How to make abbreviationsacronyms in JavaScript? - Stack Overflow

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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)

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)

• javascript

Share Improve this question Follow edited May 18, 2022 at 14:15 Kuba Nowoszyński asked May 18, 2022 at 14:13 Kuba NowoszyńskiKuba Nowoszyński 8977 bronze badges 8

• 3 "I love you".match(/\b(\S)/g).join("") – GottZ Commented May 18, 2022 at 14:15
• @GottZ Is there any simple way? Like changing something in my code? I'm not familiar with the "match" and "join" yet. – Kuba Nowoszyński Commented May 18, 2022 at 14:18
• 2 letter === '' ---> letter === ' ' and words.indexOf('', after) ---> words.indexOf(' ', after) - Note the space between single quotes. – Yousaf Commented May 18, 2022 at 14:19
• basically /\b(\S)/g/ is a regular expression that checks for word boundaries with \b so something like the start and the end of a word. \S checks for a non-whitespace character (excluding linebreaks). the (\S) ensures you want to match what ever character is matched there. /g will match more than once. .join is simply a function that joins iterable objects like arrays and match results together, using what ever is specified as delimiter. – GottZ Commented May 18, 2022 at 14:21
• 1 words.indexOf(' ', after) could also change to words.indexOf(letter, after), though this doesn't matter that much. It just reduces the amount of hardcoded values. – StackByMe Commented May 18, 2022 at 14:23

 |  Show 3 more ments

5 Answers 5

Sorted by: Reset to default 3

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'));

`

Without using arrays. But you really should learn about them.

1. Start by trimming leading and trailing whitespace.
2. Add the first character to your acronym.
3. 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.pare(char, 'a');
   const cmpZ = coll.pare(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 mented)

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)

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.

1. 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.

2. 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'));

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