var globalcounter = 0;
var concatstring = '';
for(var i=0,j=target.length;i<body.length;i++){
concatstring = body.substring(i-1,j);
if(concatstring === target){
globalcounter += 1;
concatstring = '';
}
}
return globalcounter;
};
console.log( countInstances('abcabc', 'abc') ); // ==> 2
console.log( countInstances('ababa', 'aba') ); // ==> 2
console.log( countInstances('aaabbb', 'ab') ); // ==> 1
substr_count translated to Javascript from php
Locutus (Package that translates Php to JS)
substr_count (official page, code copied below)
function substr_count (haystack, needle, offset, length) {
// eslint-disable-line camelcase
// discuss at: https://locutus.io/php/substr_count/
// original by: Kevin van Zonneveld (https://kvz.io)
// bugfixed by: Onno Marsman (https://twitter.com/onnomarsman)
// improved by: Brett Zamir (https://brett-zamir.me)
// improved by: Thomas
// example 1: substr_count('Kevin van Zonneveld', 'e')
// returns 1: 3
// example 2: substr_count('Kevin van Zonneveld', 'K', 1)
// returns 2: 0
// example 3: substr_count('Kevin van Zonneveld', 'Z', 0, 10)
// returns 3: false
var cnt = 0
haystack += ''
needle += ''
if (isNaN(offset)) {
offset = 0
}
if (isNaN(length)) {
length = 0
}
if (needle.length === 0) {
return false
}
offset--
while ((offset = haystack.indexOf(needle, offset + 1)) !== -1) {
if (length > 0 && (offset + needle.length) > length) {
return false
}
cnt++
}
return cnt
}
Check out Locutus's Translation Of Php's substr_count function
The parameters:
ustring: the superset string
countChar: the substring
A function to count substring occurrence in JavaScript:
function subStringCount(ustring, countChar){
var correspCount = 0;
var corresp = false;
var amount = 0;
var prevChar = null;
for(var i=0; i!=ustring.length; i++){
if(ustring.charAt(i) == countChar.charAt(0) && corresp == false){
corresp = true;
correspCount += 1;
if(correspCount == countChar.length){
amount+=1;
corresp = false;
correspCount = 0;
}
prevChar = 1;
}
else if(ustring.charAt(i) == countChar.charAt(prevChar) && corresp == true){
correspCount += 1;
if(correspCount == countChar.length){
amount+=1;
corresp = false;
correspCount = 0;
prevChar = null;
}else{
prevChar += 1 ;
}
}else{
corresp = false;
correspCount = 0;
}
}
return amount;
}
console.log(subStringCount('Hello World, Hello World', 'll'));
var str = 'stackoverflow';
var arr = Array.from(str);
console.log(arr);
for (let a = 0; a <= arr.length; a++) {
var temp = arr[a];
var c = 0;
for (let b = 0; b <= arr.length; b++) {
if (temp === arr[b]) {
c++;
}
}
console.log(`the ${arr[a]} is counted for ${c}`)
}
Related
Thanks to Nina I have a code to compare two sentences word by word and return the number of word matches like this:
function includeWords(wanted, seen) {
var wantedMap = wanted.split(/\s+/).reduce((m, s) => m.set(s, (m.get(s) || 0) + 1), new Map),
wantedArray = Array.from(wantedMap.keys()),
count = 0;
seen.split(/\s+/)
.forEach(s => {
var key = wantedArray.find(t => s === t || s.length > 3 && t.length > 3 && (s.startsWith(t) || t.startsWith(s)));
if (!wantedMap.get(key)) return;
console.log(s, key)
++count;
wantedMap.set(key, wantedMap.get(key) - 1);
});
return count;
}
let matches = includeWords('i was sent to earth to protect you introduced', 'they\'re were protecting him i knew that i was aware introducing');
console.log('Matched words: ' + matches);
The code works fine, but there is still one issue:
What if we want to return a match for introduced and introducing too?
If you want the program to consider the words 'introduce' and 'introducing' as a match, it would amount to a "fuzzy" match (non binary logic). One simple way of doing this would require more code, the algorithm of which would possibly resemble
Take 2 words that you wish to match, tokenize into ordered list
of letters
Compare positionally the respective letters, i.e
match a[0]==b[0]? a[1]==b[1] where a[0] represents the first letter
of the first word and b[0] represents the first tokenized
letter/character potential match candidate
KEep a rolling numeric count of such positional matches. In this case it is 8 (introduc).
divide by word length of a = 8/9 call this f
divide by word length of b = 8/11 call this g
Provide a threshold value beyond which the program will consider it a match. eg. if you say anything above 70% in BOTH f and g can be
considered a match - viola, you have your answer!
Please note that there is some normalization also needed to prevent low length words from becoming false positives. you can add a constraint that the aforementioned calculation applies to words with at least 5 letters(or something to that effect!
Hope this helps!!
Regards,
SR
You could calculate similarites for a word pair and get a relation how many characters are similar bei respecting the length of the given word and the wanted pattern.
function getSimilarity(a, b) {
var i = 0;
while (i < a.length) {
if (a[i] !== b[i]) break;
i++;
}
return i / Math.max(a.length, b.length);
}
console.log(getSimilarity('abcdefghij', 'abc')); // 0.3
console.log(getSimilarity('abcdefghij', 'abcdef')); // 0.6
console.log(getSimilarity('abcdefghij', 'abcdefghij')); // 1
console.log(getSimilarity('abcdef', 'abcdefghij')); // 0.6
console.log(getSimilarity('abcdefghij', 'abcdef')); // 0.6
console.log(getSimilarity('abcdefghij', 'xyz')); // 0
console.log(getSimilarity('introduced', 'introducing')); // 0.7272727272727273
Here's a quick fix solution.
It's not intended as a complete solution.
Since the English language has more than a few quirks that would almost require an AI to understand the language.
First add a function that can compare 2 words and returns a boolean.
It'll also make it easier to test for specific words, and adapt to what's really needed.
For example, here's a function that does the simple checks that were already used.
Plus an '...ed' versus '...ing' check.
function compareWords (word1, word2) {
if (word1 === word2)
return true;
if (word1.length > 3 && word2.length > 3) {
if (word1.startsWith(word2) || word2.startsWith(word1))
return true;
if (word1.length > 4 && word2.length > 4) {
if (/(ing|ed)$/.test(word1) && word1.replace(/(ing|ed)$/, 'inged') === word2.replace(/(ing|ed)$/, 'inged'))
return true;
}
}
return false;
}
//
// tests
//
let words = [
["same", "same"],
["different", "unsame"],
["priced", "pricing"],
["price", "priced"],
["producing", "produced"],
["produced", "producing"]
];
words.forEach( (arr, idx) => {
let word1= arr[0];
let word2= arr[1];
let isSame = compareWords(word1, word2);
console.log(`[${word1}] ≈ [${word2}] : ${isSame}`);
});
Then use it in the code you already have.
...
seen.split(/\s+/)
.forEach(s => {
var key = wantedArray.find(t => compareWords(t, s));
...
Regarding string similarity, here's f.e. an older SO post that has some methods to compare strings : Compare Strings Javascript Return %of Likely
I have implemented this, it seems to work fine. any suggestions would be appreciated..
let speechResult = "i was sent to earth to introducing protect yourself introduced seen";
let expectSt = ['they were protecting him knew introducing that you i seen was aware seen introducing'];
// Create arrays of words from above sentences
let speechResultWords = speechResult.split(/\s+/);
let expectStWords = expectSt[0].split(/\s+/);
function includeWords(){
// Declare a variable to hold the count number of matches
let arr = [];
for(let a = 0; a < speechResultWords.length; a++){
for(let b = 0; b < expectStWords.length; b++){
if(similarity(speechResultWords[a], expectStWords[b]) > 69){
arr.push(speechResultWords[a]);
console.log(speechResultWords[a] + ' includes in ' + expectStWords[b]);
}
} // End of first for loop
} // End of second for loop
let uniq = [...new Set(arr)];
return uniq.length;
};
let result = includeWords();
console.log(result)
// The algorithmn
function similarity(s1, s2) {
var longer = s1;
var shorter = s2;
if (s1.length < s2.length) {
longer = s2;
shorter = s1;
}
var longerLength = longer.length;
if (longerLength == 0) {
return 1.0;
}
return (longerLength - editDistance(longer, shorter)) / parseFloat(longerLength)*100;
}
function editDistance(s1, s2) {
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
var costs = new Array();
for (var i = 0; i <= s1.length; i++) {
var lastValue = i;
for (var j = 0; j <= s2.length; j++) {
if (i == 0)
costs[j] = j;
else {
if (j > 0) {
var newValue = costs[j - 1];
if (s1.charAt(i - 1) != s2.charAt(j - 1))
newValue = Math.min(Math.min(newValue, lastValue),
costs[j]) + 1;
costs[j - 1] = lastValue;
lastValue = newValue;
}
}
}
if (i > 0)
costs[s2.length] = lastValue;
}
return costs[s2.length];
}
the purpose of this code is to to write a function which increments a string, to create a new string. If the string already ends with a number, the number should be incremented by 1. If the string does not end with a number the number 1 should be appended to the new string. e.g. "foo123" --> "foo124" or "foo" --> "foo1".
With my code below, pretty much all my test cases are passed except a corner case for "foo999" did not print out "foo1000". I know that there should be a way to do with regex to fix my problem, but I am not too familiar with it. Can anyone please help?
function incrementString (input) {
var reg = /[0-9]/;
var result = "";
if(reg.test(input[input.length - 1]) === true){
input = input.split("");
for(var i = 0; i < input.length; i++){
if(parseInt(input[i]) === NaN){
result += input[i];
}
else if(i === input.length - 1){
result += (parseInt(input[i]) + 1).toString();
}
else{
result += input[i];
}
}
return result;
}
else if (reg.test(input[input.length - 1]) === false){
return input += 1;
}
}
You can use replace with a callback:
'foo'.replace(/(\d*)$/, function($0, $1) { return $1*1+1; });
//=> "foo1"
'foo999'.replace(/(\d*)$/, function($0, $1) { return $1*1+1; });
//=> "foo1000"
'foo123'.replace(/(\d*)$/, function($0, $1) { return $1*1+1; });
//=> "foo124"
Explanation:
/(\d*)$/ # match 0 or more digits at the end of string
function($0, $1) {...} # callback function with 2nd parameter as matched group #1
return $1*1+1; # return captured number+1. $1*1 is a trick to convert
# string to number
The most concise way that also accounts for leading zeros, non-numeric endings, and empty strings I've seen is:
''.replace(/[0-8]?9*$/, w => ++w)
//=> 1
'foo'.replace(/[0-8]?9*$/, w => ++w)
//=> foo1
'foo099'.replace(/[0-8]?9*$/, w => ++w)
//=> foo100
'foo999'.replace(/[0-8]?9*$/, w => ++w)
//=> foo1000
function pad(number, length, filler) {
number = number + "";
if (number.length < length) {
for (var i = number.length; i < length; i += 1) {
number = filler + number;
}
}
return number;
}
function incrementString (input) {
var orig = input.match(/\d+$/);
if (orig.length === 1) {
orig = pad(parseInt(orig[0]) + 1, orig[0].length, '0');
input = input.replace(/\d+$/, orig);
return input;
}
return input + "1";
}
What does it do?
It first checks if there is a trailing number. If yes, increment it and pad left it with zeros (with the function "pad" which you'd be able to sort it out yourself).
string.replace is a function which works with argument 1 the substring to search (string, regex), argument 2 the element to replace with (string, function).
In this case I've used a regex as first argument and the incremented, padded number.
The regex is pretty simple: \d means "integer" and + means "one or more of the preceeding", which means one or more digits. $ means the end of the string.
More info about regular expressions (in JavaScript): https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/RegExp (thanks #Casimir et Hippolyte for the link)
I think you can simplify your code greatly:
function incrementString(input) {
var splits = input.split(/(\d+)$/),
num = 1;
if (splits[1] !== undefined) num = parseInt(splits[1]) + 1;
return splits[0] + num;
}
This checks for any number of digits at the end of your string.
function incrementString (strng) {
//separateing number from string
let x = (strng).replace( /^\D+/g, '');
//getting the length of original number from the string
let len = x.length;
//getting the string part from strng
str = strng.split(x);
//incrementing number by 1
let number = Number(x) + 1 + '';
//padding the number with 0 to make it's length exactly to orignal number
while(number.length < len){
number = '0' + number;
}
//new string by joining string and the incremented number
str = (str + number).split(',').join('');
//return new string
return str;
}
My quick answer will be :
let newStr = string.split('');
let word = [];
let num = [];
for (let i = 0 ; i<string.length ;i++){
isNaN(string[i])? word.push(string[i]) : num.push(string[i])
}
let l=num.length-1;
let pureNum=0;
for (let i = 0 ; i<num.length ;i++){
pureNum += num[i] * Math.pow(10,l);
l--;
}
let wordNum = (pureNum+1).toString().split('');
for (let i = wordNum.length ; i<num.length ;i++){
wordNum.unshift("0");
}
return word.join("")+wordNum.join("");
}
I wonder how to write palindrome in javascript, where I input different words and program shows if word is palindrome or not. For example word noon is palindrome, while bad is not.
Thank you in advance.
function palindrome(str) {
var len = str.length;
var mid = Math.floor(len/2);
for ( var i = 0; i < mid; i++ ) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
palindrome will return if specified word is palindrome, based on boolean value (true/false)
UPDATE:
I opened bounty on this question due to performance and I've done research and here are the results:
If we are dealing with very large amount of data like
var abc = "asdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfd";
for ( var i = 0; i < 10; i++ ) {
abc += abc; // making string even more larger
}
function reverse(s) { // using this method for second half of string to be embedded
return s.split("").reverse().join("");
}
abc += reverse(abc); // adding second half string to make string true palindrome
In this example palindrome is True, just to note
Posted palindrome function gives us time from 180 to 210 Milliseconds (in current example), and the function posted below
with string == string.split('').reverse().join('') method gives us 980 to 1010 Milliseconds.
Machine Details:
System: Ubuntu 13.10
OS Type: 32 Bit
RAM: 2 Gb
CPU: 3.4 Ghz*2
Browser: Firefox 27.0.1
Try this:
var isPalindrome = function (string) {
if (string == string.split('').reverse().join('')) {
alert(string + ' is palindrome.');
}
else {
alert(string + ' is not palindrome.');
}
}
document.getElementById('form_id').onsubmit = function() {
isPalindrome(document.getElementById('your_input').value);
}
So this script alerts the result, is it palindrome or not. You need to change the your_id with your input id and form_id with your form id to get this work.
Demo!
Use something like this
function isPalindrome(s) {
return s === s.split("").reverse().join("") ? true : false;
}
alert(isPalindrome("noon"));
alternatively the above code can be optimized as [updated after rightfold's comment]
function isPalindrome(s) {
return s === s.split("").reverse().join("");
}
alert(isPalindrome("malayalam"));
alert(isPalindrome("english"));
Faster Way:
-Compute half the way in loop.
-Store length of the word in a variable instead of calculating every time.
EDIT:
Store word length/2 in a temporary variable as not to calculate every time in the loop as pointed out by (mvw) .
function isPalindrome(word){
var i,wLength = word.length-1,wLengthToCompare = wLength/2;
for (i = 0; i <= wLengthToCompare ; i++) {
if (word.charAt(i) != word.charAt(wLength-i)) {
return false;
}
}
return true;
}
Let us start from the recursive definition of a palindrome:
The empty string '' is a palindrome
The string consisting of the character c, thus 'c', is a palindrome
If the string s is a palindrome, then the string 'c' + s + 'c' for some character c is a palindrome
This definition can be coded straight into JavaScript:
function isPalindrome(s) {
var len = s.length;
// definition clauses 1. and 2.
if (len < 2) {
return true;
}
// note: len >= 2
// definition clause 3.
if (s[0] != s[len - 1]) {
return false;
}
// note: string is of form s = 'a' + t + 'a'
// note: s.length >= 2 implies t.length >= 0
var t = s.substr(1, len - 2);
return isPalindrome(t);
}
Here is some additional test code for MongoDB's mongo JavaScript shell, in a web browser with debugger replace print() with console.log()
function test(s) {
print('isPalindrome(' + s + '): ' + isPalindrome(s));
}
test('');
test('a');
test('ab');
test('aa');
test('aab');
test('aba');
test('aaa');
test('abaa');
test('neilarmstronggnortsmralien');
test('neilarmstrongxgnortsmralien');
test('neilarmstrongxsortsmralien');
I got this output:
$ mongo palindrome.js
MongoDB shell version: 2.4.8
connecting to: test
isPalindrome(): true
isPalindrome(a): true
isPalindrome(ab): false
isPalindrome(aa): true
isPalindrome(aab): false
isPalindrome(aba): true
isPalindrome(aaa): true
isPalindrome(abaa): false
isPalindrome(neilarmstronggnortsmralien): true
isPalindrome(neilarmstrongxgnortsmralien): true
isPalindrome(neilarmstrongxsortsmralien): false
An iterative solution is:
function isPalindrome(s) {
var len = s.length;
if (len < 2) {
return true;
}
var i = 0;
var j = len - 1;
while (i < j) {
if (s[i] != s[j]) {
return false;
}
i += 1;
j -= 1;
}
return true;
}
Look at this:
function isPalindrome(word){
if(word==null || word.length==0){
// up to you if you want true or false here, don't comment saying you
// would put true, I put this check here because of
// the following i < Math.ceil(word.length/2) && i< word.length
return false;
}
var lastIndex=Math.ceil(word.length/2);
for (var i = 0; i < lastIndex && i< word.length; i++) {
if (word[i] != word[word.length-1-i]) {
return false;
}
}
return true;
}
Edit: now half operation of comparison are performed since I iterate only up to half word to compare it with the last part of the word. Faster for large data!!!
Since the string is an array of char no need to use charAt functions!!!
Reference: http://wiki.answers.com/Q/Javascript_code_for_palindrome
Taking a stab at this. Kind of hard to measure performance, though.
function palin(word) {
var i = 0,
len = word.length - 1,
max = word.length / 2 | 0;
while (i < max) {
if (word.charCodeAt(i) !== word.charCodeAt(len - i)) {
return false;
}
i += 1;
}
return true;
}
My thinking is to use charCodeAt() instead charAt() with the hope that allocating a Number instead of a String will have better perf because Strings are variable length and might be more complex to allocate. Also, only iterating halfway through (as noted by sai) because that's all that's required. Also, if the length is odd (ex: 'aba'), the middle character is always ok.
Best Way to check string is palindrome with more criteria like case and special characters...
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
You can test it with following words and strings and gives you more specific result.
1. bob
2. Doc, note, I dissent. A fast never prevents a fatness. I diet on cod
For strings it ignores special characters and convert string to lower case.
String.prototype.isPalindrome = function isPalindrome() {
const cleanString = this.toLowerCase().replace(/\s+/g, '');
const cleanStringRevers = cleanString.split("").reverse().join("");
return cleanString === cleanStringRevers;
}
let nonPalindrome = 'not a palindrome';
let palindrome = 'sugus';
console.log(nonPalindrome.isPalindrome())
console.log(palindrome.isPalindrome())
The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('#while *', i, l, '...', j, r);
--j;
break;
}
console.log('#while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('#isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.
This function will remove all non-alphanumeric characters (punctuation, spaces, and symbols) and turn everything lower case in order to check for palindromes.
function palindrome(str){
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
return str == str.split('').reverse().join('') ? true : false;
}
Here is an optimal and robust solution for checking string palindrome using ES6 features.
const str="madam"
var result=[...str].reduceRight((c,v)=>((c+v)))==str?"Palindrome":"Not Palindrome";
console.log(result);
Try this
isPalindrome = (string) => {
if (string === string.split('').reverse().join('')) {
console.log('is palindrome');
}
else {
console.log('is not palindrome');
}
}
isPalindrome(string)
Here's a one-liner without using String.reverse,
const isPal = str => [...new Array(strLen = str.length)]
.reduce((acc, s, i) => acc + str[strLen - (i + 1)], '') === str;
function palindrome(str) {
var lenMinusOne = str.length - 1;
var halfLen = Math.floor(str.length / 2);
for (var i = 0; i < halfLen; ++i) {
if (str[i] != str[lenMinusOne - i]) {
return false;
}
}
return true;
}
Optimized for half string parsing and for constant value variables.
I think following function with time complexity of o(log n) will be better.
function palindrom(s){
s = s.toString();
var f = true; l = s.length/2, len = s.length -1;
for(var i=0; i < l; i++){
if(s[i] != s[len - i]){
f = false;
break;
}
}
return f;
}
console.log(palindrom(12321));
Here's another way of doing it:
function isPalin(str) {
str = str.replace(/\W/g,'').toLowerCase();
return(str==str.split('').reverse().join(''));
}
Below code tells how to get a string from textBox and tell you whether it is a palindrome are not & displays your answer in another textbox
<html>
<head>
<meta charset="UTF-8"/>
<link rel="stylesheet" href=""/>
</head>
<body>
<h1>1234</h1>
<div id="demo">Example</div>
<a accessKey="x" href="http://www.google.com" id="com" >GooGle</a>
<h1 id="tar">"This is a Example Text..."</h1>
Number1 : <input type="text" name="txtname" id="numb"/>
Number2 : <input type="text" name="txtname2" id="numb2"/>
Number2 : <input type="text" name="txtname3" id="numb3" />
<button type="submit" id="sum" onclick="myfun()" >count</button>
<button type="button" id="so2" onclick="div()" >counnt</button><br/><br/>
<ol>
<li>water</li>
<li>Mazaa</li>
</ol><br/><br/>
<button onclick="myfun()">TryMe</button>
<script>
function myfun(){
var pass = document.getElementById("numb").value;
var rev = pass.split("").reverse().join("");
var text = document.getElementById("numb3");
text.value = rev;
if(pass === rev){
alert(pass + " is a Palindrome");
}else{
alert(pass + " is Not a Palindrome")
}
}
</script>
</body>
</html>
25x faster + recursive + non-branching + terse
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
See my complete explanation here.
The code is concise quick fast and understandable.
TL;DR
Explanation :
Here isPalindrome function accepts a str parameter which is typeof string.
If the length of the str param is less than or equal to one it simply returns "false".
If the above case is false then it moves on to the second if statement and checks that if the character at 0 position of the string is same as character at the last place. It does an inequality test between the both.
str.charAt(0) // gives us the value of character in string at position 0
str.slice(-1) // gives us the value of last character in the string.
If the inequality result is true then it goes ahead and returns false.
If result from the previous statement is false then it recursively calls the isPalindrome(str) function over and over again until the final result.
function isPalindrome(str){
if (str.length <= 1) return true;
if (str.charAt(0) != str.slice(-1)) return false;
return isPalindrome(str.substring(1,str.length-1));
};
document.getElementById('submit').addEventListener('click',function(){
var str = prompt('whats the string?');
alert(isPalindrome(str))
});
document.getElementById('ispdrm').onsubmit = function(){alert(isPalindrome(document.getElementById('inputTxt').value));
}
<!DOCTYPE html>
<html>
<body>
<form id='ispdrm'><input type="text" id="inputTxt"></form>
<button id="submit">Click me</button>
</body>
</html>
function palindrome(str) {
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
var len = str.length;
for (var i = 0; i < len/2; i++) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
Or you could do it like this.
var palindrome = word => word == word.split('').reverse().join('')
How about this one?
function pall (word) {
var lowerCWord = word.toLowerCase();
var rev = lowerCWord.split('').reverse().join('');
return rev.startsWith(lowerCWord);
}
pall('Madam');
str1 is the original string with deleted non-alphanumeric characters and spaces and str2 is the original string reversed.
function palindrome(str) {
var str1 = str.toLowerCase().replace(/\s/g, '').replace(
/[^a-zA-Z 0-9]/gi, "");
var str2 = str.toLowerCase().replace(/\s/g, '').replace(
/[^a-zA-Z 0-9]/gi, "").split("").reverse().join("");
if (str1 === str2) {
return true;
}
return false;
}
palindrome("almostomla");
function isPalindrome(s) {
return s == reverseString(s);
}
console.log((isPalindrome("abcba")));
function reverseString(str){
let finalStr=""
for(let i=str.length-1;i>=0;i--){
finalStr += str[i]
}
return finalStr
}
Frist I valid this word with converting lowercase and removing whitespace and then compare with reverse word within parameter word.
function isPalindrome(input) {
const toValid = input.trim("").toLowerCase();
const reverseWord = toValid.split("").reverse().join("");
return reverseWord == input.toLowerCase().trim() ? true : false;
}
isPalindrome(" madam ");
//true
This answer is easy to read and I tried to explain by using comment. Check the code below for How to write Palindrome in JavaScript.
Step 1: Remove all non-alphanumeric characters (punctuation, spaces and symbols) from Argument string 'str' using replace() and then convert in to lowercase using toLowerCase().
Step 2: Now make string reverse. first split the string into the array using split() then reverse the array using reverse() then make the string by joining array elements using join() .
Step 3: Find the first character of nonAlphaNumeric string using charAt(0).
Step 4: Find the Last character of nonAlphaNumeric string using charAt(length of nonAlphaNumeric string - 1).
Step 5: Use If condition to chack nonAlphaNumeric string and reverse string is same or not.
Step 6: Use another If condition to chack first character of nonAlphaNumeric string is same to Last character of nonAlphaNumeric string.
function palindrome(str) {
var nonAlphaNumericStr = str.replace(/[^0-9A-Za-z]/g, "").toLowerCase(); // output - e1y1e
var reverseStr = nonAlphaNumericStr.split("").reverse().join(""); // output - e1y1e
var firstChar = nonAlphaNumericStr.charAt(0); // output - e
var lastChar = nonAlphaNumericStr.charAt(nonAlphaNumericStr.length - 1); // output - e
if(nonAlphaNumericStr === reverseStr) {
if(firstChar === lastChar) {
return `String is Palindrome`;
}
}
return `String is not Palindrome`;
}
console.log(palindrome("_eye"));
function check(txt)
{
for (var i = txt.length; i >= 0; i--)
if (txt[i] !== txt[txt.length - 1 - i])
return console.log('not palidrome');
return console.log(' palidrome');
}
check('madam');
Note: This is case sensitive
function palindrome(word)
{
for(var i=0;i<word.length/2;i++)
if(word.charAt(i)!=word.charAt(word.length-(i+1)))
return word+" is Not a Palindrome";
return word+" is Palindrome";
}
Here is the fiddle: http://jsfiddle.net/eJx4v/
I am not sure how this JSPerf check the code performance. I just tried to reverse the string & check the values. Please comment about the Pros & Cons of this method.
function palindrome(str) {
var re = str.split(''),
reArr = re.slice(0).reverse();
for (a = 0; a < re.length; a++) {
if (re[a] == reArr[a]) {
return false;
} else {
return true;
}
}
}
JS Perf test
See my code snippet below:
var list = ['one', 'two', 'three', 'four'];
var str = 'one two, one three, one four, one';
for ( var i = 0; i < list.length; i++)
{
if (str.endsWith(list[i])
{
str = str.replace(list[i], 'finish')
}
}
I want to replace the last occurrence of the word one with the word finish in the string, what I have will not work because the replace method will only replace the first occurrence of it. Does anyone know how I can amend that snippet so that it only replaces the last instance of 'one'
Well, if the string really ends with the pattern, you could do this:
str = str.replace(new RegExp(list[i] + '$'), 'finish');
You can use String#lastIndexOf to find the last occurrence of the word, and then String#substring and concatenation to build the replacement string.
n = str.lastIndexOf(list[i]);
if (n >= 0 && n + list[i].length >= str.length) {
str = str.substring(0, n) + "finish";
}
...or along those lines.
Not as elegant as the regex answers above, but easier to follow for the not-as-savvy among us:
function removeLastInstance(badtext, str) {
var charpos = str.lastIndexOf(badtext);
if (charpos<0) return str;
ptone = str.substring(0,charpos);
pttwo = str.substring(charpos+(badtext.length));
return (ptone+pttwo);
}
I realize this is likely slower and more wasteful than the regex examples, but I think it might be helpful as an illustration of how string manipulations can be done. (It can also be condensed a bit, but again, I wanted each step to be clear.)
Here's a method that only uses splitting and joining. It's a little more readable so thought it was worth sharing:
String.prototype.replaceLast = function (what, replacement) {
var pcs = this.split(what);
var lastPc = pcs.pop();
return pcs.join(what) + replacement + lastPc;
};
Thought I'd answer here since this came up first in my Google search and there's no answer (outside of Matt's creative answer :)) that generically replaces the last occurrence of a string of characters when the text to replace might not be at the end of the string.
if (!String.prototype.replaceLast) {
String.prototype.replaceLast = function(find, replace) {
var index = this.lastIndexOf(find);
if (index >= 0) {
return this.substring(0, index) + replace + this.substring(index + find.length);
}
return this.toString();
};
}
var str = 'one two, one three, one four, one';
// outputs: one two, one three, one four, finish
console.log(str.replaceLast('one', 'finish'));
// outputs: one two, one three, one four; one
console.log(str.replaceLast(',', ';'));
A simple answer without any regex would be:
str = str.substr(0, str.lastIndexOf(list[i])) + 'finish'
I did not like any of the answers above and came up with the below
function isString(variable) {
return typeof (variable) === 'string';
}
function replaceLastOccurrenceInString(input, find, replaceWith) {
if (!isString(input) || !isString(find) || !isString(replaceWith)) {
// returns input on invalid arguments
return input;
}
const lastIndex = input.lastIndexOf(find);
if (lastIndex < 0) {
return input;
}
return input.substr(0, lastIndex) + replaceWith + input.substr(lastIndex + find.length);
}
Usage:
const input = 'ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty';
const find = 'teen';
const replaceWith = 'teenhundred';
const output = replaceLastOccurrenceInString(input, find, replaceWith);
console.log(output);
// output: ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteenhundred twenty
Hope that helps!
Couldn't you just reverse the string and replace only the first occurrence of the reversed search pattern? I'm thinking . . .
var list = ['one', 'two', 'three', 'four'];
var str = 'one two, one three, one four, one';
for ( var i = 0; i < list.length; i++)
{
if (str.endsWith(list[i])
{
var reversedHaystack = str.split('').reverse().join('');
var reversedNeedle = list[i].split('').reverse().join('');
reversedHaystack = reversedHaystack.replace(reversedNeedle, 'hsinif');
str = reversedHaystack.split('').reverse().join('');
}
}
If speed is important, use this:
/**
* Replace last occurrence of a string with another string
* x - the initial string
* y - string to replace
* z - string that will replace
*/
function replaceLast(x, y, z){
var a = x.split("");
var length = y.length;
if(x.lastIndexOf(y) != -1) {
for(var i = x.lastIndexOf(y); i < x.lastIndexOf(y) + length; i++) {
if(i == x.lastIndexOf(y)) {
a[i] = z;
}
else {
delete a[i];
}
}
}
return a.join("");
}
It's faster than using RegExp.
Simple solution would be to use substring method.
Since string is ending with list element, we can use string.length and calculate end index for substring without using lastIndexOf method
str = str.substring(0, str.length - list[i].length) + "finish"
function replaceLast(text, searchValue, replaceValue) {
const lastOccurrenceIndex = text.lastIndexOf(searchValue)
return `${
text.slice(0, lastOccurrenceIndex)
}${
replaceValue
}${
text.slice(lastOccurrenceIndex + searchValue.length)
}`
}
A negative lookahead solution:
str.replace(/(one)(?!.*\1)/, 'finish')
An explanation provided by the site regex101.com,
/(one)(?!.*\1)/
1st Capturing Group (one)
one - matches the characters one literally (case sensitive)
Negative Lookahead (?!.*\1)
Assert that the Regex below does not match
. matches any character (except for line terminators)
* matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\1 matches the same text as most recently matched by the 1st capturing group
Old fashioned and big code but efficient as possible:
function replaceLast(origin,text){
textLenght = text.length;
originLen = origin.length
if(textLenght == 0)
return origin;
start = originLen-textLenght;
if(start < 0){
return origin;
}
if(start == 0){
return "";
}
for(i = start; i >= 0; i--){
k = 0;
while(origin[i+k] == text[k]){
k++
if(k == textLenght)
break;
}
if(k == textLenght)
break;
}
//not founded
if(k != textLenght)
return origin;
//founded and i starts on correct and i+k is the first char after
end = origin.substring(i+k,originLen);
if(i == 0)
return end;
else{
start = origin.substring(0,i)
return (start + end);
}
}
I would suggest using the replace-last npm package.
var str = 'one two, one three, one four, one';
var result = replaceLast(str, 'one', 'finish');
console.log(result);
<script src="https://unpkg.com/replace-last#latest/replaceLast.js"></script>
This works for string and regex replacements.
if (string.search(searchstring)>-1) {
stringnew=((text.split("").reverse().join("")).replace(searchstring,
subststring).split("").reverse().join(""))
}
//with var string= "sdgu()ert(dhfj ) he ) gfrt"
//var searchstring="f"
//var subststring="X"
//var stringnew=""
//results in
//string : sdgu()ert(dhfj ) he ) gfrt
//stringnew : sdgu()ert(dhfj ) he ) gXrt
str = (str + '?').replace(list[i] + '?', 'finish');
In Perl I can repeat a character multiple times using the syntax:
$a = "a" x 10; // results in "aaaaaaaaaa"
Is there a simple way to accomplish this in Javascript? I can obviously use a function, but I was wondering if there was any built in approach, or some other clever technique.
These days, the repeat string method is implemented almost everywhere. (It is not in Internet Explorer.) So unless you need to support older browsers, you can simply write:
"a".repeat(10)
Before repeat, we used this hack:
Array(11).join("a") // create string with 10 a's: "aaaaaaaaaa"
(Note that an array of length 11 gets you only 10 "a"s, since Array.join puts the argument between the array elements.)
Simon also points out that according to this benchmark, it appears that it's faster in Safari and Chrome (but not Firefox) to repeat a character multiple times by simply appending using a for loop (although a bit less concise).
In a new ES6 harmony, you will have native way for doing this with repeat. Also ES6 right now only experimental, this feature is already available in Edge, FF, Chrome and Safari
"abc".repeat(3) // "abcabcabc"
And surely if repeat function is not available you can use old-good Array(n + 1).join("abc")
Convenient if you repeat yourself a lot:
String.prototype.repeat = String.prototype.repeat || function(n){
n= n || 1;
return Array(n+1).join(this);
}
alert( 'Are we there yet?\nNo.\n'.repeat(10) )
Array(10).fill('a').join('')
Although the most voted answer is a bit more compact, with this approach you don't have to add an extra array item.
An alternative is:
for(var word = ''; word.length < 10; word += 'a'){}
If you need to repeat multiple chars, multiply your conditional:
for(var word = ''; word.length < 10 * 3; word += 'foo'){}
NOTE: You do not have to overshoot by 1 as with word = Array(11).join('a')
The most performance-wice way is https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
Short version is below.
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>>= 1, pattern += pattern;
}
return result + pattern;
};
var a = "a";
console.debug(a.repeat(10));
Polyfill from Mozilla:
if (!String.prototype.repeat) {
String.prototype.repeat = function(count) {
'use strict';
if (this == null) {
throw new TypeError('can\'t convert ' + this + ' to object');
}
var str = '' + this;
count = +count;
if (count != count) {
count = 0;
}
if (count < 0) {
throw new RangeError('repeat count must be non-negative');
}
if (count == Infinity) {
throw new RangeError('repeat count must be less than infinity');
}
count = Math.floor(count);
if (str.length == 0 || count == 0) {
return '';
}
// Ensuring count is a 31-bit integer allows us to heavily optimize the
// main part. But anyway, most current (August 2014) browsers can't handle
// strings 1 << 28 chars or longer, so:
if (str.length * count >= 1 << 28) {
throw new RangeError('repeat count must not overflow maximum string size');
}
var rpt = '';
for (;;) {
if ((count & 1) == 1) {
rpt += str;
}
count >>>= 1;
if (count == 0) {
break;
}
str += str;
}
// Could we try:
// return Array(count + 1).join(this);
return rpt;
}
}
If you're not opposed to including a library in your project, lodash has a repeat function.
_.repeat('*', 3);
// → '***
https://lodash.com/docs#repeat
For all browsers
The following function will perform a lot faster than the option suggested in the accepted answer:
var repeat = function(str, count) {
var array = [];
for(var i = 0; i < count;)
array[i++] = str;
return array.join('');
}
You'd use it like this :
var repeatedString = repeat("a", 10);
To compare the performance of this function with that of the option proposed in the accepted answer, see this Fiddle and this Fiddle for benchmarks.
For moderns browsers only
In modern browsers, you can now do this using String.prototype.repeat method:
var repeatedString = "a".repeat(10);
Read more about this method on MDN.
This option is even faster. Unfortunately, it doesn't work in any version of Internet explorer. The numbers in the table specify the first browser version that fully supports the method:
In ES2015/ES6 you can use "*".repeat(n)
So just add this to your projects, and your are good to go.
String.prototype.repeat = String.prototype.repeat ||
function(n) {
if (n < 0) throw new RangeError("invalid count value");
if (n == 0) return "";
return new Array(n + 1).join(this.toString())
};
String.repeat() is supported by 96.39% of browsers as of now.
function pad(text, maxLength){
return text + "0".repeat(maxLength - text.length);
}
console.log(pad('text', 7)); //text000
/**
* Repeat a string `n`-times (recursive)
* #param {String} s - The string you want to repeat.
* #param {Number} n - The times to repeat the string.
* #param {String} d - A delimiter between each string.
*/
var repeat = function (s, n, d) {
return --n ? s + (d || "") + repeat(s, n, d) : "" + s;
};
var foo = "foo";
console.log(
"%s\n%s\n%s\n%s",
repeat(foo), // "foo"
repeat(foo, 2), // "foofoo"
repeat(foo, "2"), // "foofoo"
repeat(foo, 2, "-") // "foo-foo"
);
Just for the fun of it, here is another way by using the toFixed(), used to format floating point numbers.
By doing
(0).toFixed(2)
(0).toFixed(3)
(0).toFixed(4)
we get
0.00
0.000
0.0000
If the first two characters 0. are deleted, we can use this repeating pattern to generate any repetition.
function repeat(str, nTimes) {
return (0).toFixed(nTimes).substr(2).replaceAll('0', str);
}
console.info(repeat('3', 5));
console.info(repeat('hello ', 4));
Another interesting way to quickly repeat n character is to use idea from quick exponentiation algorithm:
var repeatString = function(string, n) {
var result = '', i;
for (i = 1; i <= n; i *= 2) {
if ((n & i) === i) {
result += string;
}
string = string + string;
}
return result;
};
For repeat a value in my projects i use repeat
For example:
var n = 6;
for (i = 0; i < n; i++) {
console.log("#".repeat(i+1))
}
but be careful because this method has been added to the ECMAScript 6 specification.
function repeatString(n, string) {
var repeat = [];
repeat.length = n + 1;
return repeat.join(string);
}
repeatString(3,'x'); // => xxx
repeatString(10,'🌹'); // => "🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹"
This is how you can call a function and get the result by the helps of Array() and join()
using Typescript and arrow fun
const repeatString = (str: string, num: number) => num > 0 ?
Array(num+1).join(str) : "";
console.log(repeatString("🌷",10))
//outputs: 🌷🌷🌷🌷🌷🌷🌷🌷🌷🌷
function repeatString(str, num) {
// Array(num+1) is the string you want to repeat and the times to repeat the string
return num > 0 ? Array(num+1).join(str) : "";
}
console.log(repeatString("a",10))
// outputs: aaaaaaaaaa
console.log(repeatString("🌷",10))
//outputs: 🌷🌷🌷🌷🌷🌷🌷🌷🌷🌷
Here is what I use:
function repeat(str, num) {
var holder = [];
for(var i=0; i<num; i++) {
holder.push(str);
}
return holder.join('');
}
I realize that it's not a popular task, what if you need to repeat your string not an integer number of times?
It's possible with repeat() and slice(), here's how:
String.prototype.fracRepeat = function(n){
if(n < 0) n = 0;
var n_int = ~~n; // amount of whole times to repeat
var n_frac = n - n_int; // amount of fraction times (e.g., 0.5)
var frac_length = ~~(n_frac * this.length); // length in characters of fraction part, floored
return this.repeat(n) + this.slice(0, frac_length);
}
And below a shortened version:
String.prototype.fracRepeat = function(n){
if(n < 0) n = 0;
return this.repeat(n) + this.slice(0, ~~((n - ~~n) * this.length));
}
var s = "abcd";
console.log(s.fracRepeat(2.5))
I'm going to expand on #bonbon's answer. His method is an easy way to "append N chars to an existing string", just in case anyone needs to do that. For example since "a google" is a 1 followed by 100 zeros.
for(var google = '1'; google.length < 1 + 100; google += '0'){}
document.getElementById('el').innerText = google;
<div>This is "a google":</div>
<div id="el"></div>
NOTE: You do have to add the length of the original string to the conditional.
Lodash offers a similar functionality as the Javascript repeat() function which is not available in all browers. It is called _.repeat and available since version 3.0.0:
_.repeat('a', 10);
var stringRepeat = function(string, val) {
var newString = [];
for(var i = 0; i < val; i++) {
newString.push(string);
}
return newString.join('');
}
var repeatedString = stringRepeat("a", 1);
Can be used as a one-liner too:
function repeat(str, len) {
while (str.length < len) str += str.substr(0, len-str.length);
return str;
}
In CoffeeScript:
( 'a' for dot in [0..10]).join('')
String.prototype.repeat = function (n) { n = Math.abs(n) || 1; return Array(n + 1).join(this || ''); };
// console.log("0".repeat(3) , "0".repeat(-3))
// return: "000" "000"