How to repeat a letter inside a string?

Question

lets say the input is

let word = 'I Lo1ve Co4ding'

the output would be I Loove Coooooding

so repeating x letters after putting a number after

Still don't understand how it works and how to replace it mid string.

Answer 1

You can use the callback function argument in String.prototype.replace, and call String.prototype.repeat on the first matching group (letter) and pass the second matching group (number) plus 1.

const expandStr = (str) =>
  str.replace(/([a-z])(\d+)/gi, (g, g1, g2) => g1.repeat(+g2 + 1));

console.log(expandStr('I Lo1ve Co4ding'));

Caveat

As suggested in the comments, you may use the following:

/(\p{L})(\p{N}+)/gu

In place of:

/([a-z])(\d+)/gi

Explanation:

• \p{L} – matches a single code point in the category "letter"
• \p{N} – matches any kind of numeric character in any script
• \p{S} – symbols e.g. emoji

const expandStr = (str) =>
  str.replace(/(\p{L}|\p{S})(\p{N}+)/gu, (g, g1, g2) => g1.repeat(+g2 + 1));

console.log(expandStr('I Lo1ve Co4ding'));
console.log(expandStr('I give ️ I saw 1!')); // Works with emoji

Please refer to "Unicode Categories" to learn more.

Alternative pattern: /([^(\p{N}|\s)])(\p{N}+)/gu

Tokenizing

Here is a more traditional example that incorporates loops. It does not use regular expressions, but follows a tokenizer (parser) approach.

Note: Keep in mind that this naïve example does not account for Unicode.

const
  NUMBER_RANGE_START         = '0' .charCodeAt(0), //  48
  NUMBER_RANGE_END           = '9' .charCodeAt(0), //  57
  LETTER_UPPER_RANGE_START   = 'A' .charCodeAt(0), //  65
  LETTER_UPPER_RANGE_END     = 'Z' .charCodeAt(0), //  90
  LETTER_LOWER_RANGE_START   = 'a' .charCodeAt(0), //  97
  LETTER_LOWER_RANGE_END     = 'z' .charCodeAt(0), // 122
  WHITESPACE_CARRIAGE_RETURN = '\r'.charCodeAt(0), //  13
  WHITESPACE_LINE_FEED       = '\n'.charCodeAt(0), //  10
  WHITESPACE_SPACE           = ' ' .charCodeAt(0), //  32
  WHITESPACE_TAB             = '\t'.charCodeAt(0); //  9


const codeInRange = (code, start, end) => code >= start && code <= end;

const charInRange = (char, start, end) => codeInRange(char.charCodeAt(0), start, end);

const isNumber = (char) =>
  charInRange(char, NUMBER_RANGE_START, NUMBER_RANGE_END);

const isUpperLetter = (char) =>
  charInRange(char, LETTER_UPPER_RANGE_START, LETTER_UPPER_RANGE_END);

const isLowerLetter = (char) =>
  charInRange(char, LETTER_LOWER_RANGE_START, LETTER_LOWER_RANGE_END);

const isLetter = (char) => isLowerLetter(char) || isUpperLetter(char);

const isWhiteSpace = (char) => {
  switch (char.charCodeAt(0)) {
    case WHITESPACE_CARRIAGE_RETURN:
    case WHITESPACE_LINE_FEED:
    case WHITESPACE_SPACE:
    case WHITESPACE_TAB:
      return true;
    default:
      return false;
  }
};

const expandStr = (str) => {
  const result = [];
  let index, char, prevChar, count, step;
  for (index = 0; index < str.length; index++) {
    char = str[index];
    if (
      isNumber(char) &&
      (
        isLetter(prevChar) ||
        (
          !isWhiteSpace(prevChar) &&
          !isNumber(prevChar)
        )
      )
    ) {
      count = parseInt(char, 10);
      for (step = 0; step < count; step++) {
        result.push(prevChar);
      }
    } else {
      result.push(char);
    }
    prevChar = char;
  }
  return result.join('');
};

console.log(expandStr('I Lo1ve Co4ding')); // I Loove Coooooding

Answer 2

A little bit longer way as from Mr.Polywhirl. But I think foreach loops make for good readability and you will see how it works.

const word = 'I Lo1ve Co4ding'

function rewrite(w) {
  const arr = [...w];
  let n = [];
  arr.forEach((s,i) => {
    if(! isNaN(s) && s != ' ') {
      n.push(arr[i-1].repeat(s));
    } else {
      n.push(s);  
    }
  });
  return n.join('');
}
console.log( rewrite(word) );