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Replace a letter with its alphabet position

This looked fairly straightforward to me when I started, but for some reason I'm getting an empty array everytime I try to run the result on codewars. I'm hoping you can help me identify what the problem is.
function alphabetPosition(text) {
text.split(' ').join('');
var chari = "";
var arr = [];
var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
for(var i = 0; i < text.len; i++){
chari = text.charAt(i).toLowerCase();
if(alphabet.indexOf(chari) > -1){
arr.push(alphabet.indexOf(chari));
}
}
return arr;
}
console.log(alphabetPosition("Hello World"));
My idea is to get the text from the parameter then strip out the spaces. I made a variable for my empty array and make an alphabet string that I can later search through. In the for loop, i make each character lowercase, and if the character is found in the alphabet string, its position gets pushed into the array (arr). I appreciate your time.

The Kata works with this code. Try with this one:
function alphabetPosition(text) {
var result = "";
for (var i = 0; i < text.length; i++) {
var code = text.toUpperCase().charCodeAt(i)
if (code > 64 && code < 91) result += (code - 64) + " ";
}
return result.slice(0, result.length - 1);
}
console.log(alphabetPosition("The sunset sets at twelve o' clock."));

You need the String#length property
text.length
instead of text.len.
function alphabetPosition(text) {
var chari,
arr = [],
alphabet = "abcdefghijklmnopqrstuvwxyz",
i;
for (var i = 0; i < text.length; i++){
chari = text[i].toLowerCase();
if (alphabet.indexOf(chari) !== -1){
arr.push(alphabet.indexOf(chari));
}
}
return arr;
}
console.log(alphabetPosition("Hello World!!1"));
A solution with ES6
function alphabetPosition(text) {
return [...text].map(a => parseInt(a, 36) - 10).filter(a => a >= 0);
}
console.log(alphabetPosition("Hello World!!1"));

First : deleting space
Second : mapping each char with its alphabet rank
Third : test with the string Happy new year
var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
var alphabetPosition = text =>
text.split('').map(x => alphabet.indexOf(x) + 1);
console.log(alphabetPosition("happy new year"));

function alphabetPosition(text) {
const words = text.toLowerCase().replace(/[^a-z]/g,"");
return [...words].map(v=> v.charCodeAt() - 96);
}
First we take the text and transform it into lowercase to get rid of the capital letters using text.toLowerCase() and then we do .replace(/[^a-z]/g,"") to replace all the non a-z characters with nothing.
The next step is to spread the string out into an array using [...words] and then mapping it to get the ascii character code of each a-z character.
Since a = 97 and b = 98 etc we will subtract 96 so that we get a = 1 and b = 2 etc (the position of the letters in the alphabet)

You can make it even easier, by making use of the acii code. Because a = ascii code 97, b = 98 etc. And there is a javascript function String.charCodeAt( n ) which returns the ascii code at a specific function. You only have to alter the offset for capitals (if you want to support them).
function alphabetPosition( text ) {
var positions = [];
for ( var i = 0; i < text.length; i++ ) {
var charCode = text.charCodeAt( i );
if ( charCode >= 97 && charCode <= 122 ) {
positions.push( charCode - 96 );
} else if ( charCode >= 65 && charCode <= 90 ) { // get rid of this if you don't care about capitals
positions.push( charCode - 64 );
}
}
return positions;
}
var positions = alphabetPosition( 'Hello World' );
console.log(positions);
Checkout this working fiddle

This example will return based on a 0 based array, and uses lambda expressions with filter. I recycle the original byte array created by splitting the text passed to the method.
function alphabetPosition(text) {
var bytes = text.split('');
var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
for (var i = 0, len = text.length; i < len; i++) {
bytes[i] = alphabet.indexOf(bytes[i].toLowerCase());
}
return bytes.filter(n => { if(n > -1) return n; } ).join(' ');
}
console.log(alphabetPosition("Hello World"));
For a 1 based array result Kata Codewars Friendly
function alphabetPosition(text) {
var bytes = text.split('');
var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
for (var i = 0, len = text.length; i < len; i++) {
bytes[i] = alphabet.indexOf(bytes[i].toLowerCase()) + 1;
}
return bytes.filter(n => { if(n > 0) return n; } ).join(' ');
}
console.log(alphabetPosition("Hello World"));

You can also use .charCodeAt function:
function alphabetPosition(text) {
start = "a".charCodeAt(0);
end = "z".charCodeAt(0);
res = [];
for (var i = 0; i < text.length; i++) {
index = text.charAt(i).toLowerCase().charCodeAt(0);
if (index >= start && index <= end) {
res.push(index - start +1); // +1 assuming a is 1 instead of 0
} else {
res.push(0); // Or do w/e you like with non-characters
}
}
return res;
}
console.log(alphabetPosition("Hello World"));

You may do something like this;
var alpha = [].reduce.call("abcdefghijklmnopqrstuvwxyz0123456789 .,!",(p,c,i) => (p[c] = i,p),{}),
str = "Happy 2017 whatever..!",
coded = [].map.call(str, c => alpha[c.toLowerCase()]);
console.log(coded);

change len to length:
(var i = 0; i < text.len; i++)
// change to
(var i = 0; i < text.length; i++)

Here is a much shorter version that does the same:
function alphabetPosition(text){
return text.split('').map(function(character){ return character.charCodeAt(0) - 'a'.charCodeAt(0) + 1; })
}
console.log(alphabetPosition("Hello World"));

This was my solution on CodeWars. Works perfectly ;)
let alphabetPosition = (text) => {
let str = Array.from(text.toLowerCase().replace(/[^a-z]/g,''));
let arr = [];
for (let i = 0; i < str.length; i++) {
arr.push(str[i].charCodeAt() - 96);
}
return arr.join(' ');
}

my answer
function alphabetPosition(text) {
if (text.match(/[a-z]/gi)) { // if text is not emtpty
let justAlphabet = text.match(/[a-z]/gi).join("").toLowerCase(); //first extract characters other than letters
let alphabet = "abcdefghijklmnopqrstuvwxyz";
let a = []; // Create an empty array
for (let i = 0; i < justAlphabet.length; i++) {
a.push(alphabet.indexOf(justAlphabet[i]) + 1); //and characters index number push in this array
}
return a.join(" ");
} else {
return "";
}
}
console.log(alphabetPosition("The sunset sets at twelve o' clock."));

this one should work
const lineNumberHandler = (arr) => {
const alphabet = 'abcdefghijklmnopqrstuwxyz';
return arr.map((el) => `${alphabet.indexOf(el) + 1}:${el}`);
}

This may solve the issue too:
function alphabetPosition(text) {
const alphabet = 'abcdefghijklmnopqrstuvwxyz'
const textWithoutSpaces = text.replace(/\s/g, '');
const numbers = Array.from(textWithoutSpaces).map((letter) => {
const lowerCaseLetter = letter.toLowerCase();
const charI = alphabet.indexOf(lowerCaseLetter)+1
if(charI) return charI
return null
})
const withoutEmptyValues = numbers.filter(Boolean)
return withoutEmptyValues.join(' ')
}

function alphabetPosition(text) {
const letters = text.split('')
const result = letters.map((l, ix) => {
const code = l.toUpperCase().charCodeAt() - 64
if (l.length && code > 0 && code <= 26) {
return code
}
})
return result.join(' ').replace(/\s+/g, ' ').trim()
}
I like to do "Single Line Responsibility", so in every line you can find only one action.
The return line, delete all multiples spaces.

An optimized version of it.
function alphabetPosition(text) {
let split = text.split("");
split = split.filter((el) => /^[a-zA-Z]+$/.test(el));
let str = "";
split.forEach(
(el) => (str += el.toLowerCase().charCodeAt(0) - "96" + " ")
);
return str.trim();
}
console.log(alphabetPosition("The sunset sets at twelve o' clock."));

function alphaPosition(test){
//the object assigns the position value to every alphabet property
const lookupAlph = {a:1,b:2,c:3,d:4,e:5,f:6,g:7,h:8,i:9,j:10,k:11,l:12,m:13,n:14,o:15,p:16,q:17,r:18,s:19,t:20,u:21,v:22,w:23,x:24,y:25,z:26}
//converts our text first to lowercase, then removes every non-alphabet
// it proceeds to covert the text to an array,
// then we use the map method to convert it to its position in the alphabet
// the last thing we do is convert our array back to a string using the join method
const alphaPosition = text.toLowerCase()
.replace(/[^a-z]/g, "")
.split("")
.map((a) => lookupAlph[a])
.join(" ")
return alphaPosition;
}

Trying to square every digit of a number, but my algorithm is not working?

I'm trying to square every digit of a number;
for example:
123 should return 149
983 --> 81649 and so on
I messed it up somewhere in the following Javascript code and I'm looking for some guidance.
function splitNumber(num){
var arr = [];
while(num>0){
var c = num%10;
arr[arr.length]=c;
num=num/10;}
return arr;
}
function squareArrToNumber(arr){
var c = 0;
for(var i=arr.length-1;i>=0;i--){
arr[i]=arr[i]^2;
if(arr[i]^2>10)
c = c*100+arr[i];
else
c = c*10+arr[i];
}
return c;
}
function squareDigits(num){
squareArrToNumber(splitNumber(num));
}

Try out this code
function numToSqr(num){
var i, sqr=[],n;
num = num.toString();
for(i=0;i<num.length;i++){
n = Number(num[i]);
sqr.push(n*n);
}
return Number(sqr.join(""));
}

There are multiple things wrong with your code, starting with an overcomplication to split a string of numbers into its consituant characters just use .splt(""):
var str = "123";
var arr = str.split("");
for(var i = 0;i<arr.length;i++)
alert(arr[i]);
Next, the code num ^ 2 does not square a number. To do a square, simply multiply a number by itself (num * num)
This leaves us with a rather simple solution
function splitNumber(num){
return num.split("");
}
function joinArray(arr){
return arr.join("");
}
function squareArrToNumber(arr){
var newArr = [];
for(var i=0;i<arr.length;i++){
newArr.push(arr[i] * arr[i]);
}
return joinArray(newArr);
}
function squareDigits(num){
return squareArrToNumber(splitNumber(num));
}
alert(squareDigits("123"));
alert(squareDigits("983"));

Here is how I would do such a thing:
var num = 123;
var numArray = num.toString().split("");
var result = "";
for(var i = 0; i < numArray.length; i++){
result += parseInt(numArray[i]) * parseInt(numArray[i]);
}

function squareNum(number) {
var array = [];
// Split number into an array of numbers that make it up
array = String(number).split('');
for (let i = 0; i < array.length; i++) {
// Take each number in that array and square it (in place)
// Also can be done with forEach depending on what es version you're targetting
array[i] = Math.pow(array[i], 2);
}
// combine and return the elements of the array
return Number(array.join(''));
}
squareNum(123);
squareNum(983);

try this example
function squareDigits(n) {
return +(n.toString().split('').map(val => val * val).join(''));
}
console.log(squareDigits(4444));
here + sign is convert the string into an integer.

if(arr[i]^2>10)
should be
if(arr[i]>10)
And, as #Luaan noted, it should be
arr[i] *= arr[i]

Showing unique characters in a string only once

I have a string with repeated letters. I want letters that are repeated more than once to show only once.
Example input: aaabbbccc
Expected output: abc
I've tried to create the code myself, but so far my function has the following problems:
if the letter doesn't repeat, it's not shown (it should be)
if it's repeated once, it's show only once (i.e. aa shows a - correct)
if it's repeated twice, shows all (i.e. aaa shows aaa - should be a)
if it's repeated 3 times, it shows 6 (if aaaa it shows aaaaaa - should be a)
function unique_char(string) {
var unique = '';
var count = 0;
for (var i = 0; i < string.length; i++) {
for (var j = i+1; j < string.length; j++) {
if (string[i] == string[j]) {
count++;
unique += string[i];
}
}
}
return unique;
}
document.write(unique_char('aaabbbccc'));
The function must be with loop inside a loop; that's why the second for is inside the first.

Fill a Set with the characters and concatenate its unique entries:
function unique(str) {
return String.prototype.concat.call(...new Set(str));
}
console.log(unique('abc')); // "abc"
console.log(unique('abcabc')); // "abc"

Convert it to an array first, then use Josh Mc’s answer at How to get unique values in an array, and rejoin, like so:
var nonUnique = "ababdefegg";
var unique = Array.from(nonUnique).filter(function(item, i, ar){ return ar.indexOf(item) === i; }).join('');
All in one line. :-)

Too late may be but still my version of answer to this post:
function extractUniqCharacters(str){
var temp = {};
for(var oindex=0;oindex<str.length;oindex++){
temp[str.charAt(oindex)] = 0; //Assign any value
}
return Object.keys(temp).join("");
}

You can use a regular expression with a custom replacement function:
function unique_char(string) {
return string.replace(/(.)\1*/g, function(sequence, char) {
if (sequence.length == 1) // if the letter doesn't repeat
return ""; // its not shown
if (sequence.length == 2) // if its repeated once
return char; // its show only once (if aa shows a)
if (sequence.length == 3) // if its repeated twice
return sequence; // shows all(if aaa shows aaa)
if (sequence.length == 4) // if its repeated 3 times
return Array(7).join(char); // it shows 6( if aaaa shows aaaaaa)
// else ???
return sequence;
});
}

Using lodash:
_.uniq('aaabbbccc').join(''); // gives 'abc'

Per the actual question: "if the letter doesn't repeat its not shown"
function unique_char(str)
{
var obj = new Object();
for (var i = 0; i < str.length; i++)
{
var chr = str[i];
if (chr in obj)
{
obj[chr] += 1;
}
else
{
obj[chr] = 1;
}
}
var multiples = [];
for (key in obj)
{
// Remove this test if you just want unique chars
// But still keep the multiples.push(key)
if (obj[key] > 1)
{
multiples.push(key);
}
}
return multiples.join("");
}
var str = "aaabbbccc";
document.write(unique_char(str));

Your problem is that you are adding to unique every time you find the character in string. Really you should probably do something like this (since you specified the answer must be a nested for loop):
function unique_char(string){
var str_length=string.length;
var unique='';
for(var i=0; i<str_length; i++){
var foundIt = false;
for(var j=0; j<unique.length; j++){
if(string[i]==unique[j]){
foundIt = true;
break;
}
}
if(!foundIt){
unique+=string[i];
}
}
return unique;
}
document.write( unique_char('aaabbbccc'))
In this we only add the character found in string to unique if it isn't already there. This is really not an efficient way to do this at all ... but based on your requirements it should work.
I can't run this since I don't have anything handy to run JavaScript in ... but the theory in this method should work.

Try this if duplicate characters have to be displayed once, i.e.,
for i/p: aaabbbccc o/p: abc
var str="aaabbbccc";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 ){
return obj;
}
}
).join("");
//output: "abc"
And try this if only unique characters(String Bombarding Algo) have to be displayed, add another "and" condition to remove the characters which came more than once and display only unique characters, i.e.,
for i/p: aabbbkaha o/p: kh
var str="aabbbkaha";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){ // another and condition
return obj;
}
}
).join("");
//output: "kh"

<script>
uniqueString = "";
alert("Displays the number of a specific character in user entered string and then finds the number of unique characters:");
function countChar(testString, lookFor) {
var charCounter = 0;
document.write("Looking at this string:<br>");
for (pos = 0; pos < testString.length; pos++) {
if (testString.charAt(pos) == lookFor) {
charCounter += 1;
document.write("<B>" + lookFor + "</B>");
} else
document.write(testString.charAt(pos));
}
document.write("<br><br>");
return charCounter;
}
function findNumberOfUniqueChar(testString) {
var numChar = 0,
uniqueChar = 0;
for (pos = 0; pos < testString.length; pos++) {
var newLookFor = "";
for (pos2 = 0; pos2 <= pos; pos2++) {
if (testString.charAt(pos) == testString.charAt(pos2)) {
numChar += 1;
}
}
if (numChar == 1) {
uniqueChar += 1;
uniqueString = uniqueString + " " + testString.charAt(pos)
}
numChar = 0;
}
return uniqueChar;
}
var testString = prompt("Give me a string of characters to check", "");
var lookFor = "startvalue";
while (lookFor.length > 1) {
if (lookFor != "startvalue")
alert("Please select only one character");
lookFor = prompt(testString + "\n\nWhat should character should I look for?", "");
}
document.write("I found " + countChar(testString, lookFor) + " of the<b> " + lookFor + "</B> character");
document.write("<br><br>I counted the following " + findNumberOfUniqueChar(testString) + " unique character(s):");
document.write("<br>" + uniqueString)
</script>

Here is the simplest function to do that
function remove(text)
{
var unique= "";
for(var i = 0; i < text.length; i++)
{
if(unique.indexOf(text.charAt(i)) < 0)
{
unique += text.charAt(i);
}
}
return unique;
}

The one line solution will be to use Set. const chars = [...new Set(s.split(''))];

If you want to return values in an array, you can use this function below.
const getUniqueChar = (str) => Array.from(str)
.filter((item, index, arr) => arr.slice(index + 1).indexOf(item) === -1);
console.log(getUniqueChar("aaabbbccc"));
Alternatively, you can use the Set constructor.
const getUniqueChar = (str) => new Set(str);
console.log(getUniqueChar("aaabbbccc"));

Here is the simplest function to do that pt. 2
const showUniqChars = (text) => {
let uniqChars = "";
for (const char of text) {
if (!uniqChars.includes(char))
uniqChars += char;
}
return uniqChars;
};

const countUnique = (s1, s2) => new Set(s1 + s2).size
a shorter way based on #le_m answer

let unique=myArray.filter((item,index,array)=>array.indexOf(item)===index)

Getting a nth element from a string

How can I get a certain nth element from a string. Like if I want to get every 3rd element from the word GOOGLE how can i do that. SO far i've done this but i dont know what to type after the If
function create_string( string ) {
var string_length=string.length;
var new_string=[];
for( var i=0; i<string_length; i++) {
if(string[i]%3==0) {
}
new_string.push(string[i]);
}
return new_string;
}

Use the charAt() function of String which returns the char at a specific index passed to the function. Using charAt, I have created a script that will return every third character.
var result = "";
for(var i = 2; i < test.length; i+=3){
result += test.charAt(i);
}
If you would like to turn this script into a more reusable function:
var test = "GOOGLE";
function getEveryNthChar(n, str){
var result = "";
for(var i = (n-1); i < test.length; i+=n){
result += str.charAt(i);
}
return result;
}
alert(getEveryNthChar(1,test));
alert(getEveryNthChar(2,test));
alert(getEveryNthChar(3,test));
alert(getEveryNthChar(4,test));
Working Demo: http://jsfiddle.net/Q7Lx2/
Documentation

How about this?
function create_string( string ) {
var string_length=string.length;
var new_string=[];
for( var i=2; i<string_length; i+=3) { // instead of an if, use +=3
new_string.push(string.charAt(i));
}
return new_string.join(""); // turn your array back into a string
}
Note that if you start making this compact, you'll end up with the same answer as Kevin's ;-)
function create_string( s ) {
var new_string = '';
for( var i=2; i<s.length; i+=3) { // instead of an if, use +=3
new_string += s.charAt(i);
}
return new_string;
}

Here's a function that will work for any number, not just 3:
function stringHop(s, n) {
var result = "";
for (var i = 0; i < s.length; i+= n) {
result += s.charAt(i);
}
return result;
}
var foo = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
var bar = stringHop(foo, 2); // returns "ACEGIKMOQSUWY"
var baz = stringHop(foo, 3); // returns "ADGJMPSVY"

String.charAt(index) will return the character at the specified index, from 0 to String.length - 1. So:
String.prototype.every = function(n) {
var out = '';
for (var i = 0; i < this.length; i += n) {
out += this.charAt(i);
}
return out;
}
var str = "GOOGLE";
console.log(str.every(3)) // Outputs: GG
If you don't want to include the first character, then change the for loop to:
for (var i = n - 1; i < this.length; i += n) {

Find the characters in a string which are not duplicated

I have to make a function in JavaScript that removes all duplicated letters in a string. So far I've been able to do this: If I have the word "anaconda" it shows me as a result "anaconda" when it should show "cod". Here is my code:
function find_unique_characters( string ){
var unique='';
for(var i=0; i<string.length; i++){
if(unique.indexOf(string[i])==-1){
unique += string[i];
}
}
return unique;
}
console.log(find_unique_characters('baraban'));

We can also now clean things up using filter method:
function removeDuplicateCharacters(string) {
return string
.split('')
.filter(function(item, pos, self) {
return self.indexOf(item) == pos;
})
.join('');
}
console.log(removeDuplicateCharacters('baraban'));
Working example:

function find_unique_characters(str) {
var unique = '';
for (var i = 0; i < str.length; i++) {
if (str.lastIndexOf(str[i]) == str.indexOf(str[i])) {
unique += str[i];
}
}
return unique;
}
console.log(find_unique_characters('baraban'));
console.log(find_unique_characters('anaconda'));
If you only want to return characters that appear occur once in a string, check if their last occurrence is at the same position as their first occurrence.
Your code was returning all characters in the string at least once, instead of only returning characters that occur no more than once. but obviously you know that already, otherwise there wouldn't be a question ;-)

Just wanted to add my solution for fun:
function removeDoubles(string) {
var mapping = {};
var newString = '';
for (var i = 0; i < string.length; i++) {
if (!(string[i] in mapping)) {
newString += string[i];
mapping[string[i]] = true;
}
}
return newString;
}

With lodash:
_.uniq('baraban').join(''); // returns 'barn'

You can put character as parameter which want to remove as unique like this
function find_unique_characters(str, char){
return [...new Set(str.split(char))].join(char);
}
function find_unique_characters(str, char){
return [...new Set(str.split(char))].join(char);
}
let result = find_unique_characters("aaaha ok yet?", "a");
console.log(result);

//One simple way to remove redundecy of Char in String
var char = "aaavsvvssff"; //Input string
var rst=char.charAt(0);
for(var i=1;i<char.length;i++){
var isExist = rst.search(char.charAt(i));
isExist >=0 ?0:(rst += char.charAt(i) );
}
console.log(JSON.stringify(rst)); //output string : avsf

For strings (in one line)
removeDuplicatesStr = str => [...new Set(str)].join('');
For arrays (in one line)
removeDuplicatesArr = arr => [...new Set(arr)]

Using Set:
removeDuplicates = str => [...new Set(str)].join('');
Thanks to David comment below.

DEMO
function find_unique_characters( string ){
unique=[];
while(string.length>0){
var char = string.charAt(0);
var re = new RegExp(char,"g");
if (string.match(re).length===1) unique.push(char);
string=string.replace(re,"");
}
return unique.join("");
}
console.log(find_unique_characters('baraban')); // rn
console.log(find_unique_characters('anaconda')); //cod
​

var str = 'anaconda'.split('');
var rmDup = str.filter(function(val, i, str){
return str.lastIndexOf(val) === str.indexOf(val);
});
console.log(rmDup); //prints ["c", "o", "d"]
Please verify here: https://jsfiddle.net/jmgy8eg9/1/

Using Set() and destructuring twice is shorter:
const str = 'aaaaaaaabbbbbbbbbbbbbcdeeeeefggggg';
const unique = [...new Set([...str])].join('');
console.log(unique);

Yet another way to remove all letters that appear more than once:
function find_unique_characters( string ) {
var mapping = {};
for(var i = 0; i < string.length; i++) {
var letter = string[i].toString();
mapping[letter] = mapping[letter] + 1 || 1;
}
var unique = '';
for (var letter in mapping) {
if (mapping[letter] === 1)
unique += letter;
}
return unique;
}
Live test case.
Explanation: you loop once over all the characters in the string, mapping each character to the amount of times it occurred in the string. Then you iterate over the items (letters that appeared in the string) and pick only those which appeared only once.

function removeDup(str) {
var arOut = [];
for (var i=0; i < str.length; i++) {
var c = str.charAt(i);
if (c === '_') continue;
if (str.indexOf(c, i+1) === -1) {
arOut.push(c);
}
else {
var rx = new RegExp(c, "g");
str = str.replace(rx, '_');
}
}
return arOut.join('');
}

I have FF/Chrome, on which this works:
var h={};
"anaconda".split("").
map(function(c){h[c] |= 0; h[c]++; return c}).
filter(function(c){return h[c] == 1}).
join("")
Which you can reuse if you write a function like:
function nonRepeaters(s) {
var h={};
return s.split("").
map(function(c){h[c] |= 0; h[c]++; return c}).
filter(function(c){return h[c] == 1}).
join("");
}
For older browsers that lack map, filter etc, I'm guessing that it could be emulated by jQuery or prototype...

This code worked for me on removing duplicate(repeated) characters from a string (even if its words separated by space)
Link: Working Sample JSFiddle
/* This assumes you have trim the string and checked if it empty */
function RemoveDuplicateChars(str) {
var curr_index = 0;
var curr_char;
var strSplit;
var found_first;
while (curr_char != '') {
curr_char = str.charAt(curr_index);
/* Ignore spaces */
if (curr_char == ' ') {
curr_index++;
continue;
}
strSplit = str.split('');
found_first = false;
for (var i=0;i<strSplit.length;i++) {
if(str.charAt(i) == curr_char && !found_first)
found_first = true;
else if (str.charAt(i) == curr_char && found_first) {
/* Remove it from the string */
str = setCharAt(str,i,'');
}
}
curr_index++;
}
return str;
}
function setCharAt(str,index,chr) {
if(index > str.length-1) return str;
return str.substr(0,index) + chr + str.substr(index+1);
}

Here's what I used - haven't tested it for spaces or special characters, but should work fine for pure strings:
function uniquereduce(instring){
outstring = ''
instringarray = instring.split('')
used = {}
for (var i = 0; i < instringarray.length; i++) {
if(!used[instringarray[i]]){
used[instringarray[i]] = true
outstring += instringarray[i]
}
}
return outstring
}

Just came across a similar issue (finding the duplicates). Essentially, use a hash to keep track of the character occurrence counts, and build a new string with the "one-hit wonders":
function oneHitWonders(input) {
var a = input.split('');
var l = a.length;
var i = 0;
var h = {};
var r = "";
while (i < l) {
h[a[i]] = (h[a[i]] || 0) + 1;
i += 1;
}
for (var c in h) {
if (h[c] === 1) {
r += c;
}
}
return r;
}
Usage:
var a = "anaconda";
var b = oneHitWonders(a); // b === "cod"

Try this code, it works :)
var str="anaconda";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){
return obj;
}
}
).join("");
//output: "cod"

This should work using Regex ;
NOTE: Actually, i dont know how this regex works ,but i knew its 'shorthand' ,
so,i would have Explain to you better about meaning of this /(.+)(?=.*?\1)/g;.
this regex only return to me the duplicated character in an array ,so i looped through it to got the length of the repeated characters .but this does not work for a special characters like "#" "_" "-", but its give you expected result ; including those special characters if any
function removeDuplicates(str){
var REPEATED_CHARS_REGEX = /(.+)(?=.*?\1)/g;
var res = str.match(REPEATED_CHARS_REGEX);
var word = res.slice(0,1);
var raw = res.slice(1);
var together = new String (word+raw);
var fer = together.toString();
var length = fer.length;
// my sorted duplicate;
var result = '';
for(var i = 0; i < str.length; i++) {
if(result.indexOf(str[i]) < 0) {
result += str[i];
}
}
return {uniques: result,duplicates: length};
} removeDuplicates('anaconda')
The regular expression /([a-zA-Z])\1+$/ is looking for:
([a-zA-Z]]) - A letter which it captures in the first group; then
\1+ - immediately following it one or more copies of that letter; then
$ - the end of the string.
Changing it to /([a-zA-Z]).*?\1/ instead searches for:
([a-zA-Z]) - A letter which it captures in the first group; then
.*? - zero or more characters (the ? denotes as few as possible); until
\1 - it finds a repeat of the first matched character.

I have 3 loopless, one-line approaches to this.
Approach 1 - removes duplicates, and preserves original character order:
var str = "anaconda";
var newstr = str.replace(new RegExp("[^"+str.split("").sort().join("").replace(/(.)\1+/g, "").replace(/[.?*+^$[\]\\(){}|-]/g, "\\$&")+"]","g"),"");
//cod
Approach 2 - removes duplicates but does NOT preserve character order, but may be faster than Approach 1 because it uses less Regular Expressions:
var str = "anaconda";
var newstr = str.split("").sort().join("").replace(/(.)\1+/g, "");
//cdo
Approach 3 - removes duplicates, but keeps the unique values (also does not preserve character order):
var str = "anaconda";
var newstr = str.split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
//acdno

function removeduplicate(str) {
let map = new Map();
// n
for (let i = 0; i < str.length; i++) {
if (map.has(str[i])) {
map.set(str[i], map.get(str[i]) + 1);
} else {
map.set(str[i], 1);
}
}
let res = '';
for (let i = 0; i < str.length; i++) {
if (map.get(str[i]) === 1) {
res += str[i];
}
}
// o (2n) - > O(n)
// space o(n)
return res;
}

If you want your function to just return you a unique set of characters in your argument, this piece of code might come in handy.
Here, you can also check for non-unique values which are being recorded in 'nonUnique' titled array:
function remDups(str){
if(!str.length)
return '';
var obj = {};
var unique = [];
var notUnique = [];
for(var i = 0; i < str.length; i++){
obj[str[i]] = (obj[str[i]] || 0) + 1;
}
Object.keys(obj).filter(function(el,ind){
if(obj[el] === 1){
unique+=el;
}
else if(obj[el] > 1){
notUnique+=el;
}
});
return unique;
}
console.log(remDups('anaconda')); //prints 'cod'
If you want to return the set of characters with their just one-time occurrences in the passed string, following piece of code might come in handy:
function remDups(str){
if(!str.length)
return '';
var s = str.split('');
var obj = {};
for(var i = 0; i < s.length; i++){
obj[s[i]] = (obj[s[i]] || 0) + 1;
}
return Object.keys(obj).join('');
}
console.log(remDups('anaconda')); //prints 'ancod'

function removeDuplicates(str) {
var result = "";
var freq = {};
for(i=0;i<str.length;i++){
let char = str[i];
if(freq[char]) {
freq[char]++;
} else {
freq[char] =1
result +=char;
}
}
return result;
}

console.log(("anaconda").split('').sort().join('').replace(/(.)\1+/g, ""));
By this, you can do it in one line.
output: 'cdo'

function removeDuplicates(string){
return string.split('').filter((item, pos, self)=> self.indexOf(item) == pos).join('');
}
the filter will remove all characters has seen before using the index of item and position of the current element

Method 1 : one Simple way with just includes JS- function
var data = 'sssssddddddddddfffffff';
var ary = [];
var item = '';
for (const index in data) {
if (!ary.includes(data[index])) {
ary[index] = data[index];
item += data[index];
}
}
console.log(item);
Method 2 : Yes we can make this possible without using JavaScript function :
var name = 'sssssddddddddddfffffff';
let i = 0;
let newarry = [];
for (let singlestr of name) {
newarry[i] = singlestr;
i++;
}
// now we have new Array and length of string
length = i;
function getLocation(recArray, item, arrayLength) {
firstLaction = -1;
for (let i = 0; i < arrayLength; i++) {
if (recArray[i] === item) {
firstLaction = i;
break;
}
}
return firstLaction;
}
let finalString = '';
for (let b = 0; b < length; b++) {
const result = getLocation(newarry, newarry[b], length);
if (result === b) {
finalString += newarry[b];
}
}
console.log(finalString); // sdf

// Try this way
const str = 'anaconda';
const printUniqueChar = str => {
const strArr = str.split("");
const uniqueArray = strArr.filter(el => {
return strArr.indexOf(el) === strArr.lastIndexOf(el);
});
return uniqueArray.join("");
};
console.log(printUniqueChar(str)); // output-> cod

function RemDuplchar(str)
{
var index={},uniq='',i=0;
while(i<str.length)
{
if (!index[str[i]])
{
index[str[i]]=true;
uniq=uniq+str[i];
}
i++;
}
return uniq;
}

We can remove the duplicate or similar elements in string using for loop and extracting string methods like slice, substring, substr
Example if you want to remove duplicate elements such as aababbafabbb:
var data = document.getElementById("id").value
for(var i = 0; i < data.length; i++)
{
for(var j = i + 1; j < data.length; j++)
{
if(data.charAt(i)==data.charAt(j))
{
data = data.substring(0, j) + data.substring(j + 1);
j = j - 1;
console.log(data);
}
}
}
Please let me know if you want some additional information.