We Keep Coding

JS: given a string, change all the symbols between the first and the last letters of each word of the string if the word starts and ends with 'a'

Example:
aza asssa axxxa rrra -> a!a a!!!a a!!!a rrra
So far I've come up with this solution:
const argument = "aza asssa axxxa rrra";
const amount_of_spaces = [...argument].filter(x => x === " ").length;
let j = 0;
const argument__clone = [...argument];
const space__indices = [];
function do__stuff() {
while (j < amount_of_spaces) {
space__indices.push(argument__clone.indexOf(" ") + j);
argument__clone.splice((argument__clone.indexOf(" ")), 1);
j++;
do__stuff();
}
};
do__stuff();
const words = [];
let word = '';
for (let i = 0; i < argument.length; i++) {
if (!(space__indices.includes(i))) {
word += argument[i];
}
else {
words.push(word);
word = '';
}
}
words.push(word);
let new__word = '';
const new__words = [];
const words__static = [];
for (i of words) {
if (i[0] === 'a' && i[i.length - 1] === 'a') {
for (let j = 1; j < i.length - 1; j++) {
new__word += "!";
}
new__words.push(new__word);
new__word = '';
}
else {
words__static.push(i);
}
}
new__words.map(i => "a" + i + "a");
console.log(new__words);
console.log(words__static);
So one array stores the indices of spaces and the other one stores the words from the given string. We can separate the words because we know when one ends because we have the array with space indices. Then we check for each word whether it starts with 'a' and ends with 'a'. If the requirements are met we change all the letters within the word for "!" (excluding the very first and the very last ones). If the requirements are not met we store the word into the other array.
Eventually we have two arrays that I want to concatenate into one. The problems is if I was given something like this:
aza asssa rrra axxxa
It wouldn't have worked because of the order
Is there any better solution?

A regular expression would be simpler. Match an a after a word boundary, match more non-space characters, and finally match another a followed by a word boundary.
const input = 'aza asssa axxxa rrra';
const output = input.replace(
/(?<=\ba)\S+(?=a\b)/g,
interiorWord => '!'.repeat(interiorWord.length)
);
console.log(output);
For a more manual approach, split the input by spaces so you have an array of words, then for each word, check if it begins and ends with an a - if so, construct a new word by checking the old word's length. Then turn the array back into a single string.
const input = 'aza asssa axxxa rrra';
const words = input.split(' ');
const replacedWords = words.map(word => (
word[0] === 'a' && word[word.length - 1] === 'a' && word.length >= 3
? 'a' + '!'.repeat(word.length - 2) + 'a'
: word
));
const output = replacedWords.join(' ');
console.log(output);

How to make abbreviations/acronyms in JavaScript?

new to coding I'm trying to make a function that makes "abbreviations/acronyms" of words, e.g. 'I love you' -> 'ily'.
I've tried rewriting the code in many ways but console.log only shows me the first letter of the first given word.
function makeAbbr(words) {
let abbrev = words[0];
let after = 0;
let i = 0;
for (const letter of words) {
if (letter === '') {
i = words.indexOf('', after);
abbrev += words[i + 1];
}
after++;
}
return abbrev;
}
const words = 'a bc def';
let result = makeAbbr(words);
console.log(result)

Without using arrays. But you really should learn about them.
Start by trimming leading and trailing whitespace.
Add the first character to your acronym.
Loop over the rest of the string and add the current character to the acronym if the previous character was a space (and the current character isn't).
function makeAbbr(words) {
words = words.trim();
const length = words.length;
let acronym = words[0];
for(let i = 1; i < length; i++) {
if(words[i - 1] === ' ' && words[i] !== ' ') {
acronym += words[i];
}
}
return acronym;
}
console.log(makeAbbr('I love you'));
console.log(makeAbbr('I love you'));
console.log(makeAbbr(' I love you '));
And here's the version for GottZ
function w(char) {
char = char.toLocaleLowerCase();
const coll = Intl.Collator('en');
const cmpA = coll.compare(char, 'a');
const cmpZ = coll.compare(char, 'z');
return cmpA >= 0 && cmpZ <= 0;
}
function makeAbbr(words) {
words = words.trim();
const length = words.length;
if(!length) return '';
let acronym = words[0];
for(let i = 1; i < length; i++) {
if(!w(words[i - 1]) && w(words[i])) {
acronym += words[i];
}
}
return acronym;
}
console.log(makeAbbr('I love you'));
console.log(makeAbbr('I love you'));
console.log(makeAbbr(' I love you '));
console.log(makeAbbr(' \tI ... ! love \n\r .you '));
console.log(makeAbbr(' \tI ... ! Löve \n\r .ÿou '));

Since you wanted something using your approach, try this (code is commented)
function makeAbbr(words) {
let abbrev = "";
for (let i = 0; i < words.length - 1; i++) { // Loop through every character except the last one
if (i == 0 && words[i] != " ") { // Add the first character
abbrev += words[i];
} else if (words[i] == " " && words[i + 1] != " ") { // If current character is space and next character isn't
abbrev += words[i + 1];
}
}
return abbrev.toLowerCase();
}
const words = 'a bc def';
let result = makeAbbr(words);
console.log(result)

here is my implementation of your function:
Split the sentence into an array, get the first letter of each word and join them into one string.
const makeAbbr = string => string.split(' ').map(word => word[0]).join('');
console.log(makeAbbr('stack overflow'));
console.log(makeAbbr('i love you'));
`

If you want to use your approach exactly, you had a typo on the line specified. A character can never be "" (an empty string), but a character can be a space " ". Fixing this typo makes your solution work.
function makeAbbr(words) {
let abbrev = words[0];
let after = 0;
let i = 0;
for (const letter of words) {
if (letter === ' ') { // This line here
i = words.indexOf(' ', after);
abbrev += words[i + 1];
}
after++;
}
return abbrev.toLowerCase(); // Also added .toLowerCase()
}
const words = 'a bc def';
let result = makeAbbr(words);
console.log(result)

There are couple of things tripping you up.
let abbrev = words[0]; is just taking the first letter of the word string you passed into the function, and at some point adding something new to it.
for (const letter of words) {...}: for/of statements are used for iterating over arrays, not strings.
Here's a remixed version of your code. It still uses for/of but this time we're creating an array of words from the string and iterating over that instead.
function makeAbbr(str) {
// Initialise `abbrev`
let abbrev = '';
// `split` the string into an array of words
// using a space as the delimiter
const words = str.split(' ');
// Now we can use `for/of` to iterate
// over the array of words
for (const word of words) {
// Now concatenate the lowercase first
// letter of each word to `abbrev`
abbrev += word[0].toLowerCase();
}
return abbrev;
}
console.log(makeAbbr('I love you'));
console.log(makeAbbr('One Two Three Four Five'));

Apply text syle to a regex pattern in google sheet cells

I would like to get parts of strings (by regex) into a specific text style, but I can't manage the loop. I always get errors.
In the first row is the original text (strings separated by commas), and in the second under, is the desired text style.
Here is the sheet (french parameters) https://docs.google.com/spreadsheets/d/1vq0Ai_wEr3MamEQ-kMsXW7ZGg3RxrrkE5lITcYjO-rU/edit?usp=sharing
function NomsStyleBotanique(){
const classeur = SpreadsheetApp.getActive(); // var Feuille = classeur.getSheetByName('Feuille 1');
var ligne = classeur.getCurrentCell().getRow();
var colonne = classeur.getCurrentCell().getColumn();
var range = classeur.getActiveRange();
var richTextValues = range.getRichTextValues().map(([a]) => {
var text = a.getText();
var pos = 0;
var myregEx = /,/g;
var Noms = text.split(myregEx);
var textStyleNomPlante = SpreadsheetApp.newTextStyle()
.setFontSize(10)
.setForegroundColor("black")
.setFontFamily("Times New Roman")
.setItalic(false)
.build();
var textStyleNomAuteur = SpreadsheetApp.newTextStyle()
.setFontSize(10)
.setForegroundColor("#616399") // ("grey")
.setFontFamily("Times New Roman")
.setItalic(true)
.build();
var nbPhrases = [];
var i =0;
while (Noms){ i++; nbPhrases.push(Noms[i]); // SpreadsheetApp.getUi().alert(Noms[i]);
for (var i=0;i<nbPhrases.length;i++){
var myarr = Noms[i].split(" ");
var Espace1 = myarr[0].length+1;
var Espace2 = myarr[1].length+1;
if (Noms[i]){
if ((Noms[i].indexOf("subsp") > 1) || (Noms[i].indexOf("var.") > 1)){
var Espace3 = myarr[2].length+1;
var Espace4 = myarr[3].length+1;
pos = Espace1+Espace2+Espace3+Espace4; }
else { pos = Espace1+Espace2; } // pos = text.match(new RegExp(/\\s/, 'g'))[2];
var position = pos;
if (position > -1){
var temp = a.getTextStyle(0, position - 1);
return [
SpreadsheetApp.newRichTextValue()
.setText(Noms[i])
.setTextStyle(0, position - 1, textStyleNomPlante)
.setTextStyle(position, Noms[i].length, textStyleNomAuteur)
.build()
];
}
return [SpreadsheetApp.newRichTextValue().setText(Noms[i]).setTextStyle(Noms[i].getTextStyle()).build()];
}
}
}
} // fin boucle
);
range.setRichTextValues(richTextValues);
}

One problem here is that the author names are sometimes separated by a comma and sometimes separated by just a space. See Ten., Benth., Swart, and (Ten.) Kerguélen. However, in your comment, you said this does not happen often though and that you could just deal with this manually, so let's just assume for now that author names are never separated commas.
With the assumption, we can split the contents of each cell by , and deal with each plant name/author separately:
const plants = text.split(', ')
for (const plant of plants) {
// Find start/end of authors substring.
}
What we need is to find the indices where the "plant author" substring starts and ends.
Finding the end index of the plant author substring is easy; it's just the end of the entire plant string:
const end = plant.length
To find the start of the plant author substring, we can look for the indices of the spaces ' '. (You'll need to write your own getIndices() method for this.) If the plant contains subsp. or var., the start index is the 4th space; otherwise, it is the 2nd space:
let start
spaceIndices = getIndices(plant, ' ')
if (plant.includes('subsp.') || plant.includes('var.')) start = spaceIndices[3] + 1 // Add 1 to not include the space itself
else start = spaceIndices[1] + 1 // Add 1 to not include the space itself
Once we have the start/end indices, we can put them in an array offsets that we will use to find the startOffset and endOffset values for the setTextStyle() method.
So now we have:
const plants = text.split(', ')
let offsets = []
for (const plant of plants) {
const end = plant.length
let start
spaceIndices = getIndices(plant, ' ')
if (plant.includes('subsp.') || plant.includes('var.')) start = spaceIndices[3] + 1
else start = spaceIndices[1] + 1
offsets.push({
start,
end
})
}
Next, we have to initiate the RichTextValueBuilder object and loop through the offsets array to determine what the startOffset and endOffset values should be for the setTextStyles() method by adding where start and end values we found earlier to index
let richText = SpreadsheetApp.newRichTextValue()
.setText(text)
let authorTextStyle = SpreadsheetApp.newTextStyle()
.setBold(true)
.build()
let plantStartIndex = 0
for (const offset of offsets) {
const startOffset = plantStartIndex + offset.start
const endOffset = plantStartIndex + offset.end
richText = richText.setTextStyle(startOffset, endOffset, authorTextStyle)
plantStartIndex = plantStartIndex + offset.end + 2 // Add 2 to not include the ", " separator
}
Finally, build the RichTextValue object:
richText = richText.build()
…and tie it all together with the rest of your code:
function stylePlantNames() {
const ss = SpreadsheetApp.getActive()
const range = ss.getActiveRange()
const values = range.getValues()
let richTextValues = []
for (const row of values) {
let text = row[0]
const plants = text.split(', ')
let offsets = []
for (const plant of plants) {
const end = plant.length
let start
spaceIndices = getIndices(plant, ' ')
if (plant.includes('subsp.') || plant.includes('var.')) start = spaceIndices[3] + 1
else start = spaceIndices[1] + 1
offsets.push({
start,
end
})
}
let richText = SpreadsheetApp.newRichTextValue()
.setText(text)
let authorTextStyle = SpreadsheetApp.newTextStyle()
.setBold(true)
.build()
let plantStartIndex = 0
for (const offset of offsets) {
const startOffset = plantStartIndex + offset.start
const endOffset = plantStartIndex + offset.end
richText = richText.setTextStyle(startOffset, endOffset, authorTextStyle)
plantStartIndex = plantStartIndex + offset.end + 2
}
richText = richText.build()
richTextValues.push([richText])
}
range.setRichTextValues(richTextValues)
}
function getIndices(str, char) {
let indices = [];
for (var i = 0; i < str.length; i++) {
if (str[i] === char) indices.push(i);
}
return indices;
}
I skipped over many details of how the Apps Script APIs work for spreadsheets and rich text formatting. You'll need to set your own styles, but from your code, it seems like you already know how to do this. The tricky part of your question is figuring out how to identify the author substring, so that's what I focused on for my answer.

Regex global replace not replacing first occurrence

I'm trying to make a quick and dirty template test, but when the regular expression replaces, it only replaces the last occurrence, I thought that the \g would replace all copies. but it's only matching the last occurrence.
(How to String.match() distinct it ${SOME_TEXT} using Regex)
What I'm trying to do is prompt the user once for each unique variable name.
Title = t
name = n
result = '${Title} - n - t'
Using /\$\{([^\}]+)\}/g works, but prompts the user multiple times.
Title = t
name = n
Title = t
result = 't - n - t'
So how can I replace each token with a single value no matter how many times it appears.
<html>
<head>
<script type="application/javascript">
window.copyToClipboard = function(n) {
var u = "_hiddenCopyText_", t, i, r;
t = document.getElementById(u);
t.textContent = n;
i = document.activeElement;
t.focus();
t.setSelectionRange(0, t.value.length);
try {
r = document.execCommand("copy")
} catch (f) {
r = !1
}
return i && typeof i.focus == "function" && i.focus(),
t.textContent = "",
r
};
//${varname}
window.templateRegex = /\$\{([^\}]+)\}(?![\S\s]*\$\{\1\})/g;
window.test = " ${Title} - ${Name} - ${Title}"
window.copyTemplate = function (template) {
var result = template.replace(window.templateRegex, function(match, token){
return window.prompt("replace value for ${"+token+"}","${"+token+"}");
});
window.copyToClipboard(result);
};
</script>
</head>
<textarea id="_hiddenCopyText_"></textarea>
<button onclick="copyTemplate(window.test)">Test</button>
</html>
JsFiddle https://jsfiddle.net/ksu37c3b/

If same ${..}(${Title}) occurs several times, /\$\{([^\}]+)\}(?![\S\s]*\$\{\1\})/g will only match the last one. To match all the elements, /\$\{([^\}]+)\}/g will work, it doesn't need the negative lookahead (?![\S\s]*\$\{\1\}).
In the string "${Title} - ${Name} - ${Title}", only ${Name} and ${Title} matches the pattern, because the first one and the last one are the same, if you change the string, for example, "${Title2} - ${Name} - ${Title}" it will match all the element, because ${Title2} and ${Title} are not equal. That's what the (?![\S\s]*\$\{\1\}) try to do, if the last ${..} and the current ${..} are the same, it won't match.
update
So how can I replace each token with a single value no matter how many times it appears.
You can use the match() method to get an array of tokens occured, then replace it one by one, the code below will work:
window.templateRegex = /\$\{([^\}]+)\}(?![\S\s]*\$\{\1\})/g;
window.test = " ${Title} - ${Name} - ${Title}";
var userinput, reg, i;
var arr = window.test.match(/\$\{([^\}]+)\}(?![\S\s]*\$\{\1\})/g);
for (i = 0; i < arr.length; i++) {
arr[i] = arr[i].replace(/[\$\{\}]/g, "");
userinput = window.prompt("replace value for " + arr[i]);
reg = new RegExp("\\$\\{" + arr[i] + "\\}", 'g');
console.log(reg);
window.test = window.test.replace(reg, userinput);
}
alert(window.test);
I have modified your code as follows:
window.copyToClipboard = function(n) {
var u = "_hiddenCopyText_",
t, i, r;
t = document.getElementById(u);
t.textContent = n;
//** snip **//
};
//${varname}
window.templateRegex = /\$\{([^\}]+)\}(?![\S\s]*\$\{\1\})/g;
window.test = " ${Title} - ${Name} - ${Title}";
window.copyTemplate = function(template) {
var result = template;
var userinput, reg;
var arr = result.match(/\$\{([^\}]+)\}(?![\S\s]*\$\{\1\})/g);
for (let i = 0; i < arr.length; i++) {
arr[i] = arr[i].replace(/[\$\{\}]/g, "");
userinput = window.prompt("replace value for " + arr[i]);
reg = new RegExp("\\$\\{" + arr[i] + "\\}", 'g');
console.log(reg);
result = result.replace(reg, userinput);
}
window.copyToClipboard(result);
};
<textarea id="_hiddenCopyText_"></textarea>
<button onclick="copyTemplate(window.test)">Test</button>​

This should do what you want /\$\{([^\}]+)\}(?![\S\s]\$\{\1\})/
I just removed the * in the middle of the negative look ahead because it was causing it to match the final occurrence every time.

How do I replace a character at a particular index in JavaScript?

I have a string, let's say Hello world and I need to replace the char at index 3. How can I replace a char by specifying a index?
var str = "hello world";
I need something like
str.replaceAt(0,"h");

In JavaScript, strings are immutable, which means the best you can do is to create a new string with the changed content and assign the variable to point to it.
You'll need to define the replaceAt() function yourself:
String.prototype.replaceAt = function(index, replacement) {
return this.substring(0, index) + replacement + this.substring(index + replacement.length);
}
And use it like this:
var hello = "Hello World";
alert(hello.replaceAt(2, "!!")); // He!!o World

There is no replaceAt function in JavaScript. You can use the following code to replace any character in any string at specified position:
function rep() {
var str = 'Hello World';
str = setCharAt(str,4,'a');
alert(str);
}
function setCharAt(str,index,chr) {
if(index > str.length-1) return str;
return str.substring(0,index) + chr + str.substring(index+1);
}
<button onclick="rep();">click</button>

You can't. Take the characters before and after the position and concat into a new string:
var s = "Hello world";
var index = 3;
s = s.substring(0, index) + 'x' + s.substring(index + 1);

str = str.split('');
str[3] = 'h';
str = str.join('');

There are lot of answers here, and all of them are based on two methods:
METHOD1: split the string using two substrings and stuff the character between them
METHOD2: convert the string to character array, replace one array member and join it
Personally, I would use these two methods in different cases. Let me explain.
#FabioPhms: Your method was the one I initially used and I was afraid that it is bad on string with lots of characters. However, question is what's a lot of characters? I tested it on 10 "lorem ipsum" paragraphs and it took a few milliseconds. Then I tested it on 10 times larger string - there was really no big difference. Hm.
#vsync, #Cory Mawhorter: Your comments are unambiguous; however, again, what is a large string? I agree that for 32...100kb performance should better and one should use substring-variant for this one operation of character replacement.
But what will happen if I have to make quite a few replacements?
I needed to perform my own tests to prove what is faster in that case. Let's say we have an algorithm that will manipulate a relatively short string that consists of 1000 characters. We expect that in average each character in that string will be replaced ~100 times. So, the code to test something like this is:
var str = "... {A LARGE STRING HERE} ...";
for(var i=0; i<100000; i++)
{
var n = '' + Math.floor(Math.random() * 10);
var p = Math.floor(Math.random() * 1000);
// replace character *n* on position *p*
}
I created a fiddle for this, and it's here.
There are two tests, TEST1 (substring) and TEST2 (array conversion).
Results:
TEST1: 195ms
TEST2: 6ms
It seems that array conversion beats substring by 2 orders of magnitude! So - what the hell happened here???
What actually happens is that all operations in TEST2 are done on array itself, using assignment expression like strarr2[p] = n. Assignment is really fast compared to substring on a large string, and its clear that it's going to win.
So, it's all about choosing the right tool for the job. Again.

Work with vectors is usually most effective to contact String.
I suggest the following function:
String.prototype.replaceAt=function(index, char) {
var a = this.split("");
a[index] = char;
return a.join("");
}
Run this snippet:
String.prototype.replaceAt=function(index, char) {
var a = this.split("");
a[index] = char;
return a.join("");
}
var str = "hello world";
str = str.replaceAt(3, "#");
document.write(str);

In Javascript strings are immutable so you have to do something like
var x = "Hello world"
x = x.substring(0, i) + 'h' + x.substring(i+1);
To replace the character in x at i with 'h'

function dothis() {
var x = document.getElementById("x").value;
var index = document.getElementById("index").value;
var text = document.getElementById("text").value;
var length = document.getElementById("length").value;
var arr = x.split("");
arr.splice(index, length, text);
var result = arr.join("");
document.getElementById('output').innerHTML = result;
console.log(result);
}
dothis();
<input id="x" type="text" value="White Dog" placeholder="Enter Text" />
<input id="index" type="number" min="0"value="6" style="width:50px" placeholder="index" />
<input id="length" type="number" min="0"value="1" style="width:50px" placeholder="length" />
<input id="text" type="text" value="F" placeholder="New character" />
<br>
<button id="submit" onclick="dothis()">Run</button>
<p id="output"></p>
This method is good for small length strings but may be slow for larger text.
var x = "White Dog";
var arr = x.split(""); // ["W", "h", "i", "t", "e", " ", "D", "o", "g"]
arr.splice(6, 1, 'F');
/*
Here 6 is starting index and 1 is no. of array elements to remove and
final argument 'F' is the new character to be inserted.
*/
var result = arr.join(""); // "White Fog"

One-liner using String.replace with callback (no emoji support):
// 0 - index to replace, 'f' - replacement string
'dog'.replace(/./g, (c, i) => i == 0? 'f': c)
// "fog"
Explained:
//String.replace will call the callback on each pattern match
//in this case - each character
'dog'.replace(/./g, function (character, index) {
if (index == 0) //we want to replace the first character
return 'f'
return character //leaving other characters the same
})

Generalizing Afanasii Kurakin's answer, we have:
function replaceAt(str, index, ch) {
return str.replace(/./g, (c, i) => i == index ? ch : c);
}
let str = 'Hello World';
str = replaceAt(str, 1, 'u');
console.log(str); // Hullo World
Let's expand and explain both the regular expression and the replacer function:
function replaceAt(str, index, newChar) {
function replacer(origChar, strIndex) {
if (strIndex === index)
return newChar;
else
return origChar;
}
return str.replace(/./g, replacer);
}
let str = 'Hello World';
str = replaceAt(str, 1, 'u');
console.log(str); // Hullo World
The regular expression . matches exactly one character. The g makes it match every character in a for loop. The replacer function is called given both the original character and the index of where that character is in the string. We make a simple if statement to determine if we're going to return either origChar or newChar.

var str = "hello world";
console.log(str);
var arr = [...str];
arr[0] = "H";
str = arr.join("");
console.log(str);

This works similar to Array.splice:
String.prototype.splice = function (i, j, str) {
return this.substr(0, i) + str + this.substr(j, this.length);
};

You could try
var strArr = str.split("");
strArr[0] = 'h';
str = strArr.join("");

this is easily achievable with RegExp!
const str = 'Hello RegEx!';
const index = 11;
const replaceWith = 'p';
//'Hello RegEx!'.replace(/^(.{11})(.)/, `$1p`);
str.replace(new RegExp(`^(.{${ index }})(.)`), `$1${ replaceWith }`);
//< "Hello RegExp"

Using the spread syntax, you may convert the string to an array, assign the character at the given position, and convert back to a string:
const str = "hello world";
function replaceAt(s, i, c) {
const arr = [...s]; // Convert string to array
arr[i] = c; // Set char c at pos i
return arr.join(''); // Back to string
}
// prints "hallo world"
console.log(replaceAt(str, 1, 'a'));

You could try
var strArr = str.split("");
strArr[0] = 'h';
str = strArr.join("");
Check out this function for printing steps
steps(3)
// '# '
// '## '
// '###'
function steps(n, i = 0, arr = Array(n).fill(' ').join('')) {
if (i === n) {
return;
}
str = arr.split('');
str[i] = '#';
str = str.join('');
console.log(str);
steps(n, (i = i + 1), str);
}

#CemKalyoncu: Thanks for the great answer!
I also adapted it slightly to make it more like the Array.splice method (and took #Ates' note into consideration):
spliceString=function(string, index, numToDelete, char) {
return string.substr(0, index) + char + string.substr(index+numToDelete);
}
var myString="hello world!";
spliceString(myString,myString.lastIndexOf('l'),2,'mhole'); // "hello wormhole!"

If you want to replace characters in string, you should create mutable strings. These are essentially character arrays. You could create a factory:
function MutableString(str) {
var result = str.split("");
result.toString = function() {
return this.join("");
}
return result;
}
Then you can access the characters and the whole array converts to string when used as string:
var x = MutableString("Hello");
x[0] = "B"; // yes, we can alter the character
x.push("!"); // good performance: no new string is created
var y = "Hi, "+x; // converted to string: "Hi, Bello!"

You can extend the string type to include the inset method:
String.prototype.append = function (index,value) {
return this.slice(0,index) + value + this.slice(index);
};
var s = "New string";
alert(s.append(4,"complete "));
Then you can call the function:

You can concatenate using sub-string function at first select text before targeted index and after targeted index then concatenate with your potential char or string. This one is better
const myString = "Hello world";
const index = 3;
const stringBeforeIndex = myString.substring(0, index);
const stringAfterIndex = myString.substring(index + 1);
const replaceChar = "X";
myString = stringBeforeIndex + replaceChar + stringAfterIndex;
console.log("New string - ", myString)
or
const myString = "Hello world";
let index = 3;
myString = myString.substring(0, index) + "X" + myString.substring(index + 1);

I did a function that does something similar to what you ask, it checks if a character in string is in an array of not allowed characters if it is it replaces it with ''
var validate = function(value){
var notAllowed = [";","_",">","<","'","%","$","&","/","|",":","=","*"];
for(var i=0; i<value.length; i++){
if(notAllowed.indexOf(value.charAt(i)) > -1){
value = value.replace(value.charAt(i), "");
value = validate(value);
}
}
return value;
}

Here is a version I came up with if you want to style words or individual characters at their index in react/javascript.
replaceAt( yourArrayOfIndexes, yourString/orArrayOfStrings )
Working example: https://codesandbox.io/s/ov7zxp9mjq
function replaceAt(indexArray, [...string]) {
const replaceValue = i => string[i] = <b>{string[i]}</b>;
indexArray.forEach(replaceValue);
return string;
}
And here is another alternate method
function replaceAt(indexArray, [...string]) {
const startTag = '<b>';
const endTag = '</b>';
const tagLetter = i => string.splice(i, 1, startTag + string[i] + endTag);
indexArray.forEach(tagLetter);
return string.join('');
}
And another...
function replaceAt(indexArray, [...string]) {
for (let i = 0; i < indexArray.length; i++) {
string = Object.assign(string, {
[indexArray[i]]: <b>{string[indexArray[i]]}</b>
});
}
return string;
}

Here is my solution using the ternary and map operator. More readable, maintainable end easier to understand if you ask me.
It is more into es6 and best practices.
function replaceAt() {
const replaceAt = document.getElementById('replaceAt').value;
const str = 'ThisIsATestStringToReplaceCharAtSomePosition';
const newStr = Array.from(str).map((character, charIndex) => charIndex === (replaceAt - 1) ? '' : character).join('');
console.log(`New string: ${newStr}`);
}
<input type="number" id="replaceAt" min="1" max="44" oninput="replaceAt()"/>

My safe approach with negative indexes
/**
* #param {string} str
* #param {number} index
* #param {string} replacement
* #returns {string}
*/
static replaceAt (str, index, replacement)
{
if (index < 0) index = str.length + index
if (index < 0 || index >= str.length) throw new Error(`Index (${index}) out of bounds "${str}"`)
return str.substring(0, index) + replacement + str.substring(index + 1)
}
Use it like that:
replaceAt('my string', -1, 'G') // 'my strinG'
replaceAt('my string', 2, 'yy') // 'myyystring'
replaceAt('my string', 22, 'yy') // Uncaught Error: Index (22) out of bounds "my string"

Lets say you want to replace Kth index (0-based index) with 'Z'.
You could use Regex to do this.
var re = var re = new RegExp("((.){" + K + "})((.){1})")
str.replace(re, "$1A$`");

You can use the following function to replace Character or String at a particular position of a String. To replace all the following match cases use String.prototype.replaceAllMatches() function.
String.prototype.replaceMatch = function(matchkey, replaceStr, matchIndex) {
var retStr = this, repeatedIndex = 0;
for (var x = 0; (matchkey != null) && (retStr.indexOf(matchkey) > -1); x++) {
if (repeatedIndex == 0 && x == 0) {
repeatedIndex = retStr.indexOf(matchkey);
} else { // matchIndex > 0
repeatedIndex = retStr.indexOf(matchkey, repeatedIndex + 1);
}
if (x == matchIndex) {
retStr = retStr.substring(0, repeatedIndex) + replaceStr + retStr.substring(repeatedIndex + (matchkey.length));
matchkey = null; // To break the loop.
}
}
return retStr;
};
Test:
var str = "yash yas $dfdas.**";
console.log('Index Matched replace : ', str.replaceMatch('as', '*', 2) );
console.log('Index Matched replace : ', str.replaceMatch('y', '~', 1) );
Output:
Index Matched replace : yash yas $dfd*.**
Index Matched replace : yash ~as $dfdas.**

I se this to make a string proper case, that is, the first letter is Upper Case and all the rest are lower case:
function toProperCase(someString){
return someString.charAt(0).toUpperCase().concat(someString.toLowerCase().substring(1,someString.length));
};
This first thing done is to ensure ALL the string is lower case - someString.toLowerCase()
then it converts the very first character to upper case -someString.charAt(0).toUpperCase()
then it takes a substring of the remaining string less the first character -someString.toLowerCase().substring(1,someString.length))
then it concatenates the two and returns the new string -someString.charAt(0).toUpperCase().concat(someString.toLowerCase().substring(1,someString.length))
New parameters could be added for the replacement character index and the replacement character, then two substrings formed and the indexed character replaced then concatenated in much the same way.

The solution does not work for negative index so I add a patch to it.
String.prototype.replaceAt=function(index, character) {
if(index>-1) return this.substr(0, index) + character + this.substr(index+character.length);
else return this.substr(0, this.length+index) + character + this.substr(index+character.length);
}

"hello world".replace(/(.{3})./, "$1h")
// 'helho world'

The methods on here are complicated.
I would do it this way:
var myString = "this is my string";
myString = myString.replace(myString.charAt(number goes here), "insert replacement here");
This is as simple as it gets.